Stay away from zero

Question

Task

Given a non-negative integer n, output 1 if n is 0, and output the value of n otherwise.

Input

A non-negative integer.

• If you would like to accept the string as input, the string would match the following regex: /^(0|[1-9][0-9]*)$/, i.e. it must not have any leading zeroes, except when it is 0.
• If you accept a real integer as input, you may assume that the integer is within the handling capability of the language.

Output

A positive integer, specified above. Leading zeroes are not allowed. Your output should match the regex /^[1-9][0-9]*$/.

Testcases

input output
    0      1
    1      1
    2      2
    3      3
    4      4
    5      5
    6      6
    7      7

Scoring

This is code-golf, so shortest answer in bytes wins.

Standard loopholes apply.

Answer 1

C (gcc), 14 13 bytes

f(n){n=n?:1;}

Thanks to @betseg for reminding me of the n?:1 trick in the comments of the other C answer!

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C, 17 bytes

f(n){return!n+n;}

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C, 16 bytes

#define f(n)!n+n

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Answer 2

Japt, 2 bytes

ª1

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Explanation

ª is a shortcut for JS's || operator. Japt has implicit input, so this program calculates input||1, and the result is implicitly sent to STDOUT.

w1 would work as well, taking the maximum of the input and 1.

Answer 3

Alice, 7 bytes

1/s
o@i

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Explanation

1   Push 1. Irrelevant.
/   Reflect to SE. Switch to Ordinal.
i   Read all input as a string.
    Reflect off bottom right corner. Move back NW.
/   Reflect to W. Switch to Cardinal.
1   Push 1.
    IP wraps around to last column.
s   Sort swap: implicitly convert the input to an integer. Then, if the top stack 
    element is less than the one below, the two are swapped. It basically computes
    min and max of two values at the same time, with max on top.
/   Reflect to NW. Switch to Ordinal.
    Immediately reflect off the top boundary. Move SW.
o   Implicitly convert the result to a string and print it.
    Reflect off bottom left corner. Move back NE.
/   Reflect to S. Switch to Cardinal.
@   Terminate the program.

Answer 4

Pyth, 2 bytes

+!

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Explanation

+!
 !Q    1 if (implicit) input is 0, 0 otherwise.
+  Q   Add the (implicit) input.

Answer 5

JavaScript (ES6), 7 bytes

n=>n||1

Answer 6

Haskell, 5 bytes

max 1

Usage example: (max 1) 0 -> 1.

Nothing much to explain.

Answer 7

Retina, 4 bytes

^0
1

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If the input starts with a zero, replace that with a 1. (Works because the input is guaranteed to have no leading zeros for non-zero values.)

Answer 8

V, 4 bytes

é0À

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Abuses an non-preferred but expected behavior, so I can't really call it a bug. Explanation:

In Vim, commands accept a count. For example, <C-a> will increment a number, but 7<C-a> will increment a number by 7. However, you can't use 0 as a count, because

• 0 is already a command (go the first column), and

• In the context of a text editor, it rarely makes sense to request that a command be run 0 times.

This is fine for a text editor, but usually obnoxious for a golfing language, so V overwrites some commands so that 0 is a valid count. For example, é, ñ, Ä, and some others. However, since <C-a> is a builtin vim command, it is not overwritten, so running this with a positive input gives:

N       " N times:
 <C-a>  "   Increment

But running with 0 as input gives:

0       " Go to column one
 <C-a>  " Increment

Full explanation:

é0          " Insert a 0
  À         " Arg1 or 1 times:
   <C-a>    " Increment

Answer 9

J, 2 bytes

^*

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^ [argument] raised to the power of

* the sign of the argument (0 if 0 else 1)

Because 1=0^0 in J.

Answer 10

R, 13 bytes

max(1,scan())

reads n from stdin. With pmax, it can read in a list and return the appropriate value for each element in the list for +1 byte.

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I should note that there is another fine R solution in 13 bytes by Sven Hohenstein which allows for yet another 13 byte solution of

(n=scan())+!n

which makes me wonder if that's the lower limit for R.

Answer 11

dc, 7

?d0r^+p

Relies on the fact that dc evaluates 0⁰ to 1, but 0ⁿ to 0 for all other n.

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Answer 12

Cubix, 6 bytes

OI!1L@

Somehow managed to fit it on a unit cube... Test it online!

Explanation

Before being run, the code is arranged as a cube net:

  O
I ! 1 L
  @

The IP (instruction pointer) is then placed on the far-left face (I), facing to the right. The instructions run from there are:

I  Input a number from STDIN and push it to the stack.
!  If the top number is non-zero, skip the next instruction.
1  Push a 1 (only if the input was zero).
L  Turn left. The IP is now on the top face facing the !.
O  Output the top item as a number.

