Convert from base 10 to base 2 without built-in base conversions

Question

Background:

You have been given an assignment to convert base 10 numbers to base 2 without using any premade base conversion functions. You can't use any imported libraries either.

Problem:

Convert an input string from base 10 (decimal) to base 2 (binary). You may not use any premade base conversion code/functions/methods, or imported libraries. Since this is code-golf, the shortest answer in bytes will win.

Input will be anything from -32768 to 32767 (include sign byte handling in your code)

Answer 1

GolfScript - 17 bytes

~{.1&\2/}16*;]-1%

Not too much more verbose than the built-in ~2base.

Answer 2

JavaScript, 46

for(x=prompt(o='');x;x>>>=1)o=(x&1)+o;alert(o)

Answer 3

Brainf*ck, 98 77

Obviously this isn't for the purpose of winning but what would a competition be if it didnt have a brainfk solution

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

Since brainfk can only handle 8bit integers and no negatives I guess it doesn't fully abide by the rules but hey I was never in it to win it.

This actually does work for 16-bit input if your interpreter supports

I even got it to output in ascii values

Here is the annotated code:

++[>++++<-]                       preload 8 onto cell 1
>>,<                                input into cell 2
[-                                  iterate over cell 1
    >>++<                               put 2 in cell 3
    [->-[>+>>]>[+[-<+>]>+>>]<<<<<]      division algorithm: converts {n d} into {0 d_minus_n%d n%d n/d}
    >[-]++++++[->++++++++<]>           clears cell 4 and puts 48(ascii of 0) into cell 5
    .[-]                                output n%2 and clear it (the bit)
    >[-<<<+>>>]                         bring n/2 into cell 2 (to be used for division in next iteration)
<<<<]                               end iterate

Shorter algorithm (77):

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

This one can only handle 8bit integers.

The algorithm works by using a binary counter which is actually very short (one increment is >[->]++[-<+]-<- which then lays out the bits. The issue is that it is difficult to print out all of the bits

That last algorithm can be adapted to fit any number of bits at the expense of bytes. To be able to deal with N bit integers, it requires 53+3*N bytes to encode.

examples:

(1 bit) +>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+<]>[.>]
(2 bit) +>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+>+<<]>[.>]
(3 bit) +>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+>+>+<<<]>[.>]
etc

Answer 4

Mandatory APL answer - 21 22

"01"[1+2|⌊⎕÷2⋆⊖0,⍳15]

Examples:

      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 0
0000000000000000
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 13
0000000000001101
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 9999
0010011100001111
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: -3
1111111111111101
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 32767
0111111111111111

Answer 5

Turing Machine Code, 272 bytes

As usual, I'm using the rule table syntax defined here. You can test it on that site or, alternatively, using this java implementation.

A lot of the code is copied from my decimal-to-hex converter here.

0 * * l B
B * * l C
C * 0 r D
D * * r E
E * * r A
A _ * l 1
A * * r *
1 0 9 l 1
1 1 0 l 2
1 2 1 l 2
1 3 2 l 2
1 4 3 l 2
1 5 4 l 2
1 6 5 l 2
1 7 6 l 2
1 8 7 l 2
1 9 8 l 2
1 _ * r Y
Y * * * X
X * _ r X
X _ _ * halt
2 * * l 2
2 _ _ l 3
3 * 1 r 4
3 1 0 l 3
4 * * r 4
4 _ _ r A

Counts down from the input in base 10 while counting up from 0 in base 2. On decrementing zero, it erases the input block and terminates.

Answer 6

Javascript 59

o='';i=parseInt(prompt());do{o=(i&1)+o}while(i>>=1)alert(o)

Answer 7

Perl, 44

This is my first Perl program ever, so please forgive me if this can be easily golfed down further. Edit: Thank you @primo for taking 7 chars away from my answer.

$x=<>;do{@s=($x&1,@s)}while($x>>=1);print@s

$x=<>;do{push@s,$x&1}while($x>>=1);print reverse@s

The logic is essentially the same as my previous C solution.

Also, uses 64 bits.

Answer 8

Javascript - 56 48 and 36 28 characters

• Does not works with negative numbers.

Thanks to @Blender for shaving 8 characters.

This form takes input and shows output, 48 characters:

x=prompt();for(a="";x;x=~~(x/2))a=x%2+a;alert(a)

If just an instruction that puts in a variable a the binary form of a variable x is needed (and you don't bother in destroying the x value as a side-effect), here it is with 28 characters:

for(a="";x;x=~~(x/2))a=x%2+a

Answer 9

C, 55 chars

Prints an extra leading zero (for the sake of 2 bytes).
Recursion within printf reverses the print order, so the algorithm extracts bits right-to-left but prints left-to-right.

EDIT: Saved a char by using putchar instead of printf.

f(x){(x*=x<0?-printf("-"):1)&&f(x/2);putchar(48+x%2);}

Answer 10

Python - 61 60 characters

x=input();print"".join("01"[x>>i&1]for i in range(15,-1,-1))

Answer 11

Dyalog APL, 11 bytes

2|⌊⎕÷2*⌽⍳16

2| The division remainder when halved of
⌊ the rounded down value of
⎕ the input
÷ divided by each of
2* two to the power of each of
⍳16 {0, 1, 2, ..., 15}

Requires ⎕IO←0 which is default on many systems.

TryAPL online!

Answer 12

PowerShell, 59 87 82 70 bytes

+28 bytes for supporting negative numbers.
-12 bytes thanks to @ASCII-only

param($d)$m=$d-lt0;while($d){$n="01"[$d%2]+$n;$d=($d-$d%2)/2}'-'*$m+$n

Try it online!

Adapted from this code. Takes input through a commandline parameter -d.

