Output the sign

Question

Given a number N, output the sign of N:

• If N is positive, output 1
• If N is negative, output -1
• If N is 0, output 0

N will be an integer within the representable range of integers in your chosen language.

The Catalogue

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](https://esolangs.org/wiki/Fish), 121 bytes

/* Configuration */

var QUESTION_ID = 103822; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8478; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;

function answersUrl(index) {
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}

function commentUrl(index, answers) {
  return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
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function getAnswers() {
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        var id = +a.share_link.match(/\d+/);
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getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
  return a.owner.display_name;
}

function process() {
  var valid = [];
  
  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if(OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });
    
    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    else console.log(body);
  });
  
  valid.sort(function (a, b) {
    var aB = a.size,
        bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
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                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

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  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
    return 0;
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  for (var i = 0; i < langs.length; ++i)
  {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
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                       .replace("{{SIZE}}", lang.size)
                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
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body {
  text-align: left !important;
  display: block !important;
}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 500px;
  float: left;
}

table thead {
  font-weight: bold;
}

table td {
  padding: 5px;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f">
<div id="language-list">
  <h2>Shortest Solution by Language</h2>
  <table class="language-list">
    <thead>
      <tr><td>Language</td><td>User</td><td>Score</td></tr>
    </thead>
    <tbody id="languages">

    </tbody>
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</div>
<div id="answer-list">
  <h2>Leaderboard</h2>
  <table class="answer-list">
    <thead>
      <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>
    </thead>
    <tbody id="answers">

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<table style="display: none">
  <tbody id="answer-template">
    <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
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Answer 1

Retina, 9 bytes

[1-9].*
1

Try it online!

Replaces a non-zero digit and everything after it with 1. This leaves a potential leading - intact and changes all numbers except 0 itself to absolute value 1.

Answer 2

C (GCC), 24 23 22 18 bytes

Thanks to @aross and @Steadybox for saving a byte!

f(n){n=!!n|n>>31;}

Not guaranteed to work on all systems or compilers, works on TIO.

Answer 3

Mathematica, 4 bytes

Clip

How about not using the built-in Sign and still scoring 4 bytes? ;)

Clip with a single argument clips (or clamps) the input value between -1 and 1. Since the inputs will only be integers, this is the same as using Sign.

Answer 4

COW, 225 213 201 bytes

oomMOOmoOmoOmoOmoOMoOMoOmOomOomOoMoOMMMmoOMMMMOOMOomOo
mOoMOomoOmoOmoomOomOoMMMmoOmoOmoOMMMMOOOOOmoOMOoMOomOo
mOomOoMoOMMMmoOMMMMOOMOomOomOoMoOmoOmoOmoomoOmoomOomOo
mOomoomoOMOOmoOmoOmoOMOoMMMOOOmooMMMOOM

Try it online!

The way that this code works is that it determines the sign by alternating adding and subtracting bigger numbers, and seeing which one was the last one that worked. Given any non-zero integer, first subtract 1, then add 2, then subtract 3, etc. and you'll eventually reach 0. Keep track of your state by alternating adding and subtracting 2 to a value that starts off at 0. For example:

-5  - 1  = -6  (current state: 0 + 2 = 2)
-6  + 2  = -4  (current state: 2 - 2 = 0)
-4  - 3  = -7  (current state: 0 + 2 = 2)
-7  + 4  = -3  (current state: 2 - 2 = 0)
-3  - 5  = -8  (current state: 0 + 2 = 2)
-8  + 6  = -2  (current state: 2 - 2 = 0)
-2  - 7  = -9  (current state: 0 + 2 = 2)
-9  + 8  = -1  (current state: 2 - 2 = 0)
-1  - 9  = -10 (current state: 0 + 2 = 2)
-10 + 10 =  0  (current state: 2 - 2 = 0)
value is now at 0.  state - 1 = 0 - 1 = -1
sign of original number is -1

When you're done, subtract 1 from your state and you get the sign, positive or negative. If the original number is 0, then don't bother doing any of this and just print 0.

Detailed Explanation:

oom                                        ;Read an integer into [0]
MOO                                        ;Loop while [0] is non-empty
    moOmoOmoOmoOMoOMoOmOomOomOo            ;    Decrement [4] twice
    MoOMMMmoOMMM                           ;    Increment [1], then copy [1] to [2]
    MOO                                    ;    Loop while [2] is non-empty
        MOomOomOoMOomoOmoO                 ;        Decrement [0] and [2]
    moo                                    ;    End loop now that [2] is empty
    mOomOoMMMmoOmoOmoOMMM                  ;    Navigate to [0], and copy to [3]
    MOO                                    ;    Perform the next steps only if [3] is non-zero
        OOOmoOMOoMOomOomOomOoMoOMMMmoOMMM  ;        Clear [3], increment [4] twice, increment [1], and copy it to [2]
        MOO                                ;        Loop while [2] is non-empty
            MOomOomOoMoOmoOmoO             ;            Decrement [2] and increment [0]
        moo                                ;        End loop now that [2] is empty
    moO                                    ;        Navigate back to [3]
    moo                                    ;    End the condition
    mOomOomOo                              ;    Navigate back to [0]
moo                                        ;End loop once [0] is empty.
moO                                        ;Navigate to [1]. If [1] is 0, then input was 0.  Otherwise, [4] contains (sign of [0] + 1)
MOO                                        ;Perform the next steps only if [1] is non-zero
    moOmoOmoOMOoMMMOOO                     ;    Navigate to [4], copy it to the register, and clear [4].
moo                                        ;End condition
MMMOOM                                     ;If the register contains something (which is true iff the condition ran), paste it and print it.  Otherwise, no-op and print 0.

I'm still experimenting with golfing it (you will be shocked to discover that golfing in COW is rather difficult), so this may come down a few more bytes in the future.

Answer 5

Cubix, 10 bytes

(W0^I?>O2@

Test it online!

This code is wrapped to the following cube net:

    ( W
    0 ^
I ? > O 2 @ . .
. . . . . . . .
    . .
    . .

The code is then run with the IP (instruction pointer) starting on the I, facing east. I inputs a signed integer from STDIN, pushing it onto the stack.

The next command is ?, which changes the direction of the IP depending on the sign of the top item. If the input is 0, it keeps moving in same direction, running through the following code:

• > - Point the IP to the east. (No-op since we're already going east.)
• O - Output the top item as an integer.
• 2 - Push 2 to the stack. This is practically a no-op, because...
• @ - Terminates the program.

If the input is negative, the IP turns left at the ?; because this is a cube, the IP moves onto the 0 in the second row, heading east. 0 pushes a literal 0, then this code is run:

• ^ - Point the IP north.
• W - "Sidestep" the IP one spot to the left.
• ( - Decrement the top item.

The TOS is now -1, and the IP wraps around the cube through a bunch of no-ops . until it hits the >. This runs the same output code mentioned above, outputting -1.

If the input is positive, the same thing happens as with negative inputs, with one exception: the IP turns right instead of left at the ?, and wraps around the cube to the 2, which pushes a literal 2. This is then decremented to 1 and sent to output.

Answer 6

JavaScript (ES6), 9 bytes

Math.sign

Straightforward.

The shortest non-builtin is 13 bytes:

n=>n>0|-(n<0)

Thanks to @Neil, this can be golfed by a byte, but at the cost of only working on 32-bit integers:

n=>n>0|n>>31

Or you could do

n=>n>0?1:!n-1

which seems more golfable, but I'm not sure how.

Answer 7

APL (Dyalog APL), 1 byte

Works for complex numbers too, returning 1∠θ:

×

TryAPL online!

---

Without that built-in, for integers (as per OP):

¯1⌈1⌊⊢

¯1⌈ the largest of negative one and

1⌊ the smallest of one and

⊢ the argument

TryAPL online!

... and a general one:

>∘0-<∘0

>∘0 more-than-zero

- minus

<∘0 less-than-zero

TryAPL online!

Answer 8

Vim, 22 bytes

xVp:s/-/-1^M:s/[1-9]/1^M

Saved one byte thanks to @DJMcMayhem!

Here, ^M is a literal newline.

As @nmjcman101 pointed out in the comments, a single regex can be used (:s/\v(-)=[^0].*/\11^M, 20 bytes) instead, but since this is basically the same as a Retina answer would be, I'm sticking to my own method.

Explanation:

xVp                        Delete everything except the first character. If the number is negative, this leaves a -, a positive leaves any number between 1 and 9, and 0 leaves 0.
   :s/-/-1^M               Replace a - with a -1
            :s/[1-9]/1^M   Replace any number between 1 and 9 with 1.

Here's a gif of it running with a negative number (old version):

Here's it running with 0:

Running with positive:

Answer 9

><>, 9 8 bytes

Thanks to Sp3000 for saving a byte.

'i$-%n/

There's an unprintable 0x01 before the /.

Try it online!

Explanation

This is a port of my character code-based Labyrinth answer.

'     Push the entire program (except ' itself) onto the stack, which ends 
      with [... 1 47].
i     Read the first character of the input.
$-    Subtract the 47.
%     Take the 1 modulo this value.
n     Output the result as an integer.
0x01  Unknown command, terminates the program.

Answer 10

///, 52 36 bytes

/a/\/1\/\///2a3a4a5a6a7a8a9a10a11/1/

Ungolfed, explanation:

/2/1/
/3/1/
/4/1/
/5/1/
/6/1/
/7/1/
/8/1/
/9/1/
/10/1/
/11/1/

It's basically a MapReduce implemenatation, i.e. there are two phases:

• Replace all occurrences of digits 2-9 by 1, e.g. 1230405 -> 1110101
• Reduce pairs of 11 or 10 to 1 repeatedly, e.g. 1110101-> 1

If there was a - in front initially, it will remain and the output will be -1. A single 0 is never replaced, thus resulting in itself.

Update: Save additional 16 bytes by aliasing //1/ with a, thanks to Martin Ender.

Try it online, with test cases

Answer 11

Python 2, 17 bytes

lambda n:cmp(n,0)

Try it online!

Answer 12

Labyrinth, 10 bytes

?:+:)%:(%!

Try it online!

Explanation

Labyrinth's control flow semantics actually give you a "free" way to determine a number's sign, because the chosen path at a 3-way fork depends on whether the sign is negative, zero or positive. However, I haven't been able to fit a program with junctions into less than 12 bytes so far (although it may be possible).

Instead, here's a closed-form solution, that doesn't require any branches:

Code    Comment             Example -5      Example 0       Example 5
?       Read input.         [-5]            [0]             [5]
:+      Double.             [-10]           [0]             [10]
:)      Copy, increment.    [-10 -9]        [0 1]           [10 11]
%       Modulo.             [-1]            [0]             [10]
:(      Copy, decrement.    [-1 -2]         [0 -1]          [10 9]
%       Modulo.             [-1]            [0]             [1]
!       Print.              []              []              []

The instruction pointer then hits a dead end, turns around and terminates when % now attempts a division by zero.

Doubling the input is necessary to make this work with inputs 1 and -1, otherwise one of the two modulo operations would already attempt a division by zero.

Answer 13

Brain-Flak 74 42 40 Bytes

Saved 2 bytes thanks to 1000000000

{([({}<([()])>)]<>(())){({}())<>}}{}({})

Try it Online!

Explanation:

{                                }       # if 0 do nothing
   (          )                          # push:                           
    {}<     >                            # the input, after 
       (    )                            # pushing:
        [  ]                             # negative:
         ()                              # 1

 (                    )                  # Then push:
  [            ]                         # the negative of the input
                <>                       # on the other stack with:
                   ()                    # a 1 
                  (  )                   # pushed under it

                       {        }        # while 1: 
                        ({}())           # increment this stack and...
                              <>         # switch stacks

                                 {}      # pop the top (the counter or 0 from input)
                                   (  )  # push:
                                    {}   # the top (this is a no-op, or pushes a 0)

Answer 14

PHP, 16 bytes

Uses the new spaceship operator.

<?=$argv[1]<=>0;

Answer 15

J, 1 byte

*

Try it online (with test cases)!

Answer 16

C, 24 20 19 18 bytes

I abuse two C exploits to golf this down; This is in C (GCC).

f(a){a=a>0?:-!!a;}

Try it online!

---

Revision History:

1) f(a){return(a>0)-(a<0);} //24 bytes

2) f(a){a=(a>0)-(a<0);} //20 bytes

3) f(a){a=a>0?:-1+!a;} //19 bytes

4) f(a){a=a>0?:-!!a;} //18 bytes

---

Revision 1: First attempt. Simple logic

Revision 2: Abuses a memory/stack bug in GCC where, as far as I can tell, a non-returning function will return the last set variable in certain cases.

Revision 3: Abuses ternary behavior where undefined result will return conditional result (which is why the true return on my ternary is nil)

Revision 4: Subtract a bool cast (!!) from the ternary conditional substitution for nil referenced in revision 2.

Answer 17

Ruby, 10 bytes

->x{x<=>0}

Answer 18

Perl, 9 bytes

Requires -E at no extra cost.

say<><=>0

Usage

perl -E 'say<><=>0' <<< -9999
-1
perl -E 'say<><=>0' <<< 9999
1
perl -E 'say<><=>0' <<< -0
0

I'm happy with the fish operator!

Answer 19

MATL, 6 bytes

0>EGg-

Input may be a number or an array. The result is number or an array with the corresponding values.

Try it online! Or test several cases using array input.

Explanation

This avoids using the builtin sign function (ZS).

0>   % Take input implicitly. Push 1 if positive, 0 otherwise
E    % Multiply by 2
Gg   % Push input converted to logical: 1 if nonzero, 0 otherwise
-    % Subtract. Implicitly display

Answer 20

Stack Cats, 6 + 4 = 10 bytes

_[:I!:

+4 bytes for the ​ -nm flags. n is for numeric I/O, and since Stack Cats requires programs to be palindromic, m implicitly mirrors the source code to give the original source

_[:I!:!I:]_

Try it online! As with basically all good Stack Cats golfs, this was found by brute force, beat any manual attempts by a long shot, and can't easily be incorporated into a larger program.

Add a D flag if you'd like to see a step-by-step program trace, i.e. run with -nmD and check STDERR/debug.

---

Stack Cats uses a tape of stacks which are implicitly filled with zeroes at the bottom. At the start of the program, all input is pushed onto the input stack, with a -1 at the base to separate the input from the implicit zeroes. At the end of the program, the current stack is output, except a base -1 if present.

The relevant commands here are:

_           Perform subtraction [... y x] -> [... y y-x], where x is top of stack
[           Move left one stack, taking top of stack with you
]           Move right one stack, taking top of stack with you
:           Swap top two of stack
I           Perform [ if top is negative, ] if positive or don't move if zero. Then
                negate the top of stack.
!           Bitwise negate top of stack (n -> -n-1)

Note that all of these commands are invertible, with its inverse being the mirror of the command. This is the premise of Stack Cats — all nontrivial terminating programs are of odd length, since even length programs self-cancel.

We start with

               v
               n
              -1
...  0    0    0    0    0  ...

_ subtracts, making the top -1-n, and [ moves the result left one stack:

           v
       -1-n   -1
...  0    0    0    0    0  ...

: swaps top two and I does nothing, since the top of stack is now zero. ! then bitwise negates the top zero into a -1 and : swaps the top two back. ! then bitwise negates the top, turning -1-n back into n again:

          v
          n
         -1   -1
...  0    0    0    0    0  ...

Now we branch based on I, which is applied to our original n:

• If n is negative, we move left one stack and end with -n on an implicit zero. : swaps, putting a zero on top, and ] moves the zero on top of the -1 we just moved off. _ then subtracts, leaving the final stack like [-1 -1], and only one -1 is output since the base -1 is ignored.

• If n is zero, we don't move and : swaps, putting -1 on top. ] then moves this left -1 on top of the right -1, and _ subtracts, leaving the final stack like [-1 0], outputting the zero and ignoring the base -1.

• If n is positive, we move right one stack and end with -n on a -1. : swaps, putting the -1 on top, and ] moves this -1 right, on top of an implicit zero. _ then subtracts, giving 0 - (-1) = 1 and leaving the final stack like [1], which is output.

Answer 21

TI-Basic, 8 bytes

median({1,Ans,~1

Alternative solutions (feel free to suggest more):

max(~1,min(Ans,1               8  bytes
0:If Ans:Ans/abs(Ans           9  bytes
(Ans>0)-(Ans<0                 10 bytes

Answer 22

Jelly, 1 byte

Ṡ

TryItOnline!

The monadic sign atom, Ṡ, does exactly what is specified for an integer input, either as a full program or as a monadic link (function taking one argument).

Answer 23

Actually, 1 byte

s

Try it online!

Another case of exactly what it says on the tin - s is the sign function.

Without the builtin (4 bytes):

;A\+

Try it online!

;A\ divides the absolute value of the input by the input. This results -1 for negative inputs and 1 for positive inputs. Unfortunately, due to Actually's error handling (if something goes wrong, the command is ignored), 0 as input leaves two 0s on the stack. + rectifies this by adding them (which causes an error with anything else, so it's ignored).

Answer 24

Mathematica, 4 bytes

Sign

Exactly what it says on the tin

Answer 25

Octave, 26 24 bytes

f=@(x)real(asin(x))/pi*2

This is my first code-golf Octave answer, any golfing tips are appreciated!

Try it online!

The idea for taking the asin comes from the question where it says output the sign :)

Explanation

Note: dividing the number by pi and multiplying it by 2 is the equivalent of dividing the entire number by pi/2

Case 0:

asin(0) yields 0. Taking the real part of it and dividing it by pi/2 makes no difference to the output.

Case positive:

asin(1) yields pi/2. asin of any number bigger than 1 will give pi/2 + complex number. Taking the real part of it gives pi/2 and dividing it by pi/2 gives 1

Case negative:

asin(-1) yields -pi/2. asin of any number smaller than -1 will give -pi/2 + complex number. Taking the real part of it gives -pi/2 and dividing it by pi/2 gives -1

Answer 26

Pushy, 7 bytes

This is probably the strangest-looking program I've ever written...

&?&|/;#

Try it online!

It uses sign(x) = abs(x) / x, but with an explicit sign(0) = 0 to avoid zero division error.

          \ Take implicit input
&?   ;    \ If the input is True (not 0):
  &|      \  Push its absolute value
    /     \  Divide
      #   \ Output TOS (the sign)

This works because x / abs(x) is 1 when x is positive and -1 when x is negative. If the input is 0, the program jumps to the output command.

---

4 bytes (non-competing)

Because of holidays and having too much time, I've done a complete rewrite of the Pushy interpreter. The above program still works, but because 0 / 0 now default to 0, the following is shorter:

&|/#

Try it online!

Answer 27

Piet, 188 53 46 41 bytes

5bpjhbttttfttatraaearfjearoaearbcatsdcclq

Online interpreter available here.

This piet code does the standard (n>0)-(n<0), as there is no sign checking builtin. In fact, there is no less-than builtin, so a more accurate description of this method would be (n>0)-(0>n).

The text above represents the image. You can generate the image by pasting it into the text box on the interpreter page. For convenience I have provided the image below where the codel size is 31 pixels. The grid is there for readability and is not a part of the program. Also note that this program does not cross any white codels; follow the colored codels around the border of the image to follow the program flow.

Explanation

Instruction    Δ Hue   Δ Lightness   Stack
------------   -----   -----------   --------------------
In (Number)    4       2             n
Duplicate      4       0             n, n
Push [1]       0       1             1, n, n
Duplicate      4       0             1, 1, in, in
Subtract       1       1             0, in, in
Duplicate      4       0             0, 0, in, in
Push [4]       0       1             4, 0, 0, in, in
Push [1]       0       1             1, 4, 0, 0, in, in
Roll           4       1             0, in, in, 0
Greater        3       0             greater, in, 0
Push [3]       0       1             3, greater, in, 0
Push [1]       0       1             1, 3, greater, in, 0
Roll           4       1             in, 0, greater
Greater        3       0             less, greater
Subtract       1       1             sign
Out (Number)   5       1             [Empty]
[Exit]         [N/A]   [N/A]         [Empty]

To reduce the filesize any further, I would need to actually change the program (gasp) instead of just compressing the file as I have been doing. I would like to remove one row which would golf this down to 36. I may also develop my own interpreter which would have a much smaller input format, as actually changing the code to make it smaller is not what code golf is about.

The mods told me that the overall filesize is what counts for Piet code. As the interpreter accepts text as valid input and raw text has a much smaller byte count than any image, text is the obvious choice. I apologize for being cheeky about this but I do not make the rules. The meta discussion about this makes my opinions on the matter clear.

If you think that that goes against the spirit of Piet or would like to discuss this further for any reason, please check out the discussion on meta.

Answer 28

C#, 16 15 bytes

Improved solution thanks to Neil

n=>n>0?1:n>>31;

Alternatively, the built-in method is 1 byte longer:

n=>Math.Sign(n);

Full program with test cases:

using System;

public class P
{
    public static void Main()
    {
        Func<int,int> f =
        n=>n>0?1:n>>31;

        // test cases:
        for (int i=-5; i<= 5; i++)
            Console.WriteLine(i + " -> " + f(i));
    }
}

Answer 29

R, 25 bytes

'if'(x<-scan(),x/abs(x),0)

Takes the number to STDIN. Then checks if it's zero, if not, returns x/|x| which is either 1 of -1, and outputs 0 if x=0.

This is without using the builtin sign of course.

Answer 30

V 14 12 bytes

Thanks @DJMcMayhem for 2 bytes. Uses a reg-ex to do the substitution. Kind of fun, because it's not a built-in. I have a more fun function, but it's not working the way I expected.

ͨ-©½0]/±1

Verify Test Cases

This just translates to :%s/\v(-)=[^0].*/\11 which matches one or more - followed by anything but 0, followed by anything any number of times. It's replaced with the first match (so either a - or nothing) and a 1. The regex doesn't match 0, so that stays itself.

The More Fun Way (21 bytes)

é
Àé12|DkJòhé-òó^$/a

TryItOnline

This accepts the input as an argument rather than in the buffer.

é<CR> Insert a new line.

À run the argument as V code. a - will move the cursor to the previous line, and any number will become the count for the next command

é1 insert (count)1's

2| move to the second column

D delete everything from the second column onwards (leaving only one character)

kJ Join the two lines together.

òhé-ò translates to: "run hé- until breaking". If the 1 was on the second line, this breaks immediately after the h. If it was on the first line, it will insert a - before breaking.

ó^$/a This fixes the fact that -1,0,1 will leave a blank, and replaces a blank with the argument register.

Answer 31

05AB1E, 3 bytes

0.S

Try it online!

Answer 32

Labyrinth, 8 bytes

1,_47-%!

Try it online!

I'm posting this as a separate answer because my other Labyrinth answer is based on arithmetic on the actual numerical input value, whereas this mostly ignores the number and works with the character code of the first character instead.

Explanation

So yeah, this reads the first character code which is either 45 (-, which should yield -1), 48 (0, which should yield 0) or 49 to 57 (1-9, which should yield 1). This mapping can be accomplished via the simple formula 1 % (x - 47). To see why this works, here is the breakdown of the code for 3 different examples:

Code    Comment             Example -5      Example 0       Example 5
1       Push 1.             [1]             [1]             [1]
,       Read character.     [1 45]          [1 48]          [1 53]
_47-    Subtract 47.        [1 -2]          [1 1]           [1 6]
%       Modulo.             [-1]            [0]             [1]
!       Print.              []              []              []

The instruction pointer then hits a dead end, turns around and terminates when % now attempts a division by zero.

Another simple computation that works:

x -= 46
x %= x-1

Answer 33

Turing Machine code, 65 bytes

0 0 0 * halt
0 - - r 2
0 * 1 r 3
2 * 1 r 3
3 * _ r 3
3 _ _ * halt

Try it online.

Answer 34

Brain-Flak, 48 46 40 bytes

{([({}<([()])>)]<>(())){({}())<>}}{}({})

Try it Online

Explanation

{                                }{}({}) #Do nothing if zero
   ({}<([()])>)                          #Put a -1 under input
 ([            ]<>(()))                  #Put 1 and a negative copy of input on the off stack
                       {        }        #Until zero
                        ({}())           #Increment
                              <>         #Swap

Answer 35

Hexagony, 13 bytes

?</!~/~@$\!@

Expanded:

  ? < /
 ! ~   /
~ @ $ \ !
 @ . . .
  . . .

Try it online!

Hexagony truthy/falsiness for numbers checks based on being positive or not. This makes singling out zero a bit tricky, so we check if a number and its negation are both non-positive to check for zero. Uses the unprintable character 0x01 to literally store 1 in a memory edge to save a byte zeroing the edge first. In the expanded version it is between the ~ and the / on the second line.

Breakdown:

For positive numbers the code is very simple. We start at the top left moving eastward, then take the fork to the right. The rest of the program is "linear" along the surface of the code, giving: ?<0x01\.!@ where both \ and . are no-ops. 0x01 sets the current memory edge to 1 and then ! prints that and @ ends the program.

For negative numbers and zero, we start the same but turn left at the <. This leads us back around to the \ but this time approaching from the southwest. This time it acts as a mirror and redirects the instruction pointer westward. The $ allows us to skip the program-ending @. Next we hit ~ which negates the value that we read in. If the number was negative it is now positive, and if it was zero it is still not positive.

When hitting the edge of the hexagon we wrap to the right if the value was positive and to the left if the value was negative or zero. Negative numbers will then wrap to the right and begin moving westwards from the top right. Hitting some mirrors leads us to a familiar looking path starting with the edge being set to 1. Then ~ negates it and ! prints giving -1. We wrap around and hit the other @.

Zero will instead wrap to the bottom, which has nothing but no-ops. Then it wraps back to the middle and is printed by !. Then some mirrors redirect us to the @ to end the program.

