如何在没有boost :: timer的情况下以毫秒为单位计时

Question

我正在使用boost 1.46,它不包括boost :: timer,还有什么方法可以计算我的功能.

我目前正在这样做:

time_t now = time(0);
<some stuff>
time_t after = time(0);

cout << after - now << endl; 

但它只是在几秒钟内给出答案,所以如果函数小于1,则显示0.

谢谢

Answer 1

在linux或Windows中:

#include <ctime>
#include <iostream>

int
main(int, const char**)
{
     std::clock_t    start;

     start = std::clock();
     // your test
     std::cout << "Time: " << (std::clock() - start) / (double)(CLOCKS_PER_SEC / 1000) << " ms" << std::endl;
     return 0;
}

祝好运 ;)

Answer 2

使用std::chrono:

#include <chrono>
#include <thread>
#include <iostream>

// There are other clocks, but this is usually the one you want.
// It corresponds to CLOCK_MONOTONIC at the syscall level.
using Clock = std::chrono::steady_clock;
using std::chrono::time_point;
using std::chrono::duration_cast;
using std::chrono::milliseconds;
using namespace std::literals::chrono_literals;
using std::this_thread::sleep_for;

int main()
{
    time_point<Clock> start = Clock::now();
    sleep_for(500ms);
    time_point<Clock> end = Clock::now();
    milliseconds diff = duration_cast<milliseconds>(end - start);
    std::cout << diff.count() << "ms" << std::endl;
}

std::chrono是C++ 11,std::literals是C++ 14(否则你需要milliseconds(500)).

Answer 3

原来有一个版本的时间在boost 1.46(只是在不同的位置).感谢@jogojapan指出它.

它可以这样做:

#include <boost/timer.hpp>

timer t;
<some stuff>
std::cout << t.elapsed() << std::endl;

或者使用std libs作为@Quentin Perez指出(我会接受最初的问题)