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

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.

By: Stefanus Du Toit

Stefanus Du Toit — Sat, 30 Jul 2011 05:15:25 +0000

(and yes that is essentially no 9 – though I often don’t care about the zero case)

By: Stefanus Du Toit

Stefanus Du Toit — Sat, 30 Jul 2011 02:52:48 +0000

(x & (x – 1)) == 0

If I remember right I first read this in Hacker’s Delight or HAKMEM.

By: Ivan Tikhonov

Ivan Tikhonov — Fri, 29 Jul 2011 13:57:08 +0000

Count Ones is actually better to be renamed and rewritten as “Check the only one”

So basically you:

1. find that one exists.
2. find that no ones exist beyond first.

This way you remove increment and condition inside of loop.

int isPowerOfTwo (unsigned int x) { while(x && (x&1)==0) x>>=1; // shift till 1 found if(!x) return 0; // no ones found at all x>>=1; return x==0; // check there is no ones past first one }

Also if you unroll while loop there will be no need for “x not zero” checking at all.

By: Friday miscellany — The Endeavour

Friday miscellany — The Endeavour — Fri, 29 Jul 2011 09:55:39 +0000

[...] language fixes something 10 ways to check whether an integer is a power of 2 Version control by [...]