The IP then hits ! again, skipping the @ on the bottom face. This is not helpful, as we need to hit the @ to end the program. The IP hits the L again and goes through the middle line in reverse (L1!I) before ending up on the L one more time, which finally turns the IP onto @.

Answer 13

brainfuck, 8 bytes

+>,[>]<.

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Answer 14

V, 5 bytes

ç^0/<C-a>

Where <C-a> is 0x01.

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Explanation

ç                   " On every line
 ^0/                " that begins with a zero do:
    <C-a>           " Increment the number on that line

Answer 15

Jelly, 2 bytes

+¬

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Pretty much exactly my Pyth answer, but it's my first Jelly program.

Answer 16

Oasis, 2 bytes

Uses the following formula: a(0) = 1, a(n) = n

n1

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Answer 17

R 20 16 bytes

pryr::f(n+(n<1))

Answer 18

Brain-Flak, 22, 10 bytes

({{}}[]{})

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Explanation:

If the input is non-zero, then {{}} will pop everything off the stack and evaluate to the input. If it is zero, nothing will be popped, and it will evaluate to zero. So running ({{}}) gives

Non-zero:

n

Zero:

0
0

At this point, we'll add the height of the stack (0 for non-zero, 1 for zero) and pop one more value off the stack. (since the stack is padded with an infinite number of 0's, this will pop either the top 0 or an extra 0)

Answer 19

Brachylog, 3 bytes

∅1|

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Explanation

If we add the implicit ? (Input) and . (Output), we have:

?∅          Input is empty (that is, [] or "" or 0 or 0.0)
  1.        Output = 1
    |       Else
     ?.     Input = Output

Answer 20

MarioLANG, 12 bytes

;
=[
:<+
 =:

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How it works

Mario starts in the top left, initially walking right. He reads an int from input (;) and stores it in the current memory cell. Then he falls off the ground (=), hitting [, which makes him ignore the next command if the current cell is 0.

If the cell is not 0, he'll start walking left (<), output the current cell as an int (:), and fall to his death (end of program).

If the cell is 0, he ignores the command to turn left, and keeps walking right. He increments the current cell (+), outputs it, and falls to his death.

Answer 21

APL (Dyalog), 3 bytes

1∘⌈

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This takes the ceil of the argument and 1.

Answer 22

TI-BASIC, 7 bytes

:Prompt X
:X+not(X

Alternatively,

TI-BASIC, 7 bytes

:Prompt X
:max(X,1

Answer 23

Hexagony, 7 6 bytes

)?<@.!

Expanded:

 ) ?
< @ .
 ! .

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Saved 1 byte thanks to Martin!

If the number is nonzero print it, otherwise add one to it and print that instead.

Answer 24

Java 8, 10 bytes

i->i<1?1:i

• Thanks to @LeakyNun for saving -1 byte
  • Didn't notice it's a non-negative integer

Answer 25

Python, 15 bytes

lambda n:n or 1

Answer 26

dc, 11 bytes

[1]sf?d0=fp

[1]sf stores a macro in register f which pushes 1 to the top of the stack, ? reads input, d0=f runs macro f if input was 0, p prints the top of the stack.

Test:

$ dc -e "[1]sf?d0=fp" <<< 0
1
$ dc -e "[1]sf?d0=fp" <<< 1
1
$ dc -e "[1]sf?d0=fp" <<< 42
42

Answer 27

Excel, 10 Bytes

=A1+(A1=0)

This saves 4 Bytes over the obvious 'IF' statement solution, =IF(A1=0,1,A1).

Answer 28

R, 13 bytes

n=scan();n+!n

Here, scan is used to read the input value n. The negation of n (i.e., !n, 0 or 1) is added to n.

Answer 29

Mathematica, 9 8 bytes

Per Martin Ender:

#~Max~1&

First idea:

#/. 0->1&

Pure function with replaces 0 with 1. The space is necessary or it thinks we are dividing by .0.

Answer 30

Taxi, 517 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l 2 r.Pickup a passenger going to Addition Alley.Pickup a passenger going to Knots Landing.Go to Knots Landing:n 2 r 2 r 1 l.Pickup a passenger going to Addition Alley.Go to Addition Alley:w 1 r 1 l 2 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:n 1 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

Formatted for humans:

Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.
Pickup a passenger going to Cyclone.
Go to Cyclone:n 1 l 1 l 2 r.
Pickup a passenger going to Addition Alley.
Pickup a passenger going to Knots Landing.
Go to Knots Landing:n 2 r 2 r 1 l.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:w 1 r 1 l 2 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:n 1 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.

The key location is Knots Landing which performs the NOT operation so 0 returns 1 and all other numbers return 0. Now, you can go to Addition Alley to add the result from Knots Landing to the original input.

