Question

I’m looking for the most efficient way to calculate the minimum number of bytes needed to store an integer without losing precision.

e.g.

int: 10 = 1 byte
int: 257 = 2 bytes;
int: 18446744073709551615 (UINT64_MAX) = 8 bytes;

Thanks

P.S. This is for a hash functions which will be called many millions of times

Also the byte sizes don’t have to be a power of two

The fastest solution seems to one based on tronics answer:

    int bytes;
    if (hash <= UINT32_MAX) 
    {
        if (hash < 16777216U)
        {
            if (hash <= UINT16_MAX)
            {
                if (hash <= UINT8_MAX) bytes = 1;
                else bytes = 2;
            }
            else bytes = 3;
        }
        else bytes = 4;
    } 
    else if (hash <= UINT64_MAX) 
    {
        if (hash < 72057594000000000ULL) 
        {
            if (hash < 281474976710656ULL) 
            {
                if (hash < 1099511627776ULL) bytes = 5;
                else bytes = 6;
            }
            else bytes = 7;
        }
        else bytes = 8;
    }

The speed difference using mostly 56 bit vals was minimal (but measurable) compared to Thomas Pornin answer. Also i didn’t test the solution using __builtin_clzl which could be comparable.

Answer 1

You need just two simple ifs if you are interested on the common sizes only. Consider this (assuming that you actually have unsigned values):

if (val < 0x10000) {
    if (val < 0x100) // 8 bit
    else // 16 bit
} else {
    if (val < 0x100000000L) // 32 bit
    else // 64 bit
}

Should you need to test for other sizes, choosing a middle point and then doing nested tests will keep the number of tests very low in any case. However, in that case making the testing a recursive function might be a better option, to keep the code simple. A decent compiler will optimize away the recursive calls so that the resulting code is still just as fast.