Comments on: Ten Ways to Check if an Integer Is a Power Of Two in C

By: Rick Regan

Rick Regan — Tue, 23 Apr 2013 16:14:10 +0000

@Divanushka,

That works too. If x is a power of two, x – 1 is the inversion of x, that is, a 0 followed by all 1s. Inverting that gives a 1 followed by all 0s, that is, x.

By: Divanushka

Divanushka — Sun, 21 Apr 2013 17:49:55 +0000

return (x > 0) && ((x & ~(x – 1)) == x)

By: Rick Regan

Rick Regan — Thu, 18 Oct 2012 12:09:59 +0000

@AASHISH,

Yes, that’s method 9, “Decrement and Compare” (except you don’t check if num == 0).

By: AASHISH

AASHISH — Thu, 18 Oct 2012 10:38:42 +0000

I give u best solution.
Logic–> when u convert a decimal number to its binary number then u get series of 1 and 0
Now if a number is power of two then u will observe it has only one (bit 1) and if u do binary AND of of it with a number just previous to it ,then u will get series of Zeros
For example
8=1000(binary form ) here u can see only one (bit 1) is there and rest are zeros
Now the number just previous to 8 is 7
so, 7=0111 and if u perform AND operation on it as
1000(8)
(AND) 0111(7)
———
0000(0)
———
So if we perform AND operation between the number (which we want to check is a power of two or not) and the number just previous to it and result comes as zeros then its power of two otherwise it’s not.

PROGRAM(IN C_LANGUAGE)

#include
#include
main()
{
int num, num1,sum;
clrscr();
printf(“Enter the number”);
scanf(“%d”,&num);
printf(“\nEntered number=%d”,num);
num1=num-1;
sum=num & num1;
if(sum==0)
printf(“\n Entered number is power of two”);
else
printf(“\n Entered number is not power of two”);
getch();

Voila!!!! Isn’t it sooooo easy!!!!
Enjoy!!!

By: Lior

Lior — Wed, 25 Jan 2012 19:58:42 +0000

Another way I was today inform of (assuming word length is reasonable, n): create a Boolean array, sized 2^n. For each element, store true if the index is a power of two and false otherwise: b[0]=false, b[1]=true, b[2]=true, b[3]=false, b[4]=true, b[5]=false,…

Time complexity is now O(1).

By: Powers of two « Computing: the Science of Nearly Everything

Powers of two « Computing: the Science of Nearly Everything — Mon, 21 Nov 2011 16:55:47 +0000

[...] big thanks to Rick Regan and his in-depth Ten Ways to Check if an Integer Is a Power Of Two in C blog post) LD_AddCustomAttr("AdOpt", "1"); LD_AddCustomAttr("Origin", "other"); [...]

By: Rick Regan

Rick Regan — Wed, 31 Aug 2011 00:37:39 +0000

@AAL,

Your code is the negation of “Decrement and Compare.” You just need to negate it to get the right answer: if !nHasOnlyOne1(x) is true, the x is a power of two.

By: Krzysztof Wesołowski

Krzysztof Wesołowski — Tue, 30 Aug 2011 23:00:21 +0000

Also note that some approaches can be implemented in C as macro to be calculated with constant argument at compile time (or for example with template argument with C++ to do static_asserts etc.).

By: Harold Rabbie

Harold Rabbie — Tue, 30 Aug 2011 20:56:10 +0000

If you’re using gcc, that compiler includes a builtin function
int __builtin_popcount (unsigned int x) which returns the number of 1-bits in x.

So (__builtin_popcount (x) == 1) is true iff x is a power of 2.
Some processor architectures even include a population count instruction.

By: AAL

AAL — Tue, 30 Aug 2011 20:54:18 +0000

You can use this macro and a bit of other code to quickly determine if only one bit is set, then check if it is BIT0the odd case the number 1:

/* Returns ZERO if paramter has a single bit set high */
#define nHasOnlyOne1( p ) ( ((p)==0) || ( (p) & ((p)-1) ) )

By: Pseudonym

Pseudonym — Sun, 31 Jul 2011 02:53:31 +0000

If you’re on a POSIX.1 compliant platform, should contain the ffs() function. If you’re lucky, your system will use your CPU’s built-in priority encoder instruction to implement this, and you can do the following. Hopefully no loops, and at most one branch or conditional move:

int isPowerOfTwo(int x)
{
int bit = ffs(x);
return bit ? ((1 << (bit – 1)) == x) : 0;
}

By: Rick Regan

Rick Regan — Sat, 30 Jul 2011 15:25:37 +0000

@Ivan,

I agree. Your code is a cleaner version of my “Count Ones” (#7) — and a lot faster.