# Print the amount of ones in a binary number without using bitwise operators

## Description

Given a number, print the amount of 1s it has in binary representation.

## Input

A number >= 0 in base 10 that won't exceed the highest number your language is able to handle.

## Output

The amount of 1s in binary representation.

## Winning condition

The shortest code wins.

## Disallowed

• Bitwise operators. Other operators, like addition and multiplication, are allowed.
• Built-in base conversion functions.

## Examples

Input:     Ouput:

56432      8

Input:     Output:

45781254   11

Input:     Output:

0          0

• Are functions allowed? I want to make a Java solution, but writing full code is too tedious... :/ – HyperNeutrino Feb 13 '16 at 2:40
• I guess I won't be using Wise for this challenge... :) – MildlyMilquetoast Mar 31 '17 at 17:17

## APL, 9 12 characters

+/2|⌊⎕÷2*⍳32


This assumes that the interpreter uses 32-bit integers, and that ⎕IO is set to 0 (meaning that monadic ⍳ begins with 0, rather than 1). I used the 32-bit version of Dyalog APL.

Explanation, from right to left:

• ⍳32 generates a vector of the first 32 integers (as explained before, because ⎕IO is 0, this vector begins with 0).
• * is the power function. In this case, it generates 2 to the power of each element of the vector supplied as its right argument.
• ÷ is the divided-by function. It gives us ⎕ (evaluated user input) divided by each element of the vector to its right (each power of two).
• ⌊ floors each element of the argument to its right.
• 2| gives us the remainder of each element of to its right divided by 2.
• / reduces (folds) its right argument using the function to its left, +.

Not quite 9 characters anymore. :(

Old, rule-breaking version:

+/⎕⊤⍨32/2


Explanation, from right to left:

• 32/2: Replicate 2, 32 times.
• ⍨ commutes the dyadic function to its left, which in this case is ⊤ (i.e., X⊤⍨Y is equivalent to Y⊤X).
• ⊤ is the encode function. It encodes the integer to its right in the base given to its left. Note that, because of the commute operator, the right and left arguments are switched. The base is repeated for the number of digits required, hence 32/2.
• ⎕ is a niladic function that accepts user input and evaluates it.
• +/ reduces (folds) its right argument using +. (We add up the 0's and 1's.)
• Doesn't this break the Built-in base conversion functions contraint? – Gareth Dec 30 '11 at 21:18
• Whoops! Missed that one. – Dillon Cower Dec 30 '11 at 21:19
• Gah! Thought I'd given myself a fighting chance with my J program! :-) Nice job. – Gareth Dec 30 '11 at 23:01
• @Gareth: I didn't realize until reading your explanation just now, but my answer is pretty much identical to yours! I guess that could be expected from APL and J. :) – Dillon Cower Dec 31 '11 at 0:42

# Brainbool, 2

,.


The most reasonable interpretation, in my opinion (and what most of the answers use) of "highest number your language is able to handle" is "largest number your language natively supports". Brainbool is a brainfuck derivative that uses bits rather than bytes, and takes input and output in binary (0 and 1 characters) rather than character codes. The largest natively supported number is therefore 1, and the smallest is 0, which have Hamming weights 1 and 0 respectively.

Brainbool was created in 2010, according to Esolang.

• I knew it must have existed, but it took me an hour of sorting through Brainfuck derivatives on Esolang to find Brainbool. – lirtosiast Jul 8 '15 at 2:17

## J, 13 characters

(+ the number of digits in the number)

+/2|<.n%2^i.32


Usage: replace the n in the program with the number to be tested.

Examples:

+/2|<.56432%2^i.32
8
+/2|<.45781254%2^i.32
11
+/2|<.0%2^i.32
0


There's probably a way of rearranging this so the number can be placed at the beginning or end, but this is my first J entry and my head's hurting slightly now.