Answer 36

Python 3, 20 bytes

Since Python 3 doesn't have access to cmp like Python 2 does, it's a little longer

lambda n:(n>0)-(n<0)

Answer 37

Excel, 8 bytes

=Sign(n)

Pretty basic, but the only language I feel complete confidence in.

Without a builtin!

Excel, 23 bytes

=If(n>0,1,If(n<0,-1,0))

_{not so confident}

Crystal Reports Formula (Noncompete), 24 bytes

IIF(n>0,1,IIF(n<0,-1,0))

Answer 38

Befunge 93, 14 13 bytes

Golfed off a byte by combining the 2 1s

1~50p :0`_.@.

Try it Online!

This one is interesting, as it takes the first character of the number and alters the code accordingly.

 ~50p         Stores the first character in the space (labeled <char> here)
1    <char>   If the number is negative, it performs subtraction, giving 1 - 0 == -1
              If it is 0, 0 is on top. If it is positive, a positive # will be.

 :0`_         This checks the top number to see if it is positive.
     .@       If it is <1, it is printed. (0 or -1)
1     @.      Otherwise, the IP loops back harmlessly, and prints 1

Answer 39

Mathematica, 28 bytes

Limit[2ArcTan@x#/Pi,x->∞]&

Mathematica, 30 bytes

2HeavisideTheta@#-1/._[_]->1/2&

Mathematica, 76 bytes

Round@Integrate[E^(2#+I t#)/(2+I t)/Pi,{t,-∞,∞},PrincipalValue->True]-1&

Just to be different :)

Answer 40

Forth, 22 bytes

Golfed

: S dup 0< swap 0> - ;

Test

: S dup 0< swap 0> - ;  ok

0 S . 0  ok
1 S . 1  ok
-1 S . -1  ok
12345 S . 1  ok
-12345 S . -1  ok

Try It Online !

Answer 41

x86_64 machine language on Linux, 13 bytes

 0:   31 c0                   xor    %eax,%eax
 2:   85 ff                   test   %edi,%edi
 4:   0f 9f c0                setg   %al
 7:   c1 ef 1f                shr    $0x1f,%edi
10:   29 f8                   sub    %edi,%eax
12:   c3                      retq

The input (first function parameter) is passed into %edi. To try it out, compile and run the following C program

#include<stdio.h>
#include<stdlib.h>
#define s(x) ((int(*)(int))"\x31\xc0\x85\xff\xf\x9f\xc0\xc1\xef\x1f\x29\xf8\xc3")(x)
int main(){
  printf( "%d %d %d\n", s(-5), s(0), s(44) );
}

Answer 42

Haskell, 23 characters

A true approach, not using built-ins but function composition (the dot):

(1-).fromEnum.compare 0

(or with more space)

(1-) . fromEnum . compare 0

Explanation:

compare 0 : partially applying the compare function to
                the first argument 0 results in a function which takes
                a number and compares it to 0. 
                  (compare 0) n = compare 0 n =
                      LT    : if 0 < n
                      EQ    : if 0 = n
                      GT    : if 0 > n
fromEnum  : it maps LT to 0, EQ to 1 and GT to 2
(1-)      : n -> 1 - n

Answer 43

Plain English 101 81 bytes

To put a n number's sign into a s number:
Get the sign of the n returning the s.

Ungolfed version:

To put a number's sign into another number:
  Get the sign of the number returning the other number.

Either version can be used to golf the client code -- and make the client code more readable. For example, the following line of code displays a Windows message box containing the number -1 in the message body:

Debug -456's sign.

The Plain English IDE is available at github.com/Folds/english. The IDE runs on Windows. It compiles to 32-bit x86 code.

Answer 44

LOLCODE, 191 bytes

HAI 1.3
I HAS A J ITZ A NUMBR
GIMMEH J
BIGGR OF J AN 0, O RLY?
	YA RLY
		VISIBLE "1"
	NO WAI
		BOTH SAEM "0" AN J, O RLY?
			YA RLY
				VISIBLE "0"
			NO WAI
				VISIBLE "-1"
	OIC
OIC
KTHXBYE

Try it online!

Answer 45

Hexagony, 18 15 bytes

~.?>.)<<!!.&,.@

Try it online!

n > 0

  ~ . ?
 . . . .
< . ! . .
 , . @ .
  . . .

n == 0

  ~ . ?
 > . ) <
< . ! . &
 . . @ .
  . . .

n < 0

  ~ . ?
 > . ) <
< ! . . .
 , . @ .
  . . .

Abuses the same EOF trick used in Jo Kings answer.

Sadly I currently don't have access to these neat visual tools used in other solutions.

Answer 46

Clojure, 14 bytes

#(compare % 0)

This uses the built-in compare function of clojure: it returns a 1 if the first arg is greater than the second arg, 0 if it's equal, and -1 if it's smaller.

Usage:

(#(...) {number})

Answer 47

GNU sed, 17 15 14 bytes

-2 bytes thanks to zeppelin. -1 byte thanks to manatwork.

/^0/!s/\w\+/1/

Try it online!

Answer 48

Haskell, 6 bytes

signum

Just a boring built-in.

Answer 49

Perl 6, 9 bytes

+(*cmp 0)

Try it online!

Explanation:

+(     # turn into a number

  *    # Whatever (input)
  cmp  # compared to
  0    # 0

)

Answer 50

CJam, 1 byte

g

Try it online!

Another built-in, just for the sake of completeness.

Answer 51

Java 1.8, 11 bytes

Math.signum

In the comments, some people pointed out that this might not technically be valid, if so, here's another version at 12 bytes:

Math::signum

Answer 52

Wonder, 4 bytes

sign

Usage:

sign 1

Builtin.

Bonus solution (no builtin), 7 bytes

tt cmp0

Usage:

(tt cmp0)5

Uses a compare function with 0.

Answer 53

Javascript, 18 bytes

x=>x/Math.abs(x)|0

x/Math.abs(x) is always 1 if x is positive and -1 if x is negative. If x is 0, it returns Nan, which we transform to 0 with the |0 bit.

Answer 54

Acc!!, 50 49 bytes

N
Count i while 45/_ {
Write _
49
}
Write 48+_/49

Acc!! reads input from stdin one character code at a time. This program decides what to output simply based on the first character of the input:

• If it's -, output -1
• If it's 0, output 0
• Otherwise, output 1

Since Acc!! is a very bare-bones language, we have to use a loop for a conditional and integer division for comparison.

Commented version

# Read a character code from input and store it in _ (the accumulator)
N
# If that character was a minus sign (ASCII 45), 45/_ will be 1 and this loop will run
# If that code was a digit (ASCII 48-57), 45/_ will be 0 and the loop will be skipped
Count i while 45/_ {
  # For negative numbers, output the minus sign
  Write _
  # Set the accumulator to ASCII code of 1 so we will break out of the loop and write a 1
  49
}
# If the input was 0 (ASCII 48), _/49 will be 0 and the next line will write a 0
# Otherwise, _/49 will be 1 and the next line will write a 1
Write 48+_/49

Answer 55

Gema, 9 characters

Just a rewrite of Jordan's Sed solution.

0=0
<D>=1

Sample run:

bash-4.3$ gema '0=0;<D>=1' <<< $'-303\n-12\n-5\n0\n1\n20\n404'
-1
-1
-1
0
1
1
1

Gema, 19 characters

*=@cmpn{*;0;-1;0;1}

Posted just because Gema has a nice function for this task:

@cmpn{number;number;less-value;equal-value;greater-value}
Compare numbers.

Sample run:

bash-4.3$ gema '*=@cmpn{*;0;-1;0;1}' <<< -303
-1

bash-4.3$ gema '*=@cmpn{*;0;-1;0;1}' <<< 0
0

bash-4.3$ gema '*=@cmpn{*;0;-1;0;1}' <<< 404
1

Answer 56

Batch, 23 bytes

@cmd/cset/a"%1>>31|!!%1

>>31 evaluates to -1 if the input is negative and 0 if it is positive, while !! evaluates to 1 if it is nonzero and 0 if it is zero, so we just have to bitwise OR the two results together.

Answer 57

brainfuck, 23 bytes

-<,
[
  [
    >-[-<]
    >
  ]
  ->[>+>]
]
<.

This takes a single byte from stdin and interprets it as a signed char, printing \xff for negative, \x00 for zero, and \x01 for positive.

Try it online.

Viewed as an unsigned char, we are checking whether it is greater than 127, with 0 as a special case. We can do the comparison by decrementing from 255 twice at a time.

Answer 58

Python 3, 23 bytes

lambda n:n and n/abs(n)

I know I can make it shorter by doing (n>0) - (n<0), but everyone else seems do be doing that so I thought I would do something different.

