我的意思是:我如何衡量我的CPU花在功能执行上的时间和运行我的功能所需的挂钟时间? (我对Linux / Windows以及x86和x86_64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?
答案 0 :(得分:62)
这是一个适用于Windows和Linux以及C和C ++的复制粘贴解决方案。
正如评论中所提到的,有一个提升库可以做到这一点。但如果你不能使用boost,这应该可行:
// 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。但这是上面代码片段使用的内容:
对于Windows:
GetProcessTimes()
对于Linux:
gettimeofday()
clock()
这是一个小小的示范:
#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个主题):
Wall Time = 15.7586
CPU Time = 178.719
Sum = 2.20259e+011
答案 1 :(得分:27)
C ++ 11。写得容易多了!
使用"csrftoken=3R8dfLkO6HJatXhlzaifLI8uY96g5479"
作为挂钟,使用csrftoken=2wabMFekQdMdlWtoQ5ugCvzdnz2fPAAu;
作为cpu时钟
http://en.cppreference.com/w/cpp/chrono/system_clock
std::chrono::system_clock
Etvoilà,轻松便携!不需要#ifdef _WIN32或LINUX!
如果您需要更高的精确度,甚至可以使用std::clock
http://en.cppreference.com/w/cpp/chrono/high_resolution_clock
答案 2 :(得分:13)
给出一个具体的例子,@ lip的建议如果你能使用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;
答案 3 :(得分:-1)
使用time.h中的clock
方法:
clock_t start = clock();
/* Do stuffs */
clock_t end = clock();
float seconds = (float)(end - start) / CLOCKS_PER_SEC;
不幸的是,这个方法在Linux上返回CPU时间,但returns wall-clock time on Windows(感谢评论者提供此信息)。