Answer 13

APL(NARS), 17 chars, 34 bytes

{2∣⌊⍵÷2*(⍺-1)..0}

It is a copy and modify of Adam answer https://codegolf.stackexchange.com/a/90107 in the way one can add the parameter for the bits lenght, and ⎕IO for this function (here is ⎕IO=1) should have no importance...

  f←{2∣⌊⍵÷2*(⍺-1)..0}
  16 f 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 
  32 f 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 
  32 f ¯1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
  16 f ¯1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
  64 f ¯12345678
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 1 0 1 0 1 1 0 0 1 0 

it seeme easy handle the number of bits in this way (I cheked that last result should be right)

Answer 14

C, 81

char b[17];i=15;main(x){scanf("%d",&x);while(i+1)b[i--]=(x&1)+48,x>>=1;puts(b);}

The output has strictly 16 bits (including padding zeros)

Answer 15

Apps Script + Google Sheets, 147 144 121 bytes

Script

function j(decNumb){var str='';do{str=String(decNumb%2)+str;decNumb=decNumb/2|0;}while(decNumb>=1);return parseInt(str);}

Sheet

=j(b1)

Modified version of this script by ZygD.

Answer 16

Haskell, 66 bytes

c 0=0
c n=c(div n 2)*10+mod n 2
b('-':r)='-':b r
b r=show.c.read$r

Call with b "-1023", add main=interact b for a complete program or try it on Ideon.

c performs the conversion for positive integers.
b r=show.c.read$r converts a string to a number, applies c and converts back to string.
b('-':r)='-':b r strips a possible leading - and re-appends it to the result.

Answer 17

Java 8, 80 71 bytes

n->{String r="";for(int i=n<0?-n:n;i>0;i/=2)r=i%2+r;return n==0?"0":r;}

-9 bytes due to a rule in the comments.. Negative base-10 inputs may return the positive/absolute base-2 value as output apparently.

Explanation:

Try it online.

n->{                   // Method with integer parameter and String return-type
  String r="";         //  Result-String, starting empty
  for(int i=n<0?-n:n;  //  Start `i` at the absolute (non-negative) value of the input
      i>0;             //  Loop as long as `i` is not 0
      i/=2)            //    After every iteration: integer-divide `i` by 2
    r=i%2+r;           //   Prepend the result with `i` modulo-2
  return n==0?         //  If the input is 0:
          "0"          //   Return literal "0"
         :             //  Else:
          r;           //   Return the result-String

Answer 18

Small Basic, 133 bytes

A script that inputs from and outputs to the TextWindow console.

n=TextWindow.Read()
While n>0
c=c+1
x[c]=Math.Remainder(n,2)
n=Math.Floor(n/2)
EndWhile
For i=0To c-1
TextWindow.Write(x[c-i])
EndFor

Try it at SmallBasic.com Requires Silverlight and thus must be run in IE.

I/O is taken/given from the black console.

-22 bytes thanks to @Neil

Answer 19

C (gcc), 50 43 bytes

-7 bytes thanks to ceilingcat.

f(n){printf("-%d"+(~n?n&&f(n/2),1:0),n&1);}

Try it online!

Answer 20

Kotlin, 82 bytes

{s:String->{var i=s.toInt()
var r=""
(0..15).map{r="${i and 1}$r"
i=i shr 1}
r}()}

Try it online!

Answer 21

MATL, 15 17 bytes

t0<?16Ww+]`2&\t]x

Try it on MATL Online

TIO

(+2 bytes removing leading 0 for negative numbers, sign bit should be the first bit.)

Output on MATL Online should be read bottom-up (MSB is at the bottom).

The main part is pretty simple: `2&\t = while the value is greater than 0, divide by 2 and accumulate the remainders.

Handling negative numbers and giving them 2's complement representation was the tricky part. In the end I went with the "subtract from \$ 2^N \$" method of getting a number's two's complement. Since we're only required to handle values upto -32768, for negative numbers the code creates \$ 2^{16} = 65536 \$ with 16W, adds the input to that (eg. 65536 + (-42)), which gives something MATLAB sees as a positive number but represents the input's signed binary representation in 16-bit form.

Answer 22

PowerShell, 43 bytes

param($n)15..0|%{$r+='01'[($n-shr$_)%2]}
$r

Try it online!

Answer 23

><>,  34  33 bytes

|!/?=0:,2-}:%2:+***"  @"(0:
n{\~{

Try it online!

Answer 24

Smalltalk (Smalltalk/X), 63/78

the first version creates an intermediate string (78):

t:=Number readFrom:Stdin.
((15to:1by:-1)collect:[:i|$0+(t>>i&1)]as:String)print

actually, there is no need to create the string; just output the chars (63):

t:=Number readFrom:Stdin.
15to:1by:-1 do:[:i|($0+(t>>i&1))print]

mhmh - is there a shorter way to read to a number?

Answer 25

Python 3.x: 65 characters

b=lambda n:n<2 and'01'[n]or b(n//2)+b(n%2);print(b(int(input())))

Answer 26

C# - 104

string p(int d){var r="";long i=1;while(r.Length<=64){var g=d&i;r=(g!=0)? "1"+r:"0"+r;i=i<<1;}return r;}

This method will convert decimal to binary up to 64 bits.

When executed the above method in Linqpad - rr = p(-32768); rr.Dump();

Output: 01111111111111111111111111111111111111111111111111000000000000000

Answer 27

Bash, 44

f=b+=n/2**e%2*10**e,2**e++/n?f=b:f;echo $[f]

Pass an input value to the script through the environment variable n. The decimal representation of the binary result cannot exceed LONG_MAX.

This should also be compatible with ksh93 and zsh if b and e are initialized to 0 and proper arithmetic expansion is used.