Answer 59

GNU Sed, 19 bytes (20 counting the r flag)

/^0+$/b;s/[0-9]+/1/

(Using the -r flag)

Translation:

/^0+$/b;

If the number is zero, skip to the end of the script

s/[0-9]+/1/

Substitute any other numbers with 1. If it's negative the negative sign will remain too.

Answer 60

Befunge-93, 23 21 20 bytes

Thanks to @Mistah Figgins for saving me two three bytes

I'm sure this is further golfable. I'll look at it again in the morning.

&:#@!#._0`#@:#._1-.@

Try it online!

Only takes in one line of input for each run, but that's within spec, I guess.

Answer 61

Triangular, 26 15 bytes

$\:-|0U%<g/l0P<

Formats into this triangle:

     $ 
    \ : 
   - | 0 
  U % < g 
 / l 0 P < 

Try it online!

---

Old broken version that I understand:

$\:-%0U..g/l0P<

Try it online! Currently nonworking until Dennis pulls; found some interpreter bugs.

Formats into this triangle:

    $
   \ :
  - % 0
 U . . g
/ l 0 P <

How it works: The code, without directionals, is read as $:0gP0lU-%.

• $ reads an integer from standard input.
  stack: i
• : duplicates the top stack value.
  stack: i,i
• 0 pushes 0 to the stack.
  stack: i,i,0
• g pushes i>0 to the stack and discards both values used (thanks, Luis Mendo).
  stack: i,i>0
• P pops the top stack value into the register.
  stack: i
• 0 pushes 0 to the stack.
  stack: i,0
• l pushes i<0 to the stack and discards the values used.
  stack: i<0
• U pulls the register onto the stack.
  stack: i<0,i>0
• - computes a postfix subtract.
  stack: i<0-i>0
• % prints the top stack value as an integer.

Idea thanks to caird.

Answer 62

GolfScript, 13 bytes

~.{.abs/}{}if

Try it online!

How it Works

Divide by itself if 0 otherwise do nothing, and leaving a zero on the stack to be printed.

Answer 63

Befunge-93 (PyFunge), 12 bytes

~"1"%90p1X.@

Try it online!

Note that the X can be anything (except a new line), as it gets written over during run-time. It's just easier for explanation

Similar to my other Befunge answer, but this time it mods the first character by the ASCII for 1 first, so that a positive first digit will turn into a no-op, leaving the 1 on the top of the stack:

~               Read the first *character* of input - either a digit or "-"
 "1"%           Mod the character by the ASCII value of 1. After this step, the character
                is a '-' for negative numbers, '0' for 0, and small, unprintable
                characters for positive numbers

     90p        Puts the character in the space with the X.
        1       Pushes a 1

         X      3 different options based on the character that was put here:
         -      Negative: Subtract the 1 from the implicit 0 to get -1
         0      Zero:     Push 0
               Positive: A no-op, which leaves the 1 on top

          .     Prints out the top of the stack
           @    Ends the program

Answer 64

Minecraft Functions (18w11a, 1.13 snapshots), 131 bytes

Uses a function named a in the minecraft namespace

execute if score @p n matches 0 run say 0
execute if score @p n matches 1.. run say 1
execute if score @p n matches ..-1 run say -1

"Takes input" from a scoreboard objective named n, create it with /scoreboard objectives add n dummy and then set it using /scoreboard players set @p n 8. Then call the function using /function a

Answer 65

Hexagony, 16 bytes

...!/~.?.<.!1@.,

Try it online!

Not much smaller than the previous answer, but strangely elegant in comparison.

Expanded:

   . . .
  ! / ~ .
 ? . < . !
  1 @ . ,
   . . .

Answer 66

Japt, 2 bytes

Ug

Test it online!

U is the input number, and g is the sign function on numbers. Output is implicit.

Answer 67

Pyth, 2 bytes

._

herokuapp

Pyth's sign function.

Answer 68

QBIC, 18 8 bytes

:?sgn(a)

This utilizes Qbasic's SGN() function. : gets the input in variable a, ? prints.

---

Original version, before I learnt that QBasic has a SGN() function:

:~a=b|?a\?a/abs(a)

18 bytes. Explanation

:           Get 'a' from the command line
~a=b        If a == b (and b==0 by default)
|?a         Then print a
\?a/abs(a)  Else, print a / abs(a) --> -2/2 leaves the req. -1, 4/4 = 1

Answer 69

Matlab, 4 bytes

sign

Matlab as well has a builtin for it.

Answer 70

Octave, 4 bytes

As with many others, a built-in:

sign

---

Please, read the first sentence of this meta post before voting.

Answer 71

WolframAlpha, 3 bytes

Try it online: sgn

Answer 72

Befunge, 11 bytes

&:0`\0\`-.@

Try it online!

This is just the obvious (N > 0) - (0 < N) calculation.

&               Read N from stdin.
 :              Make a duplicate copy.
  0`            Calculate N > 0.
    \           Swap the second copy to the top of the stack.
     0\`        Calculate 0 > N.
        -       Subtract the two comparisons: (N > 0) - (0 < N)
         .@     Output the result and exit.

As Martin Ender pointed out, there's potentially a 2-byte shorter solution, using the same idea as his ><> answer:

1~"/"-%.@

Unfortunately this only works if the result of a modulo operation takes the sign of the divisor, which is not that common in Befunge implementations (in particular the reference interpreter doesn't work this way).

1               Push 1 onto the stack for later use.
 ~              Read a character of input (this will be '-' or an ASCII digit).
  "/"-          Subtract 47.
      %         Take the modulo of the 1 we pushed earlier with this difference.
       .@       Output the result and exit.

If you want to try this out you'll probably need to use one of the Python-based interpreters like PyFunge or Befungee. I suspect Fungi might work too.

Answer 73

PowerShell, 22 bytes

[math]::sign($args[0])

Boring built-in, calls the .NET function that does exactly what it says on the tin. Ho-hum.
Try it online!

---

For 26 bytes however, we get the classic greater-than less-than equation

param($b)($b-gt0)-($b-lt0)

This, at least, has a little bit of logic and thought put into it. Try it online!

---

Best yet, though is 44 bytes, where we roll our own solution.

param($b)if("$b".indexof('-')){+!!$b;exit}-1

Here we take input $b, stringify it, take the .IndexOf('-') on it, and use it in an if clause. If the negative sign isn't found, this returns -1, which is truthy in PowerShell, so we turn $b into a Boolean with !, invert the Boolean with another !, cast it as an int with +, leave it on the pipeline, and exit. This turns a positive integer (which is truthy) into $false, then $true, then 1, while turning 0 into $true, then $false, then 0. Otherwise, the .IndexOf returned 0 (meaning it was the first character in the string), which is falsey, so we skip the if and just place a -1 on the pipeline. In either case, output via implicit Write-Output happens at program completion. Try it online!

Answer 74

SAS Macro Language, 43 bytes

In the extremely wordy language...

%macro s(n);%put(%sysfunc(sign(&n)));%mend;

Answer 75

Jellyfish, 3 or 6 bytes

3 bytes with built-in:

p*i

Print (p) the sign (*) of the input (i). Automatically threads over lists.

6 bytes without built-ins:

p%S
 +i

Print (p) the division (%) of the input (i, taken from south with S) by the absolute value (+) of the input. Conveniently, division by 0 yields 0 in Jellyfish. This version also threads over lists. Try it online!

Answer 76

Clojure, 23 bytes

#(condp > % 0 -1 1 0 1)

This condp macro expands to "if less than 0 return -1, if less than 1 return 0 else 1".

(macroexpand '(condp > % 0 -1 1 0 1))
(let* [pred__7749 > expr__7750 %] (if (pred__7749 0 expr__7750) -1 (if (pred__7749 1 expr__7750) 0 1)))

Answer 77

PHP, 39 38 bytes

no comparison operators

<?=($n=$argv[1])&PHP_INT_MIN?-1:1-!$n;

should work on most systems.

PHP_INT_MIN has only one bit set: the most significant one. If this is set in the input, it is negative.
!$n (cast to integer by the subtraction) evaluates to 0 for positive values and 1 for 0.

lame solution, 30 bytes

<?=($n=$argv[1])?abs($n)/$n:0;

works also on floats.

Answer 78

Python 2, 57 bytes but no conditionals or comparitors

Just to be different, here's a solution that avoids all those ugly arithmetic functions:

def s(n):
 try:r=len([1][:n])*2-n/n
 except:r=0
 return r

Slicing a non-empty sequence [1][:n] returns [1] when n is positive and [] when negative or zero, so to distinguish these cases, n/n throws a divide by zero error for n=0.

