Unix/Linux計算程式消耗的時間（毫秒)

使用time(NULL)得到的是從1970年1月1日到目前的秒，這種精度很多時候是不夠用的。為了得到毫秒級的精度，需要使用gettimeofday:

直接上代碼:

#include <sys/time.h>#include <stdlib.h>#include <stdio.h>#include <math.h>// Return 1 if the difference is negative, otherwise 0.  int timeval_subtract(struct timeval *result, struct timeval *t2, struct timeval *t1){    long int diff = (t2->tv_usec + 1000000 * t2->tv_sec) - (t1->tv_usec + 1000000 * t1->tv_sec);    result->tv_sec = diff / 1000000;    result->tv_usec = diff % 1000000;    return (diff<0);}void timeval_print(struct timeval *tv){    char buffer[30];    time_t curtime;    printf("%ld.%06ld", tv->tv_sec, tv->tv_usec);    curtime = tv->tv_sec;    strftime(buffer, 30, "%m-%d-%Y  %T", localtime(&curtime;));    printf(" = %s.%06ld\n", buffer, tv->tv_usec);}int main(){    struct timeval tvBegin, tvEnd, tvDiff;    // begin    gettimeofday(&tvBegin;, NULL);    timeval_print(&tvBegin;);    // lengthy operation    int i,j;    for(i=0;i<999999L;++i) {        j=sqrt(i);    }    //end    gettimeofday(&tvEnd;, NULL);    timeval_print(&tvEnd;);    // diff    timeval_subtract(&tvDiff;, &tvEnd;, &tvBegin;);    printf("%ld.%06ld\n", tvDiff.tv_sec, tvDiff.tv_usec);    return 0;}