Answer 31

Perl 5, 6 + 2 bytes for the -l and -p flags

$_||=1

Takes input on separate lines from stdin. Runs with the flags -lp.

Answer 32

Braingolf, 8 bytes

!?_:1_;

Explanation:

 ?       If last element on stack > 0
!        Prevent if check from consuming last element on stack
  _      Pop last element on stack and print
   :     Else
    1    Add int literal 1 to end of stack
     _   Pop last element on stack and print
      ;  Prevent automatic pop of last element on stack

Answer 33

x86-64 Assembly, 10 9 bytes

Following the standard System V AMD64 calling convention, this function accepts a 32-bit unsigned integer parameter via the EDI register, and returns the result via the EAX register:

89 F8  |     mov   eax, edi    ; move parameter from EDI to EAX
85 C0  |     test  eax, eax    ; test input and set flags
75 01  |     jnz   Finished    ; if input is non-zero, then jump to end
FF 40  |     inc   eax         ; otherwise, input is 0, so increment it to 1
       |  Finished:
C3     |     ret               ; return, leaving result in EAX

---

x86-32 Assembly, 8 bytes

We could also write this as 32-bit code, since we're only dealing with 32-bit values.

The nice thing about this is that the INC instruction becomes only 1 byte in length. (In 32-bit mode, inc eax can be encoded simply as 40. In 64-bit mode, 40 is interpreted as a REX prefix, so the leading FF is needed. See Intel's documentation for this instruction.)

The caveat is that most 32-bit calling conventions pass parameters on the stack, rather than in registers, and loading a value from the stack (memory) takes many more bytes. If we can be allowed a __fastcall-style calling convention that passes parameters in registers (supported by virtually all C compilers, so not really cheating, just a bit less standard), then the integer parameter is passed in ECX and we get the following code, for a total of 8 bytes:

89 C8  |     mov  eax, ecx
85 C0  |     test eax, eax
75 01  |     jnz  Finished
40     |     inc  eax
       |  Finished:
C3     |     ret

Answer 34

GCC command line, 9 bytes

-Df(x)x?:

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-D in GCC checks if the following content has an =; if not, it add the code

#define <followed_content> 1

which in this case turns into

#define f(x)x?: 1

Answer 35

Ruby, 11 bytes

->n{n|1>>n}

Answer 36

PHP, 11 Bytes

<?=$argn?:1;

Online Version

Answer 37

Binary-Encoded Golfical, 12 bytes

This binary encoding can be converted back to the standard graphical representation using the encoder provided in the Golfical github repo, or run directly using the interpreter by adding the -x flag.

Hexdump of binary encoding:

00 40 02 15 17 14 00 01 23 1D 17 14

Original image:

Magnified 120x, with color labels:

Answer 38

Jelly, 2 bytes

o1

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An alternative to the other Jelly solution. o provides a default value for a zero/empty argument; in this case, we default it to 1.

Answer 39

Perl 5, 10 +1 byte for -p flag = 11 bytes

$_=$_*1||1

Run with -p flag.

Answer 40

Arnold C, 303 bytes

IT'S SHOWTIME
HEY CHRISTMAS TREE i
YOU SET US UP 1
GET YOUR ASS TO MARS i
DO IT NOW
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY
BECAUSE I'M GOING TO SAY PLEASE i
TALK TO THE HAND i
BULLSHIT
TALK TO THE HAND 1
YOU HAVE NO RESPECT FOR LOGIC
YOU HAVE BEEN TERMINATED

Trying to explain it:

IT'S SHOWTIME //main()
HEY CHRISTMAS TREE i //int i
YOU SET US UP 1 //i = 1
GET YOUR ASS TO MARS i // ? compiler told me to add that
DO IT NOW // ? compiler told me to add that
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY // something related to reading from stdin
BECAUSE I'M GOING TO SAY PLEASE i // if(i)
TALK TO THE HAND i //print i
BULLSHIT //else
TALK TO THE HAND 1 //print 1
YOU HAVE NO RESPECT FOR LOGIC //endif
YOU HAVE BEEN TERMINATED //end main()

It even beats this answer!

Answer 41

C (gcc), 11 bytes

f(n){n?:1;}

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Answer 42

x86 opcode, 6 5 bytes

       41                         INC     ECX
entry: E3 FD                      JECXZ   SHORT $-1
       91                         XCHG    EAX,ECX
       C3                         RETN

ECX -> EAX

Answer 43

Sinclair ZX81/Timex TS1000/1500 BASIC ~19 tokenized BASIC bytes

 1 INPUT A
 2 PRINT A+NOT A

This is a more efficient solution than the one below; essentially it takes the NOT value of A (which would be zero in all cases but zero) and adds that to the value of A.