Answer 79

Javascript, 37 bytes

function s(n){return n>0?1:n<0?-1:0}

Answer 80

awk, 17 bytes

!$0||$0=$0<0?-1:1

Test it:

$ echo 0 | awk '!$0||$0=$0<0?-1:1'
0
$ echo 2 | awk '!$0||$0=$0<0?-1:1'
1
$ echo -2 | awk '!$0||$0=$0<0?-1:1'
-1

Answer 81

MarioLANG, 868 bytes

;                                                    
=
  [@:
 =======================================================
                                                 ))    <
 ======================================================"
@ ((((++
=======            <
        ==========="
      @ -)+)+)+((([!))+((
      =============#====    (<
                         ===="
                       @ +)-[!)))
                       ======#==      )  <
                                 ========"
                                      >([!) 
                                      "==#        ))  <
                               @ -(-)[!+   ==========="
                               =======#:          >(([!!
                                                  "===##
                                         @ +((-))[!-
                                         =========#:

Ungolfed, with comments

;[>                                                        
==                                                                                                   output zero
    [@                                                                                            :                                                                      * start
  ==================================================================================================                  memory: [limit|limit_copy|counter_add|counter_sub|arg]
                                                    move pointer back to arg  ))    <
  =================================================================================="===
       increase counter limit
@   ((((++
=================    set counters
                              <
                  ============"     increase counter_add by one
                @ -)+)+)+ ((([! ))+((
                ==============#======
                                        reset limit
                                            (<
                                      ======="
                                    @ +)-   [! )))
                                    =========#====   try subtraction
                                                           )  <
                                                   ==========="
                                                         > ( [! ) 
                                                         "====#==
                                                 @ -(-) [!               try addition
                                                 ========#                   ))  <
                                                                  ==============="
                                                                           > (( [!  !
                                                                           "=====#==#
                                                                @  +((-)) [!
                                                                ===========#



                                                             output one       output minus one
                                                           +:                -:
                                                          ===               ===

What's going on?

The program maintains 5 memory fields (right to left):

• Input value, continuously edited in search of zero
• Counter for search in negative direction (subtraction)
• Counter for search in positive direction (addition)
• Helper to reset search radius (limit)
• Current search interval radius (limit, k)

The algorithm keeps on searching for zero in both (+ and -) directions, starting at the input value. It does k negative and k+1 positive steps on each iteration, then increases k by 2. Once zero has been found, 1 or -1 is output, depending from which side it was reached.

Detection of zero as input is a special case, handled right at the beginning.

Try it online (commented, ungolfed version)

Answer 82

Java 8, 33, 17, 14 bytes

i->i>0?1:i>>31

Does not rely on any questionable code constructs or fragments. This is a complete functional interface implementation.

Answer 83

Python 3, 13 bytes

n//abs(n-.1)

Answer 84

C, 23 bytes

A more portable (I think) 23-byte solution in C:

f(n){return(n|1)%2-!n;}

Answer 85

C#, 40 bytes

b=>System.Console.Write(b>0?1:b<0?-1:0);

Or with a built-in:

C#, 44 bytes

using System;b=>Console.Write(Math.Sign(b));

Unfortunately it's longer, then the first solution.

Answer 86

MATLAB + Octave, 15bytes

There are a few other Octave/MATLAB answers, but two of the others are simply using a built in, and the other is significantly longer.

The anonymous function:

@(a)(a>0)-(a<0)

Quite simple. If a>0, the answer will be (1-0)=1. If a<0, the answer will be (0-1)=-1. If a==0 the answer will be (0-0)=0.

You can try online here. Simply run the above code and then try with ans(input).

Answer 87

Scala, 16 bytes

n=>n compareTo 0

Answer 88

Common Lisp, 21 6 bytes

signum

Try it online!

Answer 89

dc, 22 bytes

[pq]sqd0=qdd*vr1-d*v-p

Try it online!

I don't like that 10 bytes of this is eaten up testing for zero, will continue to mull over that. The second half, dd*vr1-d*v-p uses the square root of the square to calculate the absolute value of both our value to test and that value less one. Subtracting the latter from the former yields 1 for a positive value, -1 for a negative.

Answer 90

Aceto, 4 bytes

riyp

ri reads input as integer
y puts sign on stack
p prints it

Try it online!

Answer 91

Brain-Flak, 40 38 bytes

{<>(())<>{(([{}[]]))<>([{}])<>}}<>({})

Try it online!

A bit longer at 50 bytes, but here's a version that acts only on the top value of the stack, returning 0, -1 or 1 depending on the sign of the number.

({<({}([(())])){({}([{}([{}])]))}{}{}>{}(<()>)}{})

How it Works:

{ If number is not 0
 <>(())<> Push 1 to the other stack as the sign
 { While number != 0
  (([{}[]])) Push -(number+stack height) twice
  <>([{}])<> Negate sign, alternating it between -1 and 1
 } end while
}<>({}) Switch to other stack and force a 0 if the stack is empty

Answer 92

FORTRAN 77, 87 bytes

      READ*,I
      IF(I)1,2,3
1     PRINT*,-1
2     PRINT*,0
3     PRINT*,1
      END

It is a nice use for the "harmful" arithmetic if statement. Unfortunately, this lovely feature was obsolete in Fortran 90 and posterior versions.

Curiously, gfortran can't handle with this arithmetic if, even if I save the file with .f extension. Therefore, I could not test this code.

Answer 93

Whitespace, 73 72 68 bytes

[S S S T    N
_Push_1][S N
S _Duplicate_1][S N
S _Duplicate_1][T   N
T   T   _Read_as_integer][T T   T   _Retrieve][S N
S _Duplicate_input][S N
S _Duplicate_input][N
T   T   N
_If_negative_jump_to_Label_NEGATIVE][N
T   S S N
_If_0_jump_to_Label_ZERO][S N
T   _Swap_top_two][T    N
S T _Print_as_integer][N
N
N
_Exit][N
S S N
_Create_Label_NEGATIVE][S S T   T   N
_Push_-1][T N
S T _Print_as_integer][N
N
N
_Exit][N
S S S N
_Create_Label_ZERO][T   N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Example runs:

Positive:

Command    Explanation                   Stack           Heap     STDIN    STDOUT    STDERR

SSSN       Push 1                        [1]             {}
SNS        Duplicate top (1)             [1,1]           {}
SNS        Duplicate top (1)             [1,1,1]         {}
TNTT       Read STDIN as integer         [1,1]           {1:5}    5
TTT        Retrieve at heap 1            [1,5]           {1:5}
SNS        Duplicate top (5)             [1,5,5]         {1:5}
SNS        Duplicate top (5)             [1,5,5,5]       {1:5}
NTTN       If neg.: Jump to Label_NEG    [1,5,5]         {1:5}
NTSSN      If 0: Jump to Label_ZERO      [1,5]           {1:5}
SNT        Swap top two                  [5,1]           {1:5}
TNST       Print top as integer          [5]             {1:5}             1
NNN        Exit                          [5]             {1:5}

Program stops with an error: Label does not exist
Try it online (with raw spaces, tabs and new-lines only).

Zero:

Command    Explanation                   Stack           Heap     STDIN    STDOUT    STDERR

SSSN       Push 1                        [1]             {}
SNS        Duplicate top (1)             [1,1]           {}
SNS        Duplicate top (1)             [1,1,1]         {}
TNTT       Read STDIN as integer         [1,1]           {1:0}    0
TTT        Retrieve at heap 1            [1,0]           {1:0}
SNS        Duplicate top (0)             [1,0,0]         {1:0}
SNS        Duplicate top (0)             [1,0,0,0]       {1:0}
NTTN       If neg.: Jump to Label_NEG    [1,0,0]         {1:0}
NTSSN      If 0: Jump to Label_ZERO      [1,0]           {1:0}
NSSSN      Create Label_ZERO             [1,0]           {1:0}
TNST       Print top as integer          [1]             {1:0}              0
                                                                                     error

Program stops with an error: No exit defined
Try it online (with raw spaces, tabs and new-lines only).

Negative:

Command    Explanation                   Stack           Heap     STDIN    STDOUT    STDERR

SSSN       Push 1                        [1]             {}
SNS        Duplicate top (1)             [1,1]           {}
SNS        Duplicate top (1)             [1,1,1]         {}
TNTT       Read STDIN as integer         [1,1]           {1:-5}   -5
TTT        Retrieve at heap 1            [1,-5]          {1:-5}
SNS        Duplicate top (-5)            [1,-5,-5]       {1:-5}
SNS        Duplicate top (-5)            [1,-5,-5,-5]    {1:-5}
NTTN       If neg.: Jump to Label_NEG    [1,-5,-5]       {1:-5}
NSSN       Create Label_NEG              [1,-5,-5]       {1:-5}
SSTTN      Push -1                       [1,-5,-5,-1]    {1:-5}
TNST       Print top as integer          [1,-5,-5]       {1:-5}            -1
NNN        Exit                          [1,-5,-5]       {1:-5}

Program stops with an error: Label does not exist
Try it online (with raw spaces, tabs and new-lines only).

