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

在c++ 11中，这是一个非常容易使用的方法。你必须使用std::chrono::high_resolution_clock from <chrono>头。

像这样使用它:

#include <chrono>

/* Only needed for the sake of this example. */
#include <iostream>
#include <thread>
    
void long_operation()
{
    /* Simulating a long, heavy operation. */

    using namespace std::chrono_literals;
    std::this_thread::sleep_for(150ms);
}

int main()
{
    using std::chrono::high_resolution_clock;
    using std::chrono::duration_cast;
    using std::chrono::duration;
    using std::chrono::milliseconds;

    auto t1 = high_resolution_clock::now();
    long_operation();
    auto t2 = high_resolution_clock::now();

    /* Getting number of milliseconds as an integer. */
    auto ms_int = duration_cast<milliseconds>(t2 - t1);

    /* Getting number of milliseconds as a double. */
    duration<double, std::milli> ms_double = t2 - t1;

    std::cout << ms_int.count() << "ms\n";
    std::cout << ms_double.count() << "ms\n";
    return 0;
}

这将度量函数long_operation的持续时间。

可能的输出:

150ms
150.068ms

工作示例:https://godbolt.org/z/oe5cMd

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

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

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;
}

在Scott Meyers的书中，我发现了一个通用泛型lambda表达式的例子，可以用来测量函数的执行时间。(c++ 14)

auto timeFuncInvocation = 
    [](auto&& func, auto&&... params) {
        // get time before function invocation
        const auