我正在开发一个C程序,我必须fork()
一个进程并使用getrusage()
函数来打印用户时间和子进程的内核时间。
这是我的代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/resource.h>
#include <sys/time.h>
int main(int argc, const char * argv[]) {
struct rusage child1_usage, child2_usage;
struct timeval child1_stime, child2_stime, child1_utime, child2_utime;
pid_t child1_pid, child2_pid;
switch (child1_pid = fork()) {
case -1:
perror("Fork failed!\n");
break;
case 0:
printf("Hi I am the first child!!\nGive me one number and I multiply it by 4:\n");
int num1;
scanf("%d", &num1);
printf("%d\n", num1*4);
switch (child2_pid = fork()) {
case -1:
perror("Second Fork failed!\n");
break;
case 0:
printf("Hi I am the grand child!!\nGive me one number and I multiply it by 3:\n");
int num2;
scanf("%d", &num2);
printf("%d\n", num2*4);
getrusage(RUSAGE_SELF, &child2_usage);
child2_utime = child2_usage.ru_utime;
child2_stime = child2_usage.ru_stime;
printf("Time in user mode: %lld\n", (long long) child2_utime.tv_sec);
printf("Time in kernel mode: %lld\n", (long long) child2_stime.tv_sec);
default:
sleep(10);
break;
}
getrusage(RUSAGE_SELF, &child1_usage);
child1_utime = child1_usage.ru_utime;
child1_stime = child1_usage.ru_stime;
printf("Time in user mode: %lld\n", (long long) child1_utime.tv_sec);
printf("Time in kernel mode: %lld\n", (long long) child1_stime.tv_sec);
default:
sleep(10);
break;
}
return 0;
}
我的输出都是零...可能我误解了getrusage()
是如何工作的。否则我看不出错误在哪里。
答案 0 :(得分:0)
您只打印出timeval&#39; tv_sec
,其中包含整秒数。由于程序不做任何计算密集型操作,因此其CPU时间(用户和系统)远小于1秒 - 大约0.00001,因此您只能在微秒内获得非零值({{ 1}})field。
https://www.gnu.org/software/libc/manual/html_node/Elapsed-Time.html
tv_usec
结构代表经过的时间。它在struct timeval
中声明,并具有以下成员:
sys/time.h
- 这表示经过时间的整秒数。
time_t tv_sec
- 这是剩余的经过时间(几分之一秒),表示为微秒数。它总是不到一百万。