Haskell (52 50)

k n=elem n$map(2^)[1..n] main=interact$show.k.read 

Python 2 - 23

x=input();print x&x/3<1 

This is a direct port of the golf-script solution above, that beats the current best python solution by 8 characters. The order of operations works out amazingly well here.

Bash + dc + grep, 24

dc -e2o?p|grep -c ^10\*$ 

Reads from stdin. Outputs "1" if a power of 2 and "0" otherwise:

 $ dc -e2o?p|grep -c ^10\*$ 8 1 $ dc -e2o?p|grep -c ^10\*$ 10 0 $ 

Pure Bash (only builtins), 40

[[ `read a;printf %o $a` =~ ^[124]0*$ ]] 

Reads from stdin. Returns success (0) if a power of 2 and failure (1) otherwise:

 $ [[ `read a;printf %o $a` =~ ^[124]0*$ ]] 8 $ echo $? 0 $ [[ `read a;printf %o $a` =~ ^[124]0*$ ]] 10 $ echo $? 1 $ 

Java 7, 124

This is pretty short in Java. Most of the program is just boilerplate code and to read the integer.

class H{public static void main(String[]b){int a=new java.util.Scanner(System.in).nextInt();System.out.println((a&a/3)<1);}} 

SmileBASIC, 29 bytes

This is a fairly uncreative and basic (no pun intended) answer. 0 is false, 1 is true.

INPUT N A=LOG(N,2)?(0OR A)==A 

We know that if n is a power of 2, then log2(n) will be an integer. So, this calculates the base-2 log of N, our input number, and checks if it is an integer using the following method.

In SB there are actually two number types, doubles and integers. It also has automatic coercion between doubles and ints depending on the context. When a float is coerced to an int, it simply truncates the fraction and produces an integer which represents the whole part (assuming that number is in the valid integer size–signed 32 bit.) The bitwise OR operator OR expects both of its terms to be integers, so it will automatically cast doubles to ints. Thus, 0 OR double simply produces that double's whole part, which we then check against the original value. If these two are equal, it is a power of two.

Haskell, 47 bytes

(or 21 bytes if input/output can be function argument/return)

f 0=0 f 1=1 f n=f$n/2 main=interact$show.f.read 

Returns 1 for powers of 2, and 0 otherwise.

Python 2, 29,25 bytes

x=input();print x&x*~0==x 

Try it online!

using x&x*~~x/~x is a modified version of x&(~x+1)! Since the later uses + had to use ~x which is equivalent to -x-1!

• saved 4 bytes by using x*~0 instead of x*~~x/~x!

PHP, 86 chars

function P($a,$c){if($c==$a)return 1;if($c>$a)return 0;return P($a,$c*2);}echo P(x,2); 

Replace x with the number you want to test.

PHP (58)

<?php $a=fgets(STDIN);while($a%2==0)$a/=2;echo $a==1?1:0; 

Simple divide-by-2 loop and remainder check.

ActionScript 3 (63 characters*)

var r = Math.log(prompt())/Math.log(2); return ( (r > 0) && (int(r) == r) ); 

* I'm going to ignore the fact that AS3 doesn't have a built-in prompt() method, (and hope AS3's lack of implementation doesn't get counted against me).

Extra white-space left in for readability and clarity, but not counted, since they can be removed before execution.

GTB, 17

With 1/0

`Ar?A>1:A/2→A~A=1 

With YES/NO (31)

`Ar?A>1:A/2→A@A=1$~"YES"#~"NO"& 

Scheme 40

(display(=(mod(/(log(read))(log 2))1)0)) 

It displays #t for true and #f for false (the booleans in Scheme).

For true compatibility you need to add (import(rnrs)) or (import(scheme)) (for r7rs) but it works without that in ikarus and in most REPLs.

Perl 53

perl -pe '$_=unpack"b*",pack"L",$_;s!0!!g;s!111*!no!;s!1!yes!' 

I can shave off 1 character if i use + in the re, not used for adding.

Golfscript (17 14)

Works from input, using the base builtin for binary conversion and counting ones;

~2base{1=},,1= 

Old version, 17 chars, without builtin conversions and without using the decrement operator as a substitute for -;

~2*:a 1{2*.a<}do= 

• ~ lifts the input to an integer on the stack
• 2* multiplies by 2 to solve 2^0 problem
• :a sets the variable a to the number to search for and leaves it on stack
• 1{2*.a<}do sets start value 1 and doubles as long as number<a
• = compares the number to search for with the number generated

J - 17 15 bytes - not . ceiling (0==) . mod 1 . log2 . to_i . read keyboard

Result is 1 for True and 0 for False

0=1|2^.".1!:1,1 

Yes/No version

(0=1|2^.".1!:1,1){'No',:'Yes' 

D (83 chars with boilerplate, 17 chars to get the solution)

import std.stdio; void main(){ int i; readf("%d", &i); while(!(i%2))i/2; writeln(i==1); } 

while the number is even divide by 2, when the result is 2 then it was a power of 2

Perl 5 (36 or 29 bytes)

say sprintf('%b',<>)=~/^10*$/?Yes:No 

If we don't want to strictly output "Yes" or "No", then we can go shorter:

say sprintf('%b',<>)=~/^10*$/ 

This second version outputs "1" for true, and "" for false.

autohotkey -45bytes in txt file and 48 bytes with editor and lang.

