我感兴趣的是比较CPU时间一些代码部分编写的c++和Python(运行在Linux上)。下面的方法会在两者之间产生一个"公平"的比较吗?

import resource
def cpu_time():
    return resource.getrusage(resource.RUSAGE_SELF)[0]+ # time in user mode
        resource.getrusage(resource.RUSAGE_SELF)[1] # time in system mode

允许这样计时:

def timefunc( func ):
    start=cpu_time()
    func()
    return (cpu_time()-start)

然后测试如下:

def f():
    for i in range(int(1e6)):
        pass
avg = 0
for k in range(10):
    avg += timefunc( f ) / 10.0
print avg
=> 0.002199700000000071

#include <ctime>
#include <iostream>
int main() {
    double avg = 0.0;
    int N = (int) 1e6;
    for (int k=0; k<10; k++) {
        clock_t start;
        start = clock();
        for (int i=0; i<N; i++) continue;
        avg += (double)(clock()-start) / 10.0 / CLOCKS_PER_SEC;
    }
    std::cout << avg << 'n';
    return 0;
}

生成0.002。

问题:

1. 我读过c++ clock()测量CPU时间，这是我所追求的，但我似乎找不到它是否包括用户和系统时间。
c++的
2. 结果不太精确。为什么呢?
3. 如上所述的总体比较公平性。

按照David在注释中的建议更新了c++代码:

#include <sys/resource.h>
#include <iostream>
int main() {
    double avg = 0.0;
    int N = (int) 1e6;
    int tally = 0;
    struct rusage usage;
    struct timeval ustart, ustop, sstart, sstop;
    getrusage(RUSAGE_SELF, &usage);
    ustart = usage.ru_utime;
    sstart = usage.ru_stime;
    for (int k=0; k<10; k++) {
        ustart = usage.ru_utime;
        sstart = usage.ru_stime;
        for (int i=0; i<N; i++) continue;
        getrusage(RUSAGE_SELF, &usage);
        ustop = usage.ru_utime;
        sstop = usage.ru_stime;
        avg += (
            (ustop.tv_sec+ustop.tv_usec/1e6+
            sstop.tv_sec+sstop.tv_usec/1e6)
            -
            (ustart.tv_sec+ustart.tv_usec/1e6+
            sstart.tv_sec+sstart.tv_usec/1e6)
        ) / 10.0; 
    }
    std::cout << avg << 'n';
    return 0;
}

g++ -O0 cpptimes.cpp ; ./a.out
=> 0.0020996
g++ -O1 cpptimes.cpp ; ./a.out
=> 0

所以我想getrusage给我一点更好的分辨率，但我不确定我应该读多少。