Since you did a pretty good job solving your own problem with the machine code, I figured you deserved some help with the portable version. I would leave an ifdef in where you do just use the assembly if in gnu on x86.
Anyway, here is an implementation...I'm pretty sure this is correct, but no guarantees, I just banged this out last night...you probably should get rid of the statics positive_result[] and result_negative, those are just artifacts of my unit test...
#include <stdlib.h>
#include <stdio.h>
// stdarg.h doesn't help much here because we need to call llabs()
typedef unsigned long long uint64_t;
typedef signed long long int64_t;
#define B32 0xffffffffUL
static uint64_t positive_result[2]; // used for testing
static int result_negative; // used for testing
static void mixed(uint64_t *result, uint64_t innerTerm)
{
// the high part of innerTerm is actually the easy part
result[1] += innerTerm >> 32;
// the low order a*d might carry out of the low order result
uint64_t was = result[0];
result[0] += (innerTerm & B32) << 32;
if (result[0] < was) // carry!
++result[1];
}
static uint64_t negate(uint64_t *result)
{
uint64_t t = result[0] = ~result[0];
result[1] = ~result[1];
if (++result[0] < t)
++result[1];
return result[1];
}
uint64_t higherMul(int64_t sx, int64_t sy)
{
uint64_t x, y, result[2] = { 0 }, a, b, c, d;
x = (uint64_t)llabs(sx);
y = (uint64_t)llabs(sy);
a = x >> 32;
b = x & B32;
c = y >> 32;
d = y & B32;
// the highest and lowest order terms are easy
result[1] = a * c;
result[0] = b * d;
// now have the mixed terms ad + bc to worry about
mixed(result, a * d);
mixed(result, b * c);
// now deal with the sign
positive_result[0] = result[0];
positive_result[1] = result[1];
result_negative = sx < 0 ^ sy < 0;
return result_negative ? negate(result) : result[1];
}