Old answer:

Sinclair ZX81/Timex TS1000/1500 BASIC, ~31 tokenized BASIC bytes

 1 INPUT A
 2 IF NOT A THEN LET A=NOT A
 3 PRINT A

This takes a numeric input from the user and will display 1 if zero is entered by making A NOT A; otherwise it does nothing and displays the numeric value.

Answer 44

Java 8, 9 bytes

n->n|1>>n

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C# .NET, 9 bytes

n=>n|1>>n

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Different approach (and 1 byte shorter) than the existing Java/C# .NET answers. (Also works with negative inputs.)

Explanation:

i->       // Method with integer as both parameter and return-type
  i|      //  Return the input bit-wise OR-ed with:
    1>>i  //  1 bit-wise right-shifted by the input

1>>i will be 1 when the input is 0, and 0 for every other input.
0|1 for input 0 will therefore result in 1, and n|0 for every other input will therefore result in n.

Answer 45

Keg, 4 bytes

:0=+

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Explanation

¿:    # Take 2 inputs
  0=  # Check whether the input is 0
      # (Yields 1 if the input is 0)
      # (Yields 0 otherwise)
    + # Add the input with this result
      # i.e. 0->1, 5->5

Answer 46

GolfScript, 4 3* bytes

~1|

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*thanks to user a'_'

Answer 47

Binary Lambda Calculus, 2.125 Bytes

00010110 00110001 0

Translation into lambda calculus:

\n. n (\x. n) (\x. x)

Answer 48

C, 27 Bytes

f(n){printf("%d\n",n?n:1);}

Answer 49

MATL, 3 bytes

t~+

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Explanation

t   % Implicit inupt. Duplicate
~   % Logical negation. Converts zero to 1, and nonzero to 0
+   % Add. Implicit display

Answer 50

JavaScript (ES6), 8 bytes

Beaten to the shortest solution (as usual!), so here's an alternative.

n=>n?n:1

Try It

f=
n=>n?n:1
console.log(f(0))
console.log(f(1))
console.log(f(8888))

Answer 51

PowerShell, 24 22 17 Bytes

blatantly stolen from here

!($a=$args[0])+$a

Explanation

Invert the value, returning 0 for non-0 numbers, and 1 for 0, then add the intitial to it.

this makes it basically 1/0 + value, so for 0 the first value is 1, any other numbers it's 0.

examples:
# !0+0 = 1
# !1+1 = 1
# !9+9 = 9

Answer 52

///, 11 bytes

/#0/1//#//#

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Since there is no other way to take input in ///, hard-code the input after the last #.

Works by replacing #0 with 1. Then it removes any remaining #. The # makes sure that an input of 10 would not output 11.

Version that takes input in Itflabtijtslwi:

GGaGGGGbGG/#0/1//#//#ab

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Answer 53

Ohm, 11 6 bytes

ö?┼¿1;

Uses CP-437 character encoding. Run with -c flag

Explanation:

ö?┼¿1;
       ■print(                                            )
  ┼    ■      first_input()
 ?   ; ■                   if(                     )
ö      ■                      int(first_input())!=0
   ¿   ■                                            else 
    1  ■                                                 1

Answer 54

AWK, 10 bytes

!$0{$0=1}1

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Example Usage:

awk '!$0{$0=1}1' <<< 978

Answer 55

Clojure, 12 bytes

#(get[1]% %)

The get function takes an associative data structure, a key, and a default value. Vectors are associative using sequential position as the key. So, [1] is a vector with the value 1 at position 0. If get is called with parameter 0 it will return 1, otherwise no other keys exist in the vector so it returns the default value of the parameter.

Answer 56

Universal lambda, 3 bytes (17 bits)

00010110001100010

It is a function and not a complete program. I think it should work, but didn't actually test it because it doesn't seem easy to do so. It means λx.x(λy.x)(λy.y).

Lazy K, 10 bytes

S(SIK)(KI)

It is a function, untested, too.

Answer 57

05AB1E, 7 bytes

D0›i,}1

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D           //push two inputs (implicit)
 0›         //push input greater than zero
   i        //if true
    ,}      //print input
      1     //push 1. printing is implicit if there is no previous output

Answer 58

Brain-Flak, 14 bytes

((){{}[()]}{})

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Just computes: 1 + (n ? n-1 : 0).

