Counting the number of leading zeros in a 128-bit integer

Question

How can I count the number of leading zeros in a 128-bit integer (uint128_t) efficiently?

I know GCC\'s built-in functions:

• __builti

Answer 1

Assuming a 'random' distribution, the first non-zero bit will be in the high 64 bits, with an overwhelming probability, so it makes sense to test that half first.

Have a look at the code generated for:

/* inline */ int clz_u128 (uint128_t u)
{
    unsigned long long hi, lo; /* (or uint64_t) */
    int b = 128;

    if ((hi = u >> 64) != 0) {
        b = __builtin_clzll(hi);
    }
    else if ((lo = u & ~0ULL) != 0) {
        b = __builtin_clzll(lo) + 64;
    }

    return b;
}

I would expect gcc to implement each __builtin_clzll using the bsrq instruction - bit scan reverse, i.e., most-significant bit position - in conjunction with an xor, (msb ^ 63), or sub, (63 - msb), to turn it into a leading zero count. gcc might generate lzcnt instructions with the right -march= (architecture) options.

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Edit: others have pointed out that the 'distribution' is not relevant in this case, since the HI uint64_t needs to be tested regardless.