对于旧的c++或C来说很简单:

#include <time.h> // includes clock_t and CLOCKS_PER_SEC

int main() {

    clock_t start, end;

    start = clock();
    // ...code to measure...
    end = clock();

    double duration_sec = double(end-start)/CLOCKS_PER_SEC;
    return 0;
}

计时精度(秒)为1.0/CLOCKS_PER_SEC

c++ 11清理了Jahid的回复:

#include <chrono>
#include <thread>

void long_operation(int ms)
{
    /* Simulating a long, heavy operation. */
    std::this_thread::sleep_for(std::chrono::milliseconds(ms));
}

template<typename F, typename... Args>
double funcTime(F func, Args&&... args){
    std::chrono::high_resolution_clock::time_point t1 = 
        std::chrono::high_resolution_clock::now();
    func(std::forward<Args>(args)...);
    return std::chrono::duration_cast<std::chrono::milliseconds>(
        std::chrono::high_resolution_clock::now()-t1).count();
}

int main()
{
    std::cout<<"expect 150: "<<funcTime(long_operation,150)<<"\n";

    return 0;
}

因为没有一个提供的答案是非常准确的或可复制的结果，我决定添加一个链接到我的代码，具有亚纳秒的精度和科学统计。

Note that this will only work to measure code that takes a (very) short time to run (aka, a few clock cycles to a few thousand): if they run so long that they are likely to be interrupted by some -heh- interrupt, then it is clearly not possible to give a reproducable and accurate result; the consequence of which is that the measurement never finishes: namely, it continues to measure until it is statistically 99.9% sure it has the right answer which never happens on a machine that has other processes running when the code takes too long.

https://github.com/CarloWood/cwds/blob/master/benchmark.h#L40

您可以使用一个简单的类来进行这种测量。

class duration_printer {
public:
    duration_printer() : __start(std::chrono::high_resolution_clock::now()) {}
    ~duration_printer() {
        using namespace std::chrono;
        high_resolution_clock::time_point end = high_resolution_clock::now();
        duration<double> dur = duration_cast<duration<double>>(end - __start);
        std::cout << dur.count() << " seconds" << std::endl;
    }
private:
    std::chrono::high_resolution_clock::time_point __start;
};

唯一需要做的是在函数的开始处创建一个对象

void veryLongExecutingFunction() {
    duration_calculator dc;
    for(int i = 0; i < 100000; ++i) std::cout << "Hello world" << std::endl;
}

int main() {
    veryLongExecutingFunction();
    return 0;
}

就是这样。可以修改该类以满足您的需求。

我建议使用steady_clock，它保证是单调的，不像high_resolution_clock。

#include <iostream>
#include <chrono>

using namespace std;

unsigned int stopwatch()
{
    static auto start_time = chrono::steady_clock::now();

    auto end_time = chrono::steady_clock::now();
    auto delta    = chrono::duration_cast<chrono::microseconds>(end_time - start_time);

    start_time = end_time;

    return delta.count();
}

int main() {
  stopwatch(); //Start stopwatch
  std::cout << "Hello World!\n";
  cout << stopwatch() << endl; //Time to execute last line
  for (int i=0; i<1000000; i++)
      string s = "ASDFAD";
  cout << stopwatch() << endl; //Time to execute for loop
}

输出:

Hello World!
62
163514

#include <iostream>
#include <chrono>

void function()
{
    // code here;
}

int main()
{
    auto t1 = std::chrono::high_resolution_clock::now();
    function();
    auto t2 = std::chrono::high_resolution_clock::now();

    auto duration = std::chrono::duration_cast<std::chrono::microseconds>( t2 - t1 ).count();

    std::cout << duration<<"/n";
    return 0;
}

这对我很管用。

注意:

high_resolution_clock不能在不同的标准库实现中实现一致，应该避免使用它。它通常只是std::chrono::steady_clock或std::chrono::system_clock的别名，但它是哪一个取决于库或配置。当它是system_clock时，它不是单调的(例如，时间可以倒退)。

例如，对于gcc的libstdc++，它是system_clock，对于MSVC，它是steady_clock，对于clang的libc++，它取决于配置。

通常应该直接使用std::chrono::steady_clock或std::chrono::system_clock，而不是std::chrono::high_resolution_clock:使用steady_clock测量持续时间，使用system_clock测量时钟时间。