Answer 94

TIS -n 1 1, 57 bytes

@0
MOV UP ACC
ADD 998
SUB 998
SUB 998
ADD 998
MOV ACC ANY

Try it online!

The integral data type supported by TIS ranges from -999 to 999. Any value in input, code, or calculations is silently coerced (clamped) into that range (instead of truncation, overflowing, etc.).

So, by adding 998 to any number, then subtracting the same amount, it becomes 1 is it was positive, and stays untouched otherwise. We then do the same to coerce all negative numbers to -1.

We need to subtract across two separate operations, due to the coercion mentioned above.

If you are familiar with TIS-100, you will be used to a 3x4 array of computational nodes, however, this solutions only uses one such node, nominally taking input from above and giving it below.

Answer 95

naz, 64 bytes

9a5m2x1v3a2x2v1x1f1o0m1a1o0x1x2f1o0x1x3f1r3x1v1e3x2v2e0m1a1o0x3f

Explanation (with 0x commands removed)

9a5m2x1v                 # Set variable 1 equal to 45 ("-")
3a2x2v                   # Set variable 2 equal to 48 ("0")
1x1f1o0m1a1o             # Function 1
                         # Output once, set the register equal to 1, and output again
1x2f1o                   # Function 2
                         # Output once
1x3f                     # Function 3
    1r                   # Read a byte of input
      3x1v1e             # Jump to function 1 if the register equals variable 1
            3x2v2e       # Jump to function 2 if the register equals variable 2
                  0m1a1o # Otherwise, set the register equal to 1 and output
3f                       # Call function 3

Answer 96

Keg, -hr, 13 bytes

:0<[0;|0>[1|0

Try it online!

A very simple switch like comparison happening here.

Answer 97

05AB1E, 2 bytes

A signum built-in.

.±

Try it online!

Answer 98

Scratch 3.0, 7 blocks/39 bytes

For those who are tired of the old way of introducing these answers (I'm looking at you a'_' (ಠ ͟ʖಠ) ), I'll put the ScratchBlocks Syntax first:

define f(n
set[r v]to(<(n)>(0)>-<(n)<(0

I decided to use a function instead of the old when gf clicked approach because that means I don't have to deal with having an ask() and wait next.

Try it not online but on Scratch

How This All Works

Functions in Scratch are special. Unlike functions in other programming languages, Scratch functions don't return anything. Rather, they modify global variables.

My reasoning is that if one defines a variable used solely for returning a value, then they can create a sort of "footer" elsewhere which looks something like this:

when gf clicked
ask () and wait
f (answer) 
say (r) 

This is like having a code and footer section on TIO: the function makes up the code part and is where the byte count is, while the when gf clicked part is the footer, consequently not counting towards the byte count.

Answer 99

F#, 31 9 bytes

Math.Sign

(Saved 22 bytes thanks to @pmbanka's advice :))

fun(i:int)->System.Math.Sign(i)

I was about to post a C# solution.. but it already exists :(

So I decided to have some fun with F#!!

//a solution with 24 bytes, but you have to put `open System` into the file. 
//Not sure if that would be valid...

fun(i:int)->Math.Sign(i)

Answer 100

Pyke, 5 bytes

0'<>-

Try it here!

0'<>  - 0<input, 0>input
    - - ^-^

Answer 101

Math++, 28 bytes

?>a
3+2*!a>$
a/sqrta*a
0>$
0

Answer 102

Loader, 61+44+1=106 bytes

Main module, 61 bytes

readline 0
set B =48
load a
set B =45
set K -1
load a
print 1

Module a, 44 bytes

decr B
decr *0
B:load a
~*0:print K
~*0:exit

Answer 103

PARI/GP, 4 bytes

sign

Another boring built-in.

Answer 104

Clojure, 24 bytes

#(cond(> % 0)1(< % 0)-1 0 0)

An anonymous function that checks for being greater than 0, less than 0, then defaults on 0 if neither is true. cond acts as (and is a macro resulting in) an if-else tree. Unfortunately, it requires an even number of arguments, so I had to waste 2 bytes by adding an extra number to mean true to default on.

Ungolfed:

(defn sign [n]
  (cond
    (> n 0) 1
    (< n 0) -1
    :else 0))

Answer 105

GameMaker Language, 22 bytes

return sign(argument0)

Alternative solutions:

return median(1,argument0,-1)
a=argument0 return (a>0)-(a<0)
a=argument0 return max(-1,min(a,1))
a=argument0 if a return a/abs(a)return 0

Answer 106

Go, 49 bytes

func s(x int)int{return(x>>63)|int(uint(-x)>>63)}

Note: this won't work on the playground until Go 1.8 is released due to this bug: https://github.com/golang/go/issues/16203

Answer 107

BotEngine, 7x17=119

v #  #0123456789
>Ie~>SSSSSSSSSSS
    ^ <<<<<<<<<<P
  PeeS-        1e
 Pe1-S0123456789
  0  > SSSSSSSSS
       >>>>>>>>>^

Answer 108

OCaml, 4 bytes

sign

Didn't even know there was a built-in for this. Alternative 18 bytes solution:

fun x->compare x 0

Answer 109

Stacked, 4 bytes

sign

Takes input from top of stack. Bonafide function:

$sign

Shortest non-builtin so far:

:0>\0<-

Which does:

:0>\0<-                        stack: [..., n]
:        duplicate TOS         stack: [..., n, n]
 0       push 0                stack: [..., n, n, 0]
  >      pop n,0 and push n>0  stack: [..., n, n>0]
   \     switch top two        stack: [..., n>0, n]
    0    push 0                stack: [..., n>0, n, 0]
     <   pop n,0 and push n<0  stack: [..., n>0, n<0]
      -  subtract top two      stack: [..., (n>0)-(n<0)]

Surround with [] to be a function.

Answer 110

TI-Basic, 11 bytes

:Input A
:A>0
:Ans-A<0

Reads from variable A.

:A>0            // Reads from A, returns 0 or 1
:Ans-A<0.   // Computes A<0 first, then subtracts from previous result. Implicit print.

Answer 111

JavaScript (ES6), 14 bytes

No conditionals, no Math lib usage

x=>(x>0)-(x<0)

Explanation

x > 0 is true/false for positives/non-positives and cast to number (1/0) in case for subtraction. x < 0 is analogous for negatives, so we end up with 1-0 for positives, 0-0 for zero and 0-1 for negatives.

Answer 112

Python 3, 29 bytes

lambda a:(a>>31)|a

Tested--it works. EDIT: Sigh...8 more bytes because the previous solution was taken. EDIT 2: Fixed a bug. EDIT 3: Knocked 10 bytes off. Now this is shorter than the one that was already taken!

Answer 113

Python 2, 26 Bytes

b=input()
print(b>0)-(b<0)

Answer 114

Factor, 71 bytes

Expects a string.

[ 0 swap nth { { CHAR: - [ -1 ] } { CHAR: 0 [ 0 ] } [ drop 1 ] } case ]

ungolfed:

: sign-of ( x -- y ) 0 swap nth
  {
    { CHAR: - [ -1 ] }
    { CHAR: 0 [ 0 ] }
    [ drop 1 ]
  } case ;

Answer 115

MS Excel, 9 Bytes

Takes Input from cell A1.

=SIGN(A1)

Answer 116

Excel VBA, 10 Bytes

Anonymous VBE immediate window function that takes input from cell [A1] and outputs -1,0 or 1 to the VBE immediate window

?Sgn([A1])

Answer 117

Symbolic Python, 24 bytes

Like many other solutions, this uses the formula s(x) = (x > 0) - (x < 0). Note that this solution is non-competing as the language postdates the challenge.

__={}>{}
_=(_>__)-(_<__)

Symbolic Python is a restricted source version of Python: all alphanumeric characters are banned.

The interpreter automatically puts input in the variable _. From there, the code works like so:

• {}>{} generates the value False. This is then assigned to the variable __. Although it's technically a boolean, we use this as the integer 0:
• (_>__) checks whether the input is greater than 0. (_<__) checks whether the input is smaller than 0. These booleans are then interpreted as integers, and the first is subtracted from the second.
• The result of this is put in the variable _, which is automatically printed after execution.

Answer 118

tcl 24 23

puts [expr $n<0?-1:$n>0]

Thanks to http://wiki.tcl.tk/819 page authors!

UPDATE: based on the C answer https://codegolf.stackexchange.com/a/103831/29325

puts [expr !!$n|$n>>31]

can be run on http://rextester.com/live/BKGZ8868 where i tested with -9,0,9

Answer 119

8th, 40 bytes

The following code has the same behaviour of 8th's builtin word n:sgn

: f dup 0; 0 n:> if 1 else -1 then nip ;

Explanation of word f

: f \ n -- -1|0|1
  dup     \ Duplicate input
  0;      \ Check if number is 0. If true, leave 0 on TOS and exit from word
  0 n:>   \ Check if positive
  if 1    \ Return 1 if positive
  else -1 \ Return -1 if negative
  then
  nip     \ Get rid of input
; 

Testing and Output

ok> 42 f .
1
ok> -42 f .
-1
ok> 0 f .
0

Answer 120

Ruby, 19 bytes

Not quite as golfy as my other answer.