Answer 59

Whirl, 38 bytes

01100011100011110011111100001000111100

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Explanation:

01100     op.ccw, op.intio    Mem1 = STDIN
011100    math.ccw, math.=    If (Mem1 = 0) Then (Math.Val = 1) Else (Math.Val = 0)
0111100   op.cw, op.one       Op.Val=1
11111100  math.add            Math.Val = Math.Val + Mem1
00        op.one              Op.Val=1 (Cheapest way to loop back to the Math wheel)
100       math.store          Mem1 = Math.Val
0111100   op.ccw, op.intio    STDOUT = Mem1

Answer 60

k, 2 bytes

1|

Finds the maximum of 1 and whatever number is given.

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Answer 61

Java, 29 bytes

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int f(int n){return n<1?1:n;}

Answer 62

Starry, 14 bytes

, +'      +*`.

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Explanation

Space shown as _ .

,           Read integer and push to stack
_+          Duplicate
'           If non-zero jump to branch label
______+     Push 1
*           Add
`           Mark branch label
.           Print as a number

Answer 63

Japt, 2 bytes

w1

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Returns the larger of 1 and the input.

Answer 64

C++, 22 bytes

[](int i){return!i+i;}

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Answer 65

Excel, 10 bytes

=MAX(A1,1)

Here's another 10-byte solution: link

Answer 66

C#, 11 bytes

n=>n<1?1:n;

This compiles to a Func<int, int>.

Answer 67

SpecBAS - 20 bytes

1 INPUT n: ?n OR n=0

? is shorthand for PRINT

Answer 68

Desmos, 13 bytes

f(x)=max(x,1)

Try it here

Answer 69

Clojure, 10 bytes

#(max 1%)

Not much to explain.

Answer 70

Chip, 20 18 bytes

eaABb
*`\\-!
fcCDd

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How?

 aABb
            Copy the low 4 bits from input to output
 cCDd

---

e
*           Set the higher bits of output, so that the values are ASCII digits
f

---

eaABb
*           Replicate any ASCII digits on input to output
fcCDd

---

            -
     !      Produce a high signal, but only during the first byte
            -

---

 aAB
 `\\-*      Set the lowest bit of output, if the four low bits of input are unset
  CD

---

 aAB        Set the lowest bit of output, if the four low bits
 `\\-!      of input are unset, and only on the first byte
  CD

Answer 71

Aceto, 9 8 bytes

rid0=`1p

read an integer and duplicate it, then push 0. Are they =? Then (`) push a 1. print the top element.

Answer 72

Julia, 13 12 bytes

f(n)=n<1?1:n

Unfortunately, Julia doesn't do implicit casting from int to bool, so I have to burn an entire 3 characters just to do a comparison to zero. Saved one byte by safely assuming the number isn't negative. Still too verbose for my taste, though.

Answer 73

Python 21 Bytes

int(max('1',input()))

Takes input from REPL environment

Answer 74

Groovy, 7 bytes

{it?:1}

Elvis operator; if true, return self, else return 1. Only false integer value auto-unboxxed to false in Groovy is 0. Thusly, exactly the spec.

Answer 75

SNOBOL4 (CSNOBOL4), 49 47 36 bytes

	N =INPUT
	N =EQ(N) 1
	OUTPUT =N
END

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Answer 76

Symbolic Python, 11 bytes

_+=_==(_>_)

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_+=           # Output = Input +
   _==(_>_)                       Input == 0

Answer 77

Google Sheets, 9 Bytes

Anonymous worksheet function that takes input from range A1 and outputs to the calling cell

=Max(1,A1

Answer 78

Bash, 43 bytes

if [ $1 -eq 0 ];then echo 1;else echo $1;fi

Answer 79

Julia 0.6, 10 bytes

x->x<1?1:x

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Answer 80

BitCycle, 8 7 bytes

-1 byte thanks to Jo King

?v<
!+~

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Uses the -U flag to convert decimal inputs to "signed unary," which in this case means unary for positive integers and 0 for zero.

How it works

Let's run two example inputs: 3 and 0.

An input of 3 gets converted to 111 and emerges from the source (?) moving east. It follows the arrow down to the +, where the 1 bits turn right (west). They fall into the sink (!) and are converted back to decimal 3 and output.

An input of 0 gets converted to 0. When the 0 bit reaches the +, it turns left into the dupneg (~). Here the 0 turns right (south, off the playfield) and a negated copy of it turns left (north). This negated copy, being a 1 bit, goes back around to the + and turns right. It falls into the sink, is converted to decimal 1, and is output.

Answer 81

Acc!!, 75 72 bytes

N
Count i while 0^(_%12) {
Write 49
1
}
Count i while _/48 {
Write _
N
}

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Explanation

The problem can be described as:

• If the first character read is 0, output 1 and halt.
• Otherwise, read a string of digits and echo them to output.

N reads a character and stores its code into the accumulator. The first loop triggers if the character read was 0 (i.e. ASCII 48):

• If _ == 48, then (_%12) is 0, and 0^0 is 1, which is truthy.
• If 49 <= _ <= 57, then (_%12) is between 1 and 9, and 0^(_%12) is 0, which is falsey.

Inside the first loop, we write a 1 (ASCII 49) and set the accumulator to 1. This breaks out of the loop because:

• If _ == 1, then (_%12) is 1, and 0^1 is 0, which is falsey.

So when we reach the second loop, the accumulator is either 1 or some number between 49 and 57 (inclusive). The condition this time is _/48, which is truthy for values >= 48. So if the accumulator is 1 (meaning that we went into the first loop), we skip the second loop. Otherwise, we enter it. We write whatever's in the accumulator back to output and read another character, until the ASCII code we read is less than 48 (probably 10 for newline). Then we halt.

Answer 82

><>, 6 bytes

:?!1n;

Could have been 1 byte shorter if ? had the opposite conditional behaviour.

:?       check if nonzero, then either
  !      a) skip the next instruction or
   1     b) push 1 to the stack
    n    print
     ;   terminate

Answer 83

JavaScript (Node.js), 10 bytes

n=>n&&n||1

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here is one more

JavaScript (Node.js), 8 bytes

n=>n?n:1

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and lastly

JavaScript (Node.js), 7 bytes

n=>n||1

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Answer 84

Sed, 8 bytes

Input in any number base, without any leading zeros. (It doesn't even need to be a consistent base!)

s/^0/1/

0 is the only number that begins with 0; just change that to 1.

Answer 85

Zsh, 13 bytes

<<<$[$1?$1:1]

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Bash, 15 bytes

echo $[$1?$1:1]

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Dash (or other POSIX-compliant shell), 17 bytes

echo $(($1?$1:1))

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Answer 86

Triangular, 6 bytes

$,w%1<

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Ungolfed:

  $
 , w
% 1 <
-----------------------
$              Read from input as an integer
 w,            Change directions if ToS != 0 (will go straight to print)
   <1          Push 1
     %         Print ToS as an integer

Answer 87

W, 2 bytes

1|

Explanation

a   # Take an input
 1| # Logical or with 1
    # If n is 0, this evaluates to 1
    # It evaluates to 0 otherwise

Answer 88

naz, 70 bytes

2a2x1v4a8m2x2v1x1f1r3x2v2e1o3f0x1x2f1a1o0x1x3f1r3x1v4e1o3f0x1x4f0a0x1f

Works for any input integer, provided it's passed as a file terminated with the control character STX (U+0002).

Explanation (with 0x commands removed)

2a2x1v           # Set variable 1 equal to 2
4a8m2x2v         # Set variable 2 equal to 48 ("0")
1x1f1r3x2v2e1o3f # Function 1
                 # Read a byte of input
                 # Jump to function 2 if it equals variable 2
                 # Otherwise, output it and jump to function 3
1x2f1a1o         # Function 2
                 # Add 1 to the register and output
1x3f1r3x1v4e1o3f # Function 3
                 # Read a byte of input
                 # Jump to function 4 if it equals variable 1
                 # Otherwise, output it and jump back to the start of function 3
1x4f0a           # Function 4
                 # Add 0 to the register
1f               # Call function 1

Answer 89

CJam, 5 bytes

ri1e>

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Explanation

ri   e# Read input as an integer
1    e# Push 1
e>   e# Maximum. Implictly display

Answer 90

Batch, 17 bytes

@cmd/cset/a%1+!%1

Answer 91

Math++, 3 bytes

?|1

("Body must be at least 30 characters")

Answer 92

Microscript II, 4 bytes

1sN|

Rough translation:

x=1
push x
x=readnum()
x|=pop()

x is then printed implicitly at the end.

Answer 93

braingasm, 4 bytes

;z+:

; reads a number from stdin, z+ increments that number if it is 0, : prints.

Answer 94

QBIC, 11 8 bytes

?:-(a>1)

Thanks to @l4m2 for saving me some bytes!

Explanation:

?         PRINT
 :        an integer taken from the cmd line (and store it as 'a')
  -       minus
   (a<1)  -1 if 'a' is less than 1 (can only be 0) or 0 otherwise.
          This leaves any a to be a, but turns zeroes into 1 by double negative.

---

Old code, that didn't use the inline : yet:

:?(a>0)+a+1

Explanation

:       Get an int from the cmd line, a
?       PRINT
 (a>0)    if a is greater than 0, this is -1, else 0
 +a        yields 0 for 0 and a-1 for >0
 +1        makes 1's for 0 and a's for all other values

Answer 95

CJam, 5 bytes

ri1e|

Explanation:

ri    e# Read token and convert to integer
  1e| e# OR with 0

Answer 96

Ruby, 12 bytes

->n{n<1?1:n}

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---

Alternate version that reads directly from STDIN instead of being a function. Same number of bytes after counting the -p flag (11+1 = 12 bytes). Effectively a port of the Retina solution.

sub /^0/,?1

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Answer 97

Tcl, 25

puts [expr $argv?$argv:1]

demo — How to use: In the green area, type

tclsh main.tcl $n

where $n is the input number. Do not press backspace, otherwise your browser can go back in history!

