getrusage在ru_utime.tv_usec和ru_utime.tv_sec中返回零 [英] getrusage returning zeros in ru_utime.tv_usec and ru_utime.tv_sec

问题描述

有关以下代码的getrusage返回零在 ru_utime.tv_usec 和 ru_utime.tv_sec

代码：

  #includestdlib.h
 #include stdio.h
 #includesys / time.h
 #includesys / resource.h
 
 int getr_return，who = RUSAGE_SELF; 
 struct rusage用法; 
 
 main（）
 {
 getr_return = getrusage（who，& usage）; 
 printf（getr_return =％d \\\
，getr_return）; 
 
 printf（以秒计的时间=％.61f\\\
，usage.ru_utime.tv_sec）; 
 printf（以秒计的时间=％.61f\\\
，usage.ru_utime.tv_usec）; 
 
 
 Some_Mips_consuming_code（）。 
 
 
 getr_return = getrusage（who，& usage）; 
 printf（getr_return =％d \\\
，getr_return）; 
 
 printf（以秒计的时间=％.61f\\\
，usage.ru_utime.tv_sec）; 
 printf（以秒计的时间=％.61f\\\
，usage.ru_utime.tv_usec）; 
 
出口; 
} 
  

输出：

在秒取 getr_return = 0 
时间=以秒为采取0.0000000000000000000000000000000000000000000000000000000000000 
时间采取秒= 0.0000000000000000000000000000000000000000000000000000000000000 = 0.0000000000000000000000000000000000000000000000000000000000000 
 getr_return = 0 
时间以秒计算的
时间= 0.0000000000000000000000000000000000000000000000000000000000000 
  

---

在 Linux版本2.6.18-308上编译代码

在ARM板上运行可执行文件，它的 Linux版本3.8.1-2.0

解决方案

时间测量精度，分辨率和精度有限。阅读时间（7）。不要指望计算的重要措施少于例如半秒钟。

通常，CPU时间测量是通过计算 jiffies 或定时器中断（如果您的内核中有一个与 HZ 有关）。

<由于您在开始时致电 getrusage（2），你的 main ，在它之前没有发生太多的计算（基本上，只有在 crt0.o ）。所以你应该预计它接近零。

你可以尝试使用 clock_gettime（2），提供 CLOCK_REALTIME 或 CLOCK_PROCESS_CPUTIME_ID 。

For the following code getrusage returning zeros in ru_utime.tv_usec and ru_utime.tv_sec.

Code:

#include "stdlib.h"
#include "stdio.h"
#include "sys/time.h"
#include "sys/resource.h"

int getr_return, who = RUSAGE_SELF;
struct rusage usage;

main()
{
    getr_return = getrusage(who, &usage);
    printf(" getr_return = %d\n", getr_return);

    printf(" time taken in seconds = %.61f\n", usage.ru_utime.tv_sec);
    printf(" time taken in seconds = %.61f\n", usage.ru_utime.tv_usec);


        Some_Mips_consuming_code().


        getr_return = getrusage(who, &usage);
        printf(" getr_return = %d\n", getr_return);

        printf(" time taken in seconds = %.61f\n", usage.ru_utime.tv_sec);
        printf(" time taken in seconds = %.61f\n", usage.ru_utime.tv_usec);

    exit;
 } 

output:

getr_return = 0
time taken in seconds =  0.0000000000000000000000000000000000000000000000000000000000000
time taken in seconds =  0.0000000000000000000000000000000000000000000000000000000000000
getr_return = 0
time taken in seconds =  0.0000000000000000000000000000000000000000000000000000000000000
time taken in seconds =  0.0000000000000000000000000000000000000000000000000000000000000

---

Compiled code on Linux version 2.6.18-308

Ran executable on ARM board and it's Linux version 3.8.1-2.0

解决方案

Time measurement is limited in precision, resolution and accuracy. Read time(7). Don't expect significant measures for computation less than e.g. half a second.

Often, CPU time measurement is done by counting jiffies or timer interrupts (related to HZ if you have one in your kernel).

Since you call getrusage(2) at the beginning of your main, not much computing has happened before it (basically, only pre-main initialization in e.g. crt0.o). So you should expect it to be near zero.

You might try using clock_gettime(2) with CLOCK_REALTIME or CLOCK_PROCESS_CPUTIME_ID.

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