UNIX环境高级编程 times() 疑问

Question

1. 例程

 int main(int argc, char *argv[])
{
        clock_t s_clk,e_clk;
        struct tms s_tms,e_tms;

        s_clk = times(&s_tms);
        system("ls /dev");
        system("date");
        sleep(1);
        e_clk = times(&e_tms);
        printf("e_clk %ld - s_clk %ld = %ld\n",e_clk , s_clk,e_clk-s_clk);
// pr_time(1,s_tms,e_tms);

        printf("user time e_tms %ld - s_tms %ld = %ld\n",
                        e_tms.tms_utime,s_tms.tms_stime,(e_tms.tms_utime - s_tms.tms_utime));
        printf("system time e_tms %ld - s_tms %ld = %ld\n",
                        e_tms.tms_stime,s_tms.tms_stime,(e_tms.tms_stime - s_tms.tms_stime));

        return 0;
}

./a.out
2016年 04月 03日 星期日 15:40:10 CST
e_clk 1742131978 - s_clk 1742131878 = 100
user time e_tms 0 - s_tms 0 = 0
system time e_tms 0 - s_tms 0 = 0

时间差为什么还是 0.

Answer 1

楼主要不sleep久一点试试。
如果一个进程 的运行时间很短，短到和系统的计时器间隔一个数量级，用这种方法测出来的结果必然是不够精确的，头尾都有误差。