->n{("%+b"%n)[0,2]}

Outputs -1, +0, +1

Answer 121

Japt, 1 byte (non-competing)

g

Try it online!

Answer 122

Fourier, 18 17 bytes

I~A<0{1}{`-`}A/Ao

Try it on FourIDE!

Explanation pseudocode:

A = Input
If A<0 Then
    Print "-"
End If
Print A/A

Answer 123

MY, 3 bytes

Hex:

 1A #Input integer
 2A #Sign function
 27 #Output (trailing newline)

Answer 124

Common Lisp, 39 bytes

(defun f(x)(if(= x 0)0(if(> x 0)1 -1)))

Try it online!

Answer 125

Braingolf, 1 byte

s

Try it online!

Yep.

Answer 126

Implicit, 1 byte

±

Try it online!

After implicits, this program looks like:

$±%
$    read integer input
 ±   push sign
  %  print

Answer 127

T-SQL, 36 bytes

CREATE PROC s @ INT AS PRINT SIGN(@)

Usage:

EXECUTE s -5464

Answer 128

Add++, 9 bytes

L,d0<@0>_

Try it online!

How it works

L,  - Create a lambda.
           Example argument: -8
  d - Duplicate;    STACK = [-8 -8]
  0 - Push 0;       STACK = [-8 -8 0]
  < - Less than;    STACK = [-8  0]
  @ - Reverse;      STACK = [ 0 -8]
  0 - Push 0;       STACK = [ 0 -8 0]
  > - Greater than; STACK = [ 0  1]
  _ - Subtract;     STACK = [-1]
      Implicitly return      -1

Answer 129

dc, 15 bytes

[_1*]sa?dd0>a/p

Try it online!

Explanation

This (ab)uses the fact that dc keeps the stack as is, if you're trying to divide by 0 which really helps for this challenge:

[_1*]sa          # store the macro in register a (macro negates the top)
       ?         # push input to stack
        d        # duplicate top
         d0>a    # if top is negative, execute macro
             /   # divides top two values
              p  # print the top of the stack

Answer 130

Husk, 1 byte

±

Try it online!

Answer 131

REXX, 16 bytes

say sign(arg(1))

Answer 132

Deorst, 11 bytes

i:l0<@l0>@-

Try it online!

Boring (n>0)-(n<0) solution

Answer 133

Functoid, 30 28 bytes

In functoid there is no support for negative integers since numbers are handled as Church numerals. This solution redefines numbers in a way that supports negative numbers (please refer to the explanation):

<|12K0K$BbB
r<@.1,"-"
)<@.([

Try it online! (Here is an input generator)

Explanation

The definition of negative numbers extends the Church numerals by simply introducing another lambda abstraction that encodes the sign of a number (the first argument, ie. x3). Here is how the numbers -2,-1,0,1,2 would look like in this notation:

Decimal number | extended Church numerals | absolute value as Church numeral
---------------+--------------------------+---------------------------------
            -2 |     λλλ(x3 (x2 (x2 x1))) |                   λλ(x2 (x2 x1))
            -1 |          λλλ(x3 (x2 x1)) |                        λλ(x2 x1)
             0 |                  λλλ(x1) |                           λλ(x1)
             1 |               λλλ(x2 x1) |                        λλ(x2 x1)
             2 |          λλλ(x2 (x2 x1)) |                   λλ(x2 (x2 x1))

Now for the explanation of the above code: The characters < will change the direction to left, this saves us four bytes but makes it more difficult to read.. Here's how the code would look like without doing this:

BbB$K0K21|
 @.1,"-"r<
    @.([)<

The combinator BbB ensures that K0 and K2 get grouped as one expression each, here's the more verbose code:

$(K0)(K2)1|
  @.1,"-"r<
     @.([)<

Okay now it's quite readable: First $ will apply the input so we'll have a sign-extended Church numeral as current expression. Applying K0 (constant 0) will evaluate to KK0 (which will "swallow" the next two arguments) in the case of a negative number, in the case of a non-negative number it simply vanishes. In the case of a positive number we'll have K2 which will swallow the argument 1 and else we'll finally get 1.

Summing up the function $(K0)(K2)1 gives the following numbers (regular Church numerals):

if input < 0:
    return 0
elif input > 0:
    return 2
else:
    return 1

The command | will set the direction down iff the current expression evaluates to 0 and up in the other case.

• The second line is thus only relevant if the input was negative and it simply prints -1 and terminates
• In the other case (ie. hitting the third line) it applies the current expression to [ which decrements it by 1, prints the expression (as a numeral) and terminates

Disclaimer

It might look like the above extension was done just to make this task easier, however this is not the case. In fact most definitions in the untyped (or rather uni-typed) lambda calculus are made because they simplify a lot of operations that one could be interested (such as Booleans or the Church numerals themselves).

As an example the function abs could be "implemented" by just applying the identity function to such an extended numeral which will get rid of one lambda abstraction (and in case of a negative number, keep the absolute value as is).

If you're interested, here is an article that talks more about this definition which is worth a read.

Answer 134

Yabasic, 16 bytes

Anonymous function that takes input as a number, n, and output to STDOUT

input""n
?sig(n)

Try it online!

Answer 135

Perl 5, 9 + 1 (-p) = 10 bytes

$_=$_<=>0

Try it online!

Answer 136

Pyt, 1 byte

±

Try it online!

Boring built-in answer (padding for body length)

Answer 137

Lua (51 Bytes)

n=tonumber(...) print(n<0 and -1 or n>0 and 1 or 0)

... is the variable that represents arguments in lua, tonumber(...) converts the first argument to a number.

also in lua a and b or c is a structure that returns b if a is truthy or c if a is falsy, you can nest this structure as well

Answer 138

Stax, 2 bytes

:+

Run and debug online!

Explanation

Added for completeness. It is a builtin, but all built-ins starting with a colon in Stax is actually a macro defined using other Stax operations.

Internally it is defined as c0>s0<-, which is simply (x>0)-(x<0).

Answer 139

SHELL ( 20 Bytes )

sed s/[1-9][0-9]*/1/

tests:

echo "789" | sed s/[1-9][0-9]*/1/
1

echo "-789" | sed s/[1-9][0-9]*/1/
-1

echo "0" | sed s/[1-9][0-9]*/1/
0

Answer 140

Swift, 44 bytes

func s(i:Int)->Int{return i>0 ?1:i<0 ? -1:0}

Strange spacing around the ternary options, I know, but this was the shortest way.

Try it online!

Answer 141

Brain-Flueue, 40 bytes

{([({})]){({}())<>([{}()()])<>}}(<>[]{})

Try it online!

Readable version

{
 ([({})])
 {
  ({}())<>
  ([{}()()])<>
 }
}
(<>[]{})

Answer 142

MachineCode, 26 bytes

31c085ff0f9fc0c1ef1f29f8c3

Requires the i flag for integer output. Input must be appended to the code on a separate line. This is equivalent to the C code (n>0)-(n<0). Try it online!

Alternatives with the same bytecount:

• 31c085ff0f95c0c1ff1f09f8c3 - Try it online! Equivalent to the highest voted C submission.
• 89f8c1ff1ff7d8c1e81f01f8c3 - Try it online! Equivalent to this C code.

Answer 143

Canvas, 3 2 bytes

⤢÷

Try it here!

This might be the shortest solution without a builtin that does the problem for you (I'm looking at you, Jelly.)

Explanation:

⤢÷ | *Full code*
 ÷ | Divide
   | The input (implicit)
⤢  | By its absolute value

Answer 144

Gol><>, 11 bytes

I:0(qm$0)+h

Try it online!

How it works

I:0(qm$0)+h

I            Input as number
 :           Duplicate top
  0(         Change top to "top < 0"
    q        If top is true...
     m$      Push -1 and swap top two; the stack is [-1 x]
             Otherwise, skip two commands (m$); the stack is [x]
       0)    Change top to "top > 0"
         +   Add top two
          h  Print top as number and halt

Answer 145

F#, 140 bytes

open Checked
let s n=
 if n=0 then 0
 else
  try
   let mutable p=n
   while p<>0 do
    p<-p+1
   -1
  with| :? System.OverflowException->1

Try it online!

Basically, if the number is non-zero, keep adding 1 to it until you get either 0 (so the original value was negative) or an overflow (so the original value was positive).

That's all right, isn't it?