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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 , the shortest answer in bytes will win.

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

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    \$\begingroup\$ Q: what does "sign byte handling" mean - shall I output "-xxxx" for a negative number? Then some of us are wrong, incl. me, as I output "11...11" for -1 (aka as unsigned) \$\endgroup\$
    – blabla999
    Feb 16, 2014 at 2:19
  • \$\begingroup\$ Sign byte handling - the MSB of signed variables controls if they are negative \$\endgroup\$
    – TheDoctor
    Feb 16, 2014 at 3:12
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    \$\begingroup\$ sure, but do I have to >print< them as sign '-' followed by magnitude? \$\endgroup\$
    – blabla999
    Feb 16, 2014 at 3:14
  • \$\begingroup\$ @blabla999 - No you don't \$\endgroup\$
    – TheDoctor
    Feb 16, 2014 at 3:34
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    \$\begingroup\$ the MSB of signed variables controls if they are negative - that sounds like sign bit, however as the range -32768..32767 suggests, you want 2's complement. So which do you want?.. \$\endgroup\$
    – mniip
    Feb 16, 2014 at 12:03

32 Answers 32

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Java 10, 80 71 57 bytes

n->{var r="";for(n=n<0?-n:n;n>1;n/=2)r=n%2+r;return n+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
  var r="";            //  Result-String, starting empty
  for(n=n<0?-n:n;      //  Transform the input to its absolute value
      n>1;             //  Loop as long as `n` is still >= 2:
      n/=2)            //    After every iteration: integer-divide `n` by 2
    r=n%2+r;           //   Prepend `n` modulo-2 to the result
  return n             //  Return the last `n`
         +r;           //  appended with the result
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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.

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