有没有办法将低 64 位乘法 int64_t * int64_t 存储在 C 中？

Question

This is the function I need to implement using C. Note that it is not allowed to use gcc builtins. Thank you for any help!

'''

imull(int64_t a, int64_t b, int64_t *res) {
    // a * b = ..32bits..|..32bits..
    //                       *res
    *res = a * b;
}

'''

Answer 1

When an int64_t is multiplied by an int64_t , the true arithmetic result is a 128-bit number with its sign bit at bit 127. The low 64 bits do not contain a sign bit. Therefore, the low 64 bits of the multiplication ought to be returned in a uint64_t , not an int64_t , and this can be accomplished with:

imull(int64_t a, int64_t b, uint64_t *res)
{
    *res = (uint64_t) a * (uint64_t) b;
}

(The casts avoid overflow in the multiplication, as unsigned integer arithmetic is defined to wrap, not overflow.)

If the bits must be returned in an int64_t , this can be done in a way defined by the C standard using:

imull(int64_t a, int64_t b, int64_t *res)
{
    uint64_t temporary = (uint64_t) a * (uint64_t) b;
    memcpy(res, &temporary, sizeof *res);
}

( memcpy is declared in <string.h> .)

Also note that if the true arithmetic result fits in an int64_t , then the int64_t returned by the above code will equal the true arithmetic result. (This is a property of the two's complement representation, which int64_t uses.)