Timing Quake III hack 仅在使用优化编译时才有效

问题描述

所以我刚刚发现了非常有趣的 Quake III 平方根倒数技巧。在了解了它的工作原理之后，我决定测试它。我发现该 hack 在启用优化的情况下编译时仅优于 math.h 1/sqrt(X)。

黑客的实现：

float q_sqrt(float x) {
    float x2 = x * 0.5F;
    int i = *( int* )&x;                  // evil floating point bit hack
    i = 0x5f3759df - (i >> 1);            // what the fuck?
    x = *( float* )&i;
    x = x * ( 1.5F - ( (x2 * x * x) ) );  //1st iteration
  //y = y * ( 1.5F - ( (x2 * y * y) ) );  //2nd iteration,can be removed
    return x;
}

要测试 1/sqrt(x) 与 q_sqrt(x) 相比的运行速度：

//qtest.c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>

/*
Implementation of 1/sqrt(x) used in tue quake III game
*/
float q_sqrt(float x) {
    float x2 = x * 0.5F;
    int i = *( int* )&x;                  // evil floating point bit hack
    i = 0x5f3759df - (i >> 1);            // what the fuck?
    x = *( float* )&i;
    x = x * ( 1.5F - ( (x2 * x * x) ) );  //1st iteration
  //y = y * ( 1.5F - ( (x2 * y * y) ) );  //2nd iteration,can be removed
    return x;
}


int main(int argc,char *argv[]) {
    struct timespec start,stop;
    //Will work on floats in the range [0,100]
    float maxn = 100;
    //Work on 10000 random floats or as many as user provides
    size_t num = 10000;
    //Bogus
    float ans = 0;
    //Measure nanoseconds
    size_t ns = 0;
    if (argc > 1)
        num = atoll(argv[1]);
    if (num <= 0) return -1;
    //Compute "num" random floats 
    float *vecs = malloc(num * sizeof(float));
    if (!vecs) return -1;
    for (int i = 0; i < num; i++)
        vecs[i] = maxn * ( (float)rand() / (float)RAND_MAX );

    fprintf(stderr,"Measuring 1/sqrt(x)\n");
    clock_gettime( CLOCK_REALTIME,&start);
    for (size_t i = 0; i < num; i++)
        ans += 1 / sqrt(vecs[i]);
    clock_gettime( CLOCK_REALTIME,&stop);
    ns = ( stop.tv_sec - start.tv_sec ) * 1E9 + ( stop.tv_nsec - start.tv_nsec );
    fprintf(stderr,"1/sqrt(x) took %.6f nanosecods\n",(double)ns/num );


    fprintf(stderr,"Measuring q_sqrt(x)\n");
    clock_gettime( CLOCK_REALTIME,&start);
    for (size_t i = 0; i < num; i++)
        ans += q_sqrt(vecs[i]);
    clock_gettime( CLOCK_REALTIME,"q_sqrt(x) took %.6f nanosecods\n",(double)ns/num );

    //Side by side
  //for (size_t i = 0; i < num; i++)
  //    fprintf(stdout,"%.6f\t%.6f\n",1/sqrt(vecs[i]),q_sqrt(vecs[i]));
    free(vecs);
}

在我的系统 (Ryzen 3700X) 上，我得到：

gcc -Wall -pedantic -o qtest qtest.c -lm
./qtest
Measuring 1/sqrt(x)
1/sqrt(x) took 4.470000 nanosecods
Measuring q_sqrt(x)
q_sqrt(x) took 4.859000 nanosecods


gcc -Wall -pedantic -O1 -o qtest qtest.c -lm
./qtest
Measuring 1/sqrt(x)
1/sqrt(x) took 0.378000 nanosecods
Measuring q_sqrt(x)
q_sqrt(x) took 0.497000 nanosecods


gcc -Wall -pedantic -O2 -o qtest qtest.c -lm
qtest.c: In function ‘q_sqrt’:
qtest.c:11:14: warning: dereferencing type-punned pointer will break strict-aliasing rules [-Wstrict-aliasing]
  11 |     int i = *( int* )&x;                  // evil floating point bit hack
     |
qtest.c:13:10: warning: dereferencing type-punned pointer will break strict-aliasing rules [-Wstrict-aliasing]
  13 |     x = *( float* )&i;
     |
./qtest
Measuring 1/sqrt(x)
1/sqrt(x) took 0.500000 nanosecods
Measuring q_sqrt(x)
q_sqrt(x) took 0.002000 nanosecods

我的期望是 q_sqrt(x) 比 1/sqrt(X) 开箱即用。阅读更多内容后，我现在知道要么libm 优化得更好，要么我的CPU 配备了sqrt(X) 的硬件解决方案。毕竟，自从快速逆根黑客开发以来，CPU 已经发生了突飞猛进的变化。

我不明白的是编译器会应用什么类型的优化来使它更快。当然，也许我的基准是构思错误的？

感谢您的帮助！！

解决方法

正如您所说，大多数现代 CPU 都包含一个浮点单元，该单元通常提供计算平方根的硬件指令。 FPU 还提供除法指令，因此我希望您的处理器（尽管我不知道）能够仅用几条汇编指令来计算逆 sqrt。您的结果有点令人惊讶：您应该检查是否真的使用了 FPU。我不了解锐龙，但在 ARM 处理器上，您可以编译软件以使用硬件浮点指令或软件库。

现在回答您的问题：GCC 优化是一个复杂的故事，通常不可能准确预测给定级别对性能的影响。所以像你一样运行一些测试，或者看看here的理论。

,

CLang/LLVM 的具体区别是这些。

没有优化（-O0）：

q_sqrt(float):                             # @q_sqrt(float)
        push    rbp
        mov     rbp,rsp
        movss   dword ptr [rbp - 4],xmm0
        movss   xmm0,dword ptr [rip + .LCPI0_1] # xmm0 = mem[0],zero,zero
        mulss   xmm0,dword ptr [rbp - 4]
        movss   dword ptr [rbp - 8],xmm0
        mov     eax,dword ptr [rbp - 4]
        mov     dword ptr [rbp - 12],eax
        mov     ecx,dword ptr [rbp - 12]
        sar     ecx,1
        mov     eax,1597463007
        sub     eax,ecx
        mov     dword ptr [rbp - 12],eax
        movss   xmm0,dword ptr [rbp - 12]      # xmm0 = mem[0],zero
        movss   dword ptr [rbp - 4],dword ptr [rbp - 4]       # xmm0 = mem[0],zero
        movss   xmm2,dword ptr [rbp - 8]       # xmm2 = mem[0],zero
        mulss   xmm2,dword ptr [rbp - 4]
        mulss   xmm2,dword ptr [rbp - 4]
        movss   xmm1,dword ptr [rip + .LCPI0_0] # xmm1 = mem[0],zero
        subss   xmm1,xmm2
        mulss   xmm0,xmm1
        movss   dword ptr [rbp - 4],zero
        pop     rbp
        ret

优化（-Ofast）：

q_sqrt(float):                             # @q_sqrt(float)
        movd    eax,xmm0
        sar     eax
        mov     ecx,1597463007
        sub     ecx,eax
        movd    xmm1,ecx
        mulss   xmm0,dword ptr [rip + .LCPI0_0]
        movdqa  xmm2,xmm1
        mulss   xmm2,xmm1
        mulss   xmm0,xmm2
        addss   xmm0,dword ptr [rip + .LCPI0_1]
        mulss   xmm0,xmm1
        ret

您可以使用 https://godbolt.org/ 来检查编译器的汇编输出，使用各种不同的标志并检查它如何影响输出。