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I want to test whether any single bit is set in a C int. I don't care which bit it is.

Equivalent to

if (1 == count_bits (n))

I'm curious if there is a non-looping algorithm for this which does not require special hardware support.

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  • 9
    Check if n is non-zero? Commented Oct 29, 2018 at 10:19
  • 3
    Your text might indicate, that you are looking for a == 0, but your code snippet indicates, that you want to check if exactly one bit is set. Can you please specify it? Commented Oct 29, 2018 at 10:19
  • 2
    In an int? How do you want negative values handled? Commented Oct 29, 2018 at 10:31

2 Answers 2

Reset to default
5

If only one bit is set, then the number is a power of two. 

unsigned int IsPowerOfTwoNoLoop(unsigned int input) { unsigned int temp = input; if(0 == input) { return 0; } temp -= 1; /*Example of the logic:*/ /* 0100 (4 DEC) - 1 = 0011 0011 & 0100 = 0000 */ if (temp & input) /* if not zero - not pow of two */ { printf("%u Is NOT the power of two !\n", input); return 0; } printf("%u Is the power of two !\n", input); return 1; }
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3 Comments

For the explanation see stackoverflow.com/questions/600293/…
I would suggest to use uint32_t instead of unsigned int. Why don't you accept any digit and reinterpret them as an unsigned value in your function?
Review: it would be much clearer (=better) to return bool from this function.
1

The simplest way is probably to convert to an unsigned type.

Then we are simply testing whether we have an exact power of two.

We can use a simple observation that these are the only non-zero numbers that have no bits in common with n-1:

bool is_power_of_two(unsigned int n) { return n && ! (n & (n-1)); }
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2 Comments

"This code falsely returns one when input is zero. – Eric Postpischil"
Oops - I mistakenly thought I'd taken that into account, and failed.

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