轻松测量经过时间 [英] Easily measure elapsed time

问题描述

我试图使用时间（）来衡量不同点我的计划。

我不明白的是，为什么在之前和之后的值是一样的吗？我明白这不是来分析我的计划的最好方式，我只是想看到的东西多长时间。

 的printf（** MyProgram ::之前的时间=％LD \\ N，时间（NULL））;doSomthing（）;
doSomthingLong（）;的printf（** MyProgram ::经过时间=％LD \\ N，时间（NULL））;
 

我曾尝试：

  timeval结构差异，startTV，endTV;函数gettimeofday（安培; startTV，NULL）;做一点事（）;
doSomethingLong（）;函数gettimeofday（安培; endTV，NULL）;timersub（安培; endTV，＆安培; startTV，＆安培;差异）;的printf（**花费时间=％LD％LD \\ N，diff.tv_sec，diff.tv_usec）;
 

做我读的结果**如何拍摄时间= 0 26339 ？这是否意味着26339纳秒= 26.3毫秒？

什么 **所用的时间= 4 45025 ，这是否意味着4秒，25毫秒？

解决方案

 的#include＆LT;＆的ctime GT;空隙f（）的{
  使用命名空间std;
  clock_t表示开始=时钟（）;  code_to_time（）;  clock_t表示结束=时钟（）;
  双elapsed_secs =双（结束 - 开始）/ CLOCKS_PER_SEC;
}
 

的时间（）功能只精确到秒之内，但也有的 CLOCKS_PER_SEC 一秒钟内的时钟。这是一个简单的，便携式测量，即使它过于简单化了。

I am trying to use time() to measure various points of my program.

What I don't understand is why the values in the before and after are the same? I understand this is not the best way to profile my program, I just want to see how long something take.

printf("**MyProgram::before time= %ld\n", time(NULL));

doSomthing();
doSomthingLong();

printf("**MyProgram::after time= %ld\n", time(NULL));

I have tried:

struct timeval diff, startTV, endTV;

gettimeofday(&startTV, NULL); 

doSomething();
doSomethingLong();

gettimeofday(&endTV, NULL); 

timersub(&endTV, &startTV, &diff);

printf("**time taken = %ld %ld\n", diff.tv_sec, diff.tv_usec);

How do I read a result of **time taken = 0 26339? Does that mean 26,339 nanoseconds = 26.3 msec?

What about **time taken = 4 45025, does that mean 4 seconds and 25 msec?

解决方案

#include <ctime>

void f() {
  using namespace std;
  clock_t begin = clock();

  code_to_time();

  clock_t end = clock();
  double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
}

The time() function is only accurate to within a second, but there are CLOCKS_PER_SEC "clocks" within a second. This is an easy, portable measurement, even though it's over-simplified.

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