Reasonably portable way to get top 64-bits from 64x64 bit multiply? [duplicate]

Answer 1

This answer shows how to get the (exact) top 64-bits from a 64x64 bit multiply on a system that doesn't support 128-bit integers. The answer by @amdn will give better performance on systems that do support 128-bit integers.

The diagram below shows one method for computing a 128-bit product from two 64-bit numbers. Each black rectangle represents a 64-bit number. The 64-bit inputs to the method, X and Y, are divided into 32-bits chunks labeled a, b, c, and d. Then four 32x32 bit multiplications are performed, giving four 64-bit products labeled a*c, b*c, a*d, and b*d. The four products must be shifted and added to compute the final answer.

Note that the lower 32-bits of the 128-bit product are solely determined by the lower 32-bits of partial product b*d. The next 32-bits are determined by the lower 32-bits of the following

mid34 = ((b*c) & 0xffffffff) + ((a*d) & 0xffffffff) + ((b*d) >> 32);

Note that mid34 is the sum of three 32-bit numbers and therefore is in fact a 34-bit sum. The upper two bits of mid34 act as a carry into the top 64-bits of the 64x64 bit multiply.

Which brings us to the demo code. The top64 function computes the upper 64-bits of a 64x64 multiply. It's a little verbose to allow for the calculation of the lower 64-bits to be shown in a comment. The main function takes advantage of 128-bit integers to verify the results with a simple test case. Further testing is left as an exercise for the reader.

#include <stdio.h>
#include <stdint.h>

typedef unsigned __int128 uint128_t;

uint64_t top64( uint64_t x, uint64_t y )
{
    uint64_t a = x >> 32, b = x & 0xffffffff;
    uint64_t c = y >> 32, d = y & 0xffffffff;

    uint64_t ac = a * c;
    uint64_t bc = b * c;
    uint64_t ad = a * d;
    uint64_t bd = b * d;

    uint64_t mid34 = (bd >> 32) + (bc & 0xffffffff) + (ad & 0xffffffff);

    uint64_t upper64 = ac + (bc >> 32) + (ad >> 32) + (mid34 >> 32);
//  uint64_t lower64 = (mid34 << 32) | (bd & 0xffffffff);

    return upper64;
}

int main( void )
{
    uint64_t x = 0x0000000100000003;
    uint64_t y = 0x55555555ffffffff;

    uint128_t m = x, n = y;
    uint128_t p = m * n;

    uint64_t top = p >> 64;
    printf( "%016llx %016llx\n", top, top64( x, y ) );
}

Answer 2

A little algebra never hurt:

#include <stdint.h>

uint64_t top64(uint64_t x, uint64_t y) {
    uint64_t a = x >> 32, b = x & 0xFFFFFFFF;
    uint64_t c = y >> 32, d = y & 0xFFFFFFFF;

    return a * c + ((b * d >> 32) + (a * d) + (b * c)) >> 32 +
    ((((a * d) & 0xFFFFFFFF) + ((b *c) & 0xFFFFFFFF) + ((b * d) >> 32)) >> 32);
}

Answer 3

Both gcc and clang support 128-bit integers as an extension.

Here's one way to do it demo

#include <iostream>
#include <cstdint>

//https://gcc.gnu.org/onlinedocs/gcc-4.8.1/gcc/_005f_005fint128.html#_005f_005fint128
using u128 = unsigned __int128;
using u64  = uint64_t;

void mul64x64( u64 a, u64 b, u64 & hi, u64 & lo ) {
    u128 product = u128(a) * b;
    lo = product;
    hi = product >> 64;
}

int main()
{
    u64 hi, lo;
    mul64x64( 40282220, u64{1} << 63, hi, lo );
    std::cout << hi << std::endl;
}

Output is

set -x ; clang++ -std=c++11 -O0 -Wall -Werror main.cpp && ./a.out
+ clang++ -std=c++11 -O0 -Wall -Werror main.cpp
+ ./a.out
20141110