Computing high 64 bits of a 64x64 int product in C

问题:

I would like my C function to efficiently compute the high 64 bits of the product of two 64 bit signed ints. I know how to do this in x86-64 assembly, with imulq and pulling the result out of %rdx. But I'm at a loss for how to write this in C at all, let alone coax the compiler to do it efficiently.

Does anyone have any suggestions for writing this in C? This is performance sensitive, so "manual methods" (like Russian Peasant, or bignum libraries) are out.

This dorky inline assembly function I wrote works and is roughly the codegen I'm after:

static long mull_hi(long inp1, long inp2) {
    long output = -1;
    __asm__("movq %[inp1], %%rax;"
            "imulq %[inp2];"
            "movq %%rdx, %[output];"
            : [output] "=r" (output)
            : [inp1] "r" (inp1), [inp2] "r" (inp2)
            :"%rax", "%rdx");
    return output;
}

回答1:

If you're using a relatively recent GCC on x86_64:

int64_t mulHi(int64_t x, int64_t y) {
    return (int64_t)((__int128_t)x*y >> 64);
}

At -O1 and higher, this compiles to what you want:

_mulHi:
0000000000000000    movq    %rsi,%rax
0000000000000003    imulq   %rdi
0000000000000006    movq    %rdx,%rax
0000000000000009    ret

I believe that clang and VC++ also have support for the __int128_t type, so this should also work on those platforms, with the usual caveats about trying it yourself.

回答2:

The general answer is that x * y can be broken down into (a + b) * (c + d), where a and c are the high order parts.

First, expand to ac + ad + bc + bd

Now, you multiply the terms as 32 bit numbers stored as long long (or better yet, uint64_t), and you just remember that when you multiplied a higher order number, you need to scale by 32 bits. Then you do the adds, remembering to detect carry. Keep track of the sign. Naturally, you need to do the adds in pieces.

For code implementing the above, see my other answer.

回答3:

With regard to your assembly solution, don't hard-code the mov instructions! Let the compiler do it for you. Here's a modified version of your code:

static long mull_hi(long inp1, long inp2) {
    long output;
    __asm__("imulq %2"
            : "=d" (output)
            : "a" (inp1), "r" (inp2));
    return output;
}

Helpful reference: Machine Constraints

回答4:

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 based on my general answer. 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 artefacts 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];
}

回答5:

Wait, you have a perfectly good, optimized assembly solution already working for this, and you want to back it out and try to write it in an environment that doesn't support 128 bit math? I'm not following.

As you're obviously aware, this operation is a single instruction on x86-64. Obviously nothing you do is going to make it work any better. If you really want portable C, you'll need to do something like DigitalRoss's code above and hope that your optimizer figures out what you're doing.

If you need architecture portability but are willing to limit yourself to gcc platforms, there are __int128_t (and __uint128_t) types in the compiler intrinsics that will do what you want.