如何打印時間在linux

Question

所以我用rusage打印出它從用戶需要多長時間，以及系統處理的指令，沿着線的東西，

  //DO STUFF HERE 
      printf(" TOTAL TIMES: "); 
      tusage.ru_utime.tv_sec = rusage.ru_utime.tv_sec + rusage.ru_stime.tv_sec; 

Answer 1

試試這個：

#include <stdio.h> 
#include <sys/resource.h> 
int main() { 
    struct rusage rusage; 
    struct rusage tusage; 
    int i, j, r=0; 
    for (i = 0; i < 10000; i++) { 
    for (j = 1; j < 100000; j++) { 
     r = i % j + i/j; 
    } 
    } 
    getrusage(RUSAGE_SELF, &rusage); 
    printf("TOTAL TIME \n"); 
    tusage.ru_utime.tv_sec = rusage.ru_utime.tv_sec + rusage.ru_stime.tv_sec; 
    tusage.ru_utime.tv_usec = rusage.ru_utime.tv_usec + rusage.ru_stime.tv_usec; 
    tusage.ru_utime.tv_sec += tusage.ru_utime.tv_usec/1000000; 
    tusage.ru_utime.tv_usec = tusage.ru_utime.tv_usec % 1000000; 
    printf("%ld.%06ld\n", tusage.ru_utime.tv_sec, tusage.ru_utime.tv_usec); 

    return r; 
}