C语言运行100三次的测试代码 windows版本
/*
* for windows only
* modify by benjiam.
*/
#include <stdio.h>
#include <windows.h>
#define BAILOUT 16
#define MAX_ITERATIONS 1000
int mandelbrot(float x, float y)
{
float cr = y - 0.5;
float ci = x;
float zi = 0.0;
float zr = 0.0;
int i = 0;
while(1) {
i ++;
float temp = zr * zi;
float zr2 = zr * zr;
float zi2 = zi * zi;
zr = zr2 - zi2 + cr;
zi = temp + temp + ci;
if (zi2 + zr2 > BAILOUT)
return i;
if (i > MAX_ITERATIONS)
return 0;
}
}
void run2() {
int x,y;
for (y = -39; y < 39; y++) {
;//fputs( "\n ", stderr);
for (x = -39; x < 39; x++) {
int i = mandelbrot(x/40.0, y/40.0);
if (i==0)
;//fputs( "* ", stderr);
else
;//fputs( " ", stderr);
}
}
;//fputs( "\n ", stderr);
}
void run() {
/*
struct timeval aTv;
gettimeofday(&aTv, NULL);
long init_time = aTv.tv_sec;
long init_usec = aTv.tv_usec;
*/
FILETIME time1;
FILETIME time2;
GetSystemTimeAsFileTime(&time1);
int i;
for (i = 0; i < 100; i++)
run2();
//gettimeofday(&aTv,NULL);
//double query_time = (aTv.tv_sec - init_time) + (double)(aTv.tv_usec - init_usec)/1000000.0;
GetSystemTimeAsFileTime(&time2);
__int64 t1=time1.dwHighDateTime;
t1 < <=32;
t1|=time1.dwLowDateTime;
__int64 t2=time2.dwHighDateTime;
t2 < <=32;
t2|=time2.dwLowDateTime;
printf ( "C Elapsed %0.2d\n ", t2-t1);
}
int main (int argc, const char * argv[]) {
run();
run();
run();
return 0;
}
原来的代码 本质变成了io测试，这是不对的。对c 还是java 都是不公平的。所以我去掉了io操作。
[解决办法]
想得出什么结论
