52

我的意思是:我如何测量我的 CPU 花费在函数执行上的时间和运行我的函数所需的挂钟时间?(我对 Linux/Windows 以及 x86 和 x86_64 都感兴趣)。看看我想做什么(我在这里使用 C++,但我更喜欢 C 解决方案):

int startcputime, endcputime, wcts, wcte;

startcputime = cputime();
function(args);
endcputime = cputime();

std::cout << "it took " << endcputime - startcputime << " s of CPU to execute this\n";

wcts = wallclocktime();
function(args);
wcte = wallclocktime();

std::cout << "it took " << wcte - wcts << " s of real time to execute this\n";

另一个重要问题:这种时间测量架构是否独立?

4

4 回答 4

70

这是一个适用于 Windows 和 Linux 以及 C 和 C++ 的复制粘贴解决方案。

正如评论中提到的,有一个 boost 库可以做到这一点。但是如果你不能使用 boost,这应该可以工作:

//  Windows
#ifdef _WIN32
#include <Windows.h>
double get_wall_time(){
    LARGE_INTEGER time,freq;
    if (!QueryPerformanceFrequency(&freq)){
        //  Handle error
        return 0;
    }
    if (!QueryPerformanceCounter(&time)){
        //  Handle error
        return 0;
    }
    return (double)time.QuadPart / freq.QuadPart;
}
double get_cpu_time(){
    FILETIME a,b,c,d;
    if (GetProcessTimes(GetCurrentProcess(),&a,&b,&c,&d) != 0){
        //  Returns total user time.
        //  Can be tweaked to include kernel times as well.
        return
            (double)(d.dwLowDateTime |
            ((unsigned long long)d.dwHighDateTime << 32)) * 0.0000001;
    }else{
        //  Handle error
        return 0;
    }
}

//  Posix/Linux
#else
#include <time.h>
#include <sys/time.h>
double get_wall_time(){
    struct timeval time;
    if (gettimeofday(&time,NULL)){
        //  Handle error
        return 0;
    }
    return (double)time.tv_sec + (double)time.tv_usec * .000001;
}
double get_cpu_time(){
    return (double)clock() / CLOCKS_PER_SEC;
}
#endif

有很多方法可以实现这些时钟。但这是上面的代码片段使用的:

对于 Windows:

对于 Linux:


这是一个小演示:

#include <math.h>
#include <iostream>
using namespace std;

int main(){

    //  Start Timers
    double wall0 = get_wall_time();
    double cpu0  = get_cpu_time();

    //  Perform some computation.
    double sum = 0;
#pragma omp parallel for reduction(+ : sum)
    for (long long i = 1; i < 10000000000; i++){
        sum += log((double)i);
    }

    //  Stop timers
    double wall1 = get_wall_time();
    double cpu1  = get_cpu_time();

    cout << "Wall Time = " << wall1 - wall0 << endl;
    cout << "CPU Time  = " << cpu1  - cpu0  << endl;

    //  Prevent Code Elimination
    cout << endl;
    cout << "Sum = " << sum << endl;

}

输出(12 个线程):

Wall Time = 15.7586
CPU Time  = 178.719

Sum = 2.20259e+011
于 2013-07-03T05:55:28.460 回答
36

C++11。写起来容易多了!

用于std::chrono::system_clock挂钟和std::clockCPU 时钟 http://en.cppreference.com/w/cpp/chrono/system_clock

#include <cstdio>
#include <ctime>
#include <chrono>

.... 

std::clock_t startcputime = std::clock();
do_some_fancy_stuff();
double cpu_duration = (std::clock() - startcputime) / (double)CLOCKS_PER_SEC;
std::cout << "Finished in " << cpu_duration << " seconds [CPU Clock] " << std::endl;


auto wcts = std::chrono::system_clock::now();
do_some_fancy_stuff();
std::chrono::duration<double> wctduration = (std::chrono::system_clock::now() - wcts);
std::cout << "Finished in " << wctduration.count() << " seconds [Wall Clock]" << std::endl;

等等,简单便携!不需要#ifdef _WIN32 或LINUX!

chrono::high_resolution_clock如果您需要更高的精度, 您甚至可以使用http://en.cppreference.com/w/cpp/chrono/high_resolution_clock

于 2015-08-31T18:11:34.917 回答
16

如果可以的话,举一个@lip建议使用的具体例子boost::timer(用Boost 1.51测试):

#include <boost/timer/timer.hpp>

// this is wallclock AND cpu time
boost::timer::cpu_timer timer;

... run some computation ...

boost::timer::cpu_times elapsed = timer.elapsed();
std::cout << " CPU TIME: " << (elapsed.user + elapsed.system) / 1e9 << " seconds"
          << " WALLCLOCK TIME: " << elapsed.wall / 1e9 << " seconds"
          << std::endl;
于 2013-08-13T08:46:36.457 回答
0

使用time.hclock中的方法:

clock_t start = clock();
/* Do stuffs */
clock_t end = clock();
float seconds = (float)(end - start) / CLOCKS_PER_SEC;

不幸的是,此方法在 Linux 上返回 CPU 时间,但在 Windows 上返回挂钟时间(感谢评论者提供此信息)。

于 2013-07-02T18:07:11.150 回答