Answer 98

Beam, 40 bytes

This reads the input as a string as it doesn't have any other option. Basically it checks if the first character is 0 by removing 48 from it. If 0 increment and output as a number, otherwise output it as a number, then read the rest of the input outputting it as characters.

 r'''''>`----\
 :+n++(------/
/<:<
r@
\u

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Answer 99

Befunge-98, 5 bytes

&:!+.

Try it Online! (Warning - will take 60 seconds per test because of how TIO treats &)

Explanation

&:       Push 2 copies of the input
  !      Logical not the 1st copy - i.e. push 1 if 0, push 0 otherwise
   +     Add them together - this results in +0 normally, but +1 in the case of 0
    .    Print that value
         Loops back around to the first command
&        Because there's no more input, the TIO interpreter stalls for a minute and ends.

Befunge-93 Variant, 6 bytes

It's trivial to implement Befunge-93 in a similar way, except we can't use the & to end the program this time. Thus, the code is 1 byte longer, for 6 bytes:

&:!+.@

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Answer 100

Clojure(Script), 10 bytes

#(max 1 %)

Answer 101

Silberjoder, 42 bytes

0+b1,-CB+b1:BC<. +iB-b1+CB-CA[+CA.,-CA]1

How it works:

0

The first three characters are all data and are not executed. They are a newline (10), a zero (48), and a DC3 (19).

+b1

Point b at the "0" character. Note that a is still pointing at the newline character.

,

Read the first character of input.

-CB

Subtract what b is pointing to (the "0") from the first digit.

+b1

Point b at the DC3 character, which has value 19

:BC

Jump to position 19+3=22 if the c is pointing to anything other than zero. This would happen if the first digit of the number was anything other than "0". Otherwise...

<.

Move c on top of the "1" at the end of the program and print it.

 +iB

b is still pointing at 19, so we add 19 to the instruction pointer, jumping to the "1" at the end of the program, causing the program to halt after one more cycle. (The extra space is ignored, but we need it there to position this instruction so that the instruction pointer jumps beyond the "]" at the end of the program. If we don't do this, we will enter the loop at the end, and print an extraneous semicolon whenever 0 is input.)

-b1

This is position 22, so we jump here whenever the number didn't start with "0". We move b back to point at the "0".

+CB

Add the 48 back to the first digit of the number, restoring it to its proper character value.

-CA

Subtract the newline from the digit.

[+CA.

If it's not zero, restore it to its original value and print it.

,-CA]

Repeat reading digits, comparing them with newline, and printing them until newline is seen.

1

Data. Ignored. Program halts.

Answer 102

JavaScript, 19 Bytes

(x)=>{return x?x:1}

Ungolefed, with example:

function f(n) {
  if(n)
    return n;
  else
    return 1;
}

f(0); // 1
f(1); // 1
f(2); // 2

Answer 103

Swift, 25 bytes

max(1, Int(readLine()!)!)

If you just have the body of a closure, then it is just 10 bytes:

max(1, $0)

Takes in standard input and then returns the max int, which would either be 1 or higher.

Answer 104

Charcoal, 4 bytes

Ｉ∨Ｎ¹

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Explanation

Ｉ    Cast (number is casted to string)
 ∨Ｎ¹ input number logical-or 1

Answer 105

Groovy, 9 bytes

f={it?:1}

Inside a groovy script file you can run with "groovy -D n=0 ":

println System.properties.n?:1

As a closure:

f={it?:1}

1) "?:" is groovy's "elvis-operator"

One instance of where this is handy is for returning a 'sensible default' value if an expression resolves to false-ish

2) "it" refers to a single anonymous parameter the closure is called with

3) 0 evaluates to false in groovy-truth

Answer 106

Ruby, 13 bytes

->i{i==0?1:i}

Answer 107

JS ES5, 26 bytes

function (n){return n?n:1}

Answer 108

T-SQL, 21 bytes

SELECT MAX(1,a)FROM t

SQL input is allowed via a pre-existing named table (table t with INT field a).

Answer 109

Excel VBA, ₁₇ 13 Bytes

Anonymous VBE Immediate function that takes input as expected type unsigned integer and then outputs to the VBE immediate window

?[Max(A1,1)]

Previous Version

?[If(A1,A1,1)]

Answer 110

Husk, 3 2 bytes

|1

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Ungolfed/Explanation

    -- implicit input N
|   -- if N is truthy: N
 1  -- else (N==0): 1

Thanks @Zgarb for -1 byte!

