这是一个既能在Windows上工作,又能在C和C++上工作的复制粘贴解决方案.
正如 comments 中提到的,有一个boost库可以实现这一点.但如果你不能使用boost,这应该是可行的:
// 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.但下面是上面的代码片段使用的内容:
对于Windows:
对于Linux:
下面是一个小演示:
#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