Explanation(mainly so that I understand it in the future)

i.32 - creates an array of the numbers 1 to 32

2^ - turns the list into the powers of two 1 to 4294967296

n% - divides the input number by each element in the list

<. - rounds all the divison results down to the next integer

2| - same as %2 in most languages - returns 0 if even and 1 if odd

+/ - totals the items in the list (which are now just 1s or 0s)

• I'll be happy to upvote this once it reads from stdin (or whatever equivalent J has). – Steven Rumbalski Dec 30 '11 at 18:51
• The best I could do I think (maybe, depending on figuring out how) is move the input to the end of the program. Standard input isn't mentioned in the question though? – Gareth Dec 30 '11 at 21:18
• I'm sorry for not specifying the way of input. It would be unfair to change the rules now, so I'll accept this one. I will mention it next time! – pimvdb Dec 30 '11 at 22:21
• @pimvdb No problem, it wasn't a complaint. I think with J programs though all you can do is define a verb that operates on the input given it. Not sure how I'd rearrange this to do that though. Maybe JB or one of the other J experts could help me out with that... – Gareth Dec 30 '11 at 22:59
• ...and having read some more I now see that I was completely wrong about standard input. – Gareth Dec 31 '11 at 10:02

# Brainfuck, 53 characters

This was missing an obligatory Brainfuck solution, so I made this one:

[[->+<[->->>>+<<]>[->>>>+<<]<<<]>>>>[-<<<<+>>>>]<<<<]


Takes number from cell 1 and puts the result into cell 6.

Unenrolled and commented version:

[  while n != 0
[  div 2 loop
-
>+<  marker for if/else
[->->>>+<<]  if n != 0 inc n/2
>
[->>>>+<<]  else inc m
<<<
]
>>>>  move n/2 back to n
[-<<<<+>>>>]
<<<<
]


# Python 2.6, 41 characters

t,n=0,input()
while n:t+=n%2;n/=2
print t


note: My other answer uses lambda and recursion and this one uses a while loop. I think they are different enough to warrant two answers.

## Ruby, 38 characters

f=->u{u<1?0:u%2+f[u/2]}
p f[gets.to_i]


Another solution using ruby and the same recursive approach as Steven.

## GolfScript, 17 16 characters

~{.2%\2/.}do]0-,


Edit: new version saves 1 character by using list operation instead of fold (original version was ~{.2%\2/.}do]{+}*, direct count version: ~0\{.2%@+\2/.}do;).

## C, 45

f(n,c){for(c=0;n;n/=2)c+=n%2;printf("%d",c);}


Nothing really special here for golfing in C: implicit return type, implicit integer type for parameters.

## Python 2.6, 45 characters

b=lambda n:n and n%2+b(n/2)
print b(input())

• Can be shortened by two characters by using def instead of a lambda. – Konrad Rudolph Dec 31 '11 at 11:51
• @KonradRudolph: Actually, you lose the advantage once you include the return statement. – Steven Rumbalski Dec 31 '11 at 18:02
• Oops, I forgot that. Stupid. – Konrad Rudolph Dec 31 '11 at 18:06
• You don't need the print b(input()). It is acceptable to return the value and take "input" as arguments for functions. – caird coinheringaahing Mar 31 '17 at 14:27

7

## PHP, 55 (alternative solution)

function b($i){return$i|0?($i%2)+b($i/2):0;}echo b($n);  Again, this assumes that $n holds the value to be tested. This is an alternative because it uses the or-operator to floor the input.

Both solutions work and do not cause notices.

# Ocaml, 45 characters

Based on @Leah Xue's solution. Three spaces could be removed and it's sligthly shorter (~3 characters) to use function instead of if-then-else.

let rec o=function 0->0|x->(x mod 2)+(o(x/2))


# Mathematica 26

Count[n~IntegerDigits~2,1]


## Scala, 86 characters

object O extends App{def f(i:Int):Int=if(i>0)i%2+f(i/2)else 0
print(f(args(0).toInt))}


Usage: scala O 56432

## D (70 chars)

int f(string i){int k=to!int(i),r;while(k){if(k%2)r++;k/=2;}return r;}


# R, 53 characters

o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())

Examples:

> o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())
1: 56432
2:
[1] 8
> o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())
1: 45781254
2:
[1] 11
> o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0};o(scan())
1: 0
2:
[1] 0


If inputting the number is not part of the character count, then it is 43 characters:

o=function(n){h=n%/%2;n%%2+if(h)o(h)else 0}

with test cases

> o(56432)
[1] 8
> o(45781254)
[1] 11
> o(0)
[1] 0


# OCaml, 52 characters

let rec o x=if x=0 then 0 else (x mod 2) + (o (x/2))


# Scheme

I polished the rules a bit to add to the challenge. The function doesn't care about the base of the number because it uses its own binary scale. I was inspired by the way analog to numeric conversion works. I just use plain recursion for this:

(define (find-ones n)
(define (nbits n)
(let nbits ([i 2])
(if (< i n) (nbits (* i 2)) i)))
(let f ([half (/ (nbits n) 2)] [i 0] [n n])
(cond [(< half 2) i]
[(< n i) (f (/ half 2) i (/ n 2))]
[else (f (/ half 2) (+ i 1) (/ n 2))])))


Isn't reading a number into binary or printing the number from binary a "builtin base conversion function", thus invalidating every answer above that prints an integer? If you permit reading and printing an integer, like almost all the above answers do, then I'll make claims using a builtin popcount function :

There was a popCount routine added to the Data.Bits module for GHC v7.2.1/v7.4.1 this summer (see tickets concerning the primop and binding).
import Data.Bits
main=interact$show.popCount.read  I cannot beat the above Python and Perl scores using their GMPY or GMP::Mpz modules for GMP sadly, although GMP does offer a popcount function too. ## JavaScript, 494745 42 bytes for(n=prompt(o=0);n=n/2|0;o+=n%2);alert(o)  Edit 1: remove q and use ~~ for rounding, save 2 chars. Edit 2: use |0 rounding operator instead of ~~ to save parentheses (2 chars). Edit 3: simplify n>0 to n and combine with n=n/2|0 to make entire condition; now have wasted statement space :( • Isn't the |0 a bitwise operator? – Vilx- Jan 2 '12 at 15:36 • Technically yes. But i'm using it purely to round to the nearest int, so I'm not getting any bit-wise benefit :) – mellamokb Jan 3 '12 at 16:36 • Smells like bending the rules to me... but I'm not the judge. – Vilx- Jan 4 '12 at 8:46 • Input 1 gives output 0. – Atreys Jan 31 '12 at 16:50 • | is bitwise operator... it is disallowed. Time to do Math.round :-) – Jamie Jul 8 '15 at 4:02 # Java 7, 36 bytes int b(Long a){return a.bitCount(a);}  Because of course this, of all things, is something that Java has a builtin for... • Doesn't this fit under "built-in base-conversion functions", which are banned? – FlipTack Dec 16 '16 at 20:22 • @Flp.Tkc I'm not actually doing base-conversion. I have no idea how bitCount operates under the hood. – Poke Dec 16 '16 at 20:29 • this seems like just using a bulitin to do the job, but ok... – FlipTack Dec 16 '16 at 20:30 • @Flp.Tkc That's... exactly what it is? I'm even including all required libraries (there aren't any). This is demonstrating the strength of the language! related meta – Poke Dec 16 '16 at 20:35 # TI-Basic (TI-84 Plus CE), 30 bytes Prompt X 0→S While X S+remainder(2,X→S int(X/2→X End S  TI-Basic is a tokenized language, all tokens but remainder( are one-byte, remainder is two • Welcome to the site! – DJMcMayhem Mar 31 '17 at 16:59 # PHP, 36 bytes while($n){$o+=$n%2;$n/=2*1;}echo$o;

Assumes $n is the number to be tested, shows a PHP Notice for $o, and doesn't exactly work when $n is 0 (outputs nothing). # PHP, 53 bytes $n=$argv[1];$o=0;while($n){$o+=$n%2;$n/=2*1;}echo \$o;