Answer 111

,,,, 2 bytes

1∨

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,,, is on TIO now, so that's cool. Computes input or 1 (logical OR) and implicitly outputs the result.

Answer 112

Element, 8 bytes

_2:'!"+`

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Answer 113

Pushy, 4 bytes

&n+#

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   #  \ print:
  +   \   input + ...
&n    \   not(input)

Answer 114

Rust, 27 bytes

|x:u64|if x==0{1}else{x};

Anonymous function, or lambda, taking input from x.

First time golfing with Rust, and i must say im quite impressed.

Answer 115

Befunge, 6 Bytes

&:!+.@

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& Gets input as number
  :! Duplicates and inverts
    +.@ Adds the inverted input to the original, prints and ends the program

Answer 116

Implicit, 7 bytes

$!{.

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$      read input
 !{    if falsy
   .   increment
       implicit output

Answer 117

J, 6 Bytes

(>.1:)

Standard solution: Return the max of 1 and the argument.

The parenthesis ensure it's evaluated as a monadic hook:

(>.1:) 0
0 >. (1: 0)    NB. Definition of a monadic hook.
0 >. 1         NB. 1: is a constant function, always returns 1.
1              NB. >. returns the max of its two arguments.

Answer 118

Clean, 9 bytes

?0=1
?n=n

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Answer 119

tinylisp, 14 bytes

(q((n)(i n n 1

This is a lambda function. In order to be able to call it, you either need to give it a name using d or call it directly (which would require explicitly closing the parentheses before specifying the argument). Try it online!

The function takes one argument, n. If n is truthy (all positive integers), return n. Otherwise (zero), return 1.

Answer 120

Whitespace, 41 bytes

[S S S N
_Push_0][S N
S _Duplicate][T N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][S N
S _Duplicate_input][N
T   S N
_If_0_jump_to_Label_0][T    N
S T _Print_as_integer][N
N
N
_Exit][N
S S N
_Create_Label_0][S S S T    N
_Push_1][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.

Explanation in pseudo-code:

Integer i = STDIN as integer
If i == 0:
  Call function Label_0
Print i
Exit program

function Label_0:
  Print 1
  Exit implicitly with error: Exit not defined

Example runs:

Input: 0

Command    Explanation              Stack    Heap     STDIN    STDOUT    STDERR

SSSN       Push 0                   [0]
SNS        Duplicate top (0)        [0,0]
TNTT       Read STDIN as integer    [0]      {0:0}    0
TTT        Retrieve                 [0]      {0:0}
SNS        Duplicate top (0)        [0,0]    {0:0}
NTSN       If 0: Jump to Label_0    [0]      {0:0}
NSSN       Create Label_0           [0]      {0:0}
SSSTN      Push 1                   [0,1]    {0:0}
TNST       Print as integer         [0]      {0:0}             1
                                                                         error

Try it online (with raw spaces, tabs and new-lines only).
Stops with error: Exit not defined.

Example runs:

Input: 5

Command    Explanation              Stack    Heap     STDIN    STDOUT    STDERR

SSSN       Push 0                   [0]
SNS        Duplicate top (0)        [0,0]
TNTT       Read STDIN as integer    [0]      {0:5}    5
TTT        Retrieve                 [5]      {0:5}
SNS        Duplicate top (5)        [5,5]    {0:5}
NTSN       If 0: Jump to Label_0    [5]      {0:5}
TNST       Print as integer         []       {0:5}             5
NNN        Exit program             []       {0:5}

Try it online (with raw spaces, tabs and new-lines only).

Answer 121

Gol><>, 5 bytes

I:z+h

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Given n, calculate n + !n, print as int and halt. Unfortunately Gol><> doesn't have implicit input option, so the bytes are the same as regular ><>.

Answer 122

05AB1E, 2 bytes

_+

Try it online or verify some more test cases.

2 bytes alternative by @Adnan:

$M

Try it online or verify some more test cases.

Explanation:

_   # Check if the (implicit) input is 0 (0 becomes 1; everything else becomes 0)
 +  # Add it to the (implicit) input (0 becomes 1; everything else stays the same)
    # (and output the result implicitly)

$   # Push both 1 and the input to the stack
 M  # Push the largest number of the stack (without changing the rest of the stack)
    # (and output the top of the stack implicitly as result)

Answer 123

Gaia, 2 bytes

1Ṁ

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Takes the Ṁax of 1 and the input.

Answer 124

Excel VBA 17 bytes

[a1]=IIF(n=0,1,n)

Simple if else statement you can run it from vba immediate window