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Asked: 2026-05-26T22:37:24+00:00

The following code rounds up the argument to an int size boundary number of

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The following code rounds up the argument to an int size boundary number of bytes.

  #define _INTSIZE(n) ((sizeof(n) + sizeof(int) - 1) & ~(sizeof(int) - 1))

On my machine int is 4 bytes, so — correct me if I’m wrong — this should be the same as looking for the next multiple of 4 of an integer (on my machine). By next multiple of 4, I mean the number should be rounded up to a multiple of 4 if not a multiple of 4. If already a multiple of 4, it should be left alone.

I’ve been playing around with this code. The long and short of it is: why does this code work? (Maybe it doesn’t, but it seems to.) I would like some reason to think it will work for ALL cases, not just the ones I’ve tried out.

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1 Answer

  1. Editorial Team

    The code first adds three to the number.

    Then it zeroes out the last two bits to round down to a multiple of fours. Just like you can round down to the nearest multiple of 100 in decimal by replacing the last two digits with zeroes.)

    If the number is already a multiple of four, adding three to it and then rounding down to the nearest multiple of four leaves it alone, as desired. If the number is 1, 2, or 3 more than a multiple of 4, adding 3 to it raises it above the next multiple of 4, which it then rounds down to, exactly as desired.

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