c - Timing Quake III hack 仅在优化编译时有效

Question

所以我刚刚发现了非常有趣的 Quake III 逆平方根 hack。在了解了它的工作原理和所有内容之后，我决定对其进行测试。我发现，当启用优化进行编译时，该 hack 的性能仅优于 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, &stop);
    ns = ( stop.tv_sec - start.tv_sec ) * 1E9 + ( stop.tv_nsec - start.tv_nsec );
    fprintf(stderr, "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) 开箱即用会比 1/sqrt(X) 更好。阅读更多内容后，我现在知道要么 libm 得到了更好的优化，要么我的 CPU 配备了 sqrt(X) 的硬件解决方案。毕竟，自从快速逆root hack发展以来，CPU已经发生了突飞猛进的变化。

我不明白编译器将应用什么类型的优化来使其速度更快。当然，也许我的基准考虑不周？

感谢您的帮助!!

Answer 1

正如您所说，大多数现代 CPU 都包含浮点单元，通常提供硬件指令来计算平方根。 FPU 还提供除法指令，因此我希望您的处理器(尽管我不知道)能够仅用几个汇编指令来计算逆 sqrt。您的结果有点令人惊讶:您应该检查 FPU 是否真正被使用。我不了解 Ryzen，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。因此，像您一样运行一些测试，或者看看 here理论。