MsgBox % mod(log(k)/log(2), 1) ? "no" : "yes" 

C99 94 92 71

int main(){int a,i=1;scanf("%d",&a);while(i){if(a==i)return 1;i<<=1;}} 

readable

#include <stdio.h> int main() { int a, i=1; scanf("%d",&a); while(i){ if(a==i) return 1; i<<=1; } } 

What it does

It XORs the input value with a power of 2, until the bit of variable i is shifted out and the value becomes 0. Only if the input value is also a power of 2 (with other words, only one bit in the input was set) the XOR resolves to zero.

The input will be compared with a power of 2 value, that is left shifted in every iteration (in other words multiplied by 2) and returns if both are equal, thus the input is also a power of 2.

On success 1 is returned, 0 otherwise.

Why C99?

Because the return value of a C program is implicitly defined as 0, if not specified, since C99. Compile with: gcc -std=c99 ...

JavaScript (43 characters when compacted, 35 for some browsers)

Readable version:

var r = Math.log(prompt()) / Math.LN2; alert( (r > 0) && (~~r == r) ); 

Compacted version (credit goes to FireFly for helping me shave off even more characters):

r=Math.log(prompt())/Math.LN2;alert(~~r==r) 

JS Fiddle: http://jsfiddle.net/IQAndreas/LFYg7/2/

EDIT: On the off chance that you use FireFox 25+ or any other browser with Math.log2(), only 35 characters are required (thanks to xfix for this suggestion):

r=Math.log2(prompt());alert(~~r==r) 

Python (40 char)

p=lambda n:n>0and(n<2or n%2<1and p(n/2)) 

Works for negative number also (returning False)

C (39 chars)

p(n){return n>0&&(n<2||n%2<1&&p(n/2));} 

or with trick, (work on old compilers without optimizing, withoput return operator returns last evaluated expression value) (32 chars)

p(n){n>0&&(n<2||n%2<1&&p(n/2));} 

PHP, 55

<? if(fmod(log($argv[1],2),1)==0) echo 1; else echo 0; 

Eg:

# php pow2.php 8 1 # php pow2.php 9 0 

PHP w/ interactive I/O, 107

<? $h=fopen('php://stdin','r');$n=trim(fgets($h));fclose($h);if(fmod(log($n,2),1)==0) echo 1; else echo 0; 

BASH, 48

x=`echo "obase=2;$1"|bc|rev|bc`;echo "$x==1"|bc 

Takes input as a command line argument and returns 1 for true and 0 for false. You can easily modify it to echo Yes/No, but the bash if-construct makes it much longer.

~-~! - 37

'=|*;~~==%['&*/~~]*|:'&^==~[@|1|]@|0| 

I had to improvise with the number input...it's the Unicode value of the character you enter. Teehee.

This uses a recursive /2 solution.

Java 143

Thinking binary ( a power of 2 has always only one bit set so the compare with the integer with ionly highest bit set can only return true on powers of 2 (including the ominous 0 :P)

Update: shaved 2 bytes off due to a derp

enum P{public static void main(String[]a){int i=new java.util.Scanner(System.in).nextInt();System.out.print(i==(Integer.highestOneBit(i)*1));}} 

Ungolfed

enum P; { public static void main(String[]a) { int i = new java.util.Scanner(System.in).nextInt(); System.out.print(i==(Integer.highestOneBit(i)*1)); } } 

JavaScript (EcmaScript 6) - 13 Characters

f=x=>!(x&x/3) 

Creates a function f that takes a value and returns true (or false) if it is a power of 2 (or not) and will output the answer to the console.

Based on Ilmari Karonen's answer (so I included a few others of my own, below).

EcmaScript 6 can be run on FireFox or SpiderMonkey - no other browser/JavaScript engine yet supports EcmaScript 6 arrow functions. If you want to prompt for input then you can call it as an anonymous function (23 characters):

(x=>!(x&x/3))(prompt()) 

But it's less characters skip the function bit and just do:

JavaScript - 19 Characters

!((a=prompt())&a/3) 

Will do the same as above but should run in all browsers and will prompt for input and output to the console (or in SpiderMonkey, you can replace prompt() [which doesn't exist as SpiderMonkey is command-line based] with readline()).

JavaScript (EcmaScript 6) - 20 Characters

f=x=>x>1?f(x/2):x==1 

Creates a function f and, when called, will output the answer to the console.

JavaScript - 29 Characters

for(a=prompt();a>1;)a/=2;a==1 

Same as above but will prompt the user for input and output to the console.

JavaScript - 32 Characters

for(a=prompt();!(a%2);)a/=2;a==1 

JavaScript - 34 Characters

for(a=prompt();a==(a|0);)a/=2;a==1 

dc, 13 bytes

?[2~0=m]dsmxp 

Reads input from STDIN. Input is not explicitly stored in a variable, though it is stored on the stack - not sure if this too much rule-bending.

Successively divides by 2 until a 1 remainder is found. Since dc doesn't have any native concept of truthy/falsey, I'm taking the liberty of defining my own here. 0 means true, any +ve integer means false:

$ for i in 1 2 3 15 16 17 65535 65536; do dc pow2.dc <<< $i; done 0 0 1 7 0 8 32767 0 $ 

Batch - 24 bytes

Stealing Ilmari's winning GolfScript answer...

@cmd/c"set/a!(%1/3^&%1)" 

See explanation here.

Note: This question is older than Pyth, so this answer is not eligible to win.

Pyth, 5 bytes

!%lQ1 

This is essentially identical to @marinus's APL answer. We take the log base 2 of the input, take it mod 1, and take the logical not.

Pyth, 6 bytes

!stjQ2 

This answer converts the input to base 2, removes the first digit, sums the remainder, and takes the logical not.

Python (23)

int(bin(input())[3:])<1