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I need to multiply two signed 64-bit integers a
and b
together, then shift the (128-bit) result to a signed 64-bit integer. What's the fastest way to do that?
My 64-bit integers actually represent fixed-point numbers with fmt
fractional bits. fmt
is chosen so that a * b >> fmt
should not overflow, for instance abs(a) < 64<<fmt
and abs(b) < 2<<fmt
with fmt==56
would never overflow in 64-bits as the final result would be < 128<<fmt
and therefore fit in an int64.
The reason I want to do that is to quickly and precisely evaluate quintic polynomials of the form ((((c5*x + c4)*x + c3)*x + c2)*x + c1)*x + c0
in fixed point format, with every number a signed 64-bit fixed-point number with fmt
fractional bits. I'm looking for the most efficient way to achieve that.
As a commenter on the question pointed out, this is most easily accomplished efficiently by machine-dependent code, rather than by portable code. The asker states that the main platform is x86_64, and that has a built-in instruction for performing 64 ✕ 64 → 128 bit multiplication. This is easily accessed using a small piece of inline assembly. Note that details of inline assembly may differ somewhat with compiler, the code below was built with the Intel C/C++ compiler.
#include <stdint.h>
/* compute mul_wide (a, b) >> s, for s in [0,63] */
int64_t mulshift (int64_t a, int64_t b, int s)
{
int64_t res;
__asm__ (
"movq %1, %%rax;\n\t" // rax = a
"movl %3, %%ecx;\n\t" // ecx = s
"imulq %2;\n\t" // rdx:rax = a * b
"shrdq %%cl, %%rdx, %%rax;\n\t" // rax = int64_t (rdx:rax >> s)
"movq %%rax, %0;\n\t" // res = rax
: "=rm" (res)
: "rm"(a), "rm"(b), "rm"(s)
: "%rax", "%rdx", "%ecx");
return res;
}
A portable C99 equivalent to the above code is shown below. I have tested this extensively against the inline assembly version and no mismatches were found.
void umul64wide (uint64_t a, uint64_t b, uint64_t *hi, uint64_t *lo)
{
uint64_t a_lo = (uint64_t)(uint32_t)a;
uint64_t a_hi = a >> 32;
uint64_t b_lo = (uint64_t)(uint32_t)b;
uint64_t b_hi = b >> 32;
uint64_t p0 = a_lo * b_lo;
uint64_t p1 = a_lo * b_hi;
uint64_t p2 = a_hi * b_lo;
uint64_t p3 = a_hi * b_hi;
uint32_t cy = (uint32_t)(((p0 >> 32) + (uint32_t)p1 + (uint32_t)p2) >> 32);
*lo = p0 + (p1 << 32) + (p2 << 32);
*hi = p3 + (p1 >> 32) + (p2 >> 32) + cy;
}
void mul64wide (int64_t a, int64_t b, int64_t *hi, int64_t *lo)
{
umul64wide ((uint64_t)a, (uint64_t)b, (uint64_t *)hi, (uint64_t *)lo);
if (a < 0LL) *hi -= b;
if (b < 0LL) *hi -= a;
}
/* compute mul_wide (a, b) >> s, for s in [0,63] */
int64_t mulshift (int64_t a, int64_t b, int s)
{
int64_t res;
int64_t hi, lo;
mul64wide (a, b, &hi, &lo);
if (s) {
res = ((uint64_t)hi << (64 - s)) | ((uint64_t)lo >> s);
} else {
res = lo;
}
return res;
}
来源:https://stackoverflow.com/questions/31652875/fastest-way-to-multiply-two-64-bit-ints-to-128-bit-then-to-64-bit