我只想知道为什么快速反演算法比math.h sqrt函数慢。这是我的代码示例
代码试图演示比较慢反转和快反转在调试时,我看到1秒用于慢速反转,4秒用于快速反转。问题在哪里?
#include<stdio.h>
#include<time.h>
#include<math.h>
#include"inverse.h"
#define SIZE 256
int main()
{
char buffer[SIZE];
time_t curtime;
time_t curtime2;
struct tm *loctime;
int i = 0;
float x = 0;
curtime = time(NULL);
loctime = localtime (&curtime);
fputs (asctime (loctime), stdout);
while(i < 100000000)
{
i++;
//x = 1/sqrt(465464.015465);
x = inverse_square_root(465464.015465);
}
curtime = time(NULL);
loctime = localtime (&curtime);
fputs (asctime (loctime), stdout);
getchar();
return 0;
}
float inverse_square_root(float number)
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the heck?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
最佳答案
“问题”可能是您现在有了实现sqrt()
的硬件,使得它比软件方法更快。如果没有更多关于系统的详细信息,或者是一些分析和反汇编数据,很难判断。
See this answer以获取有关x86fsqrt
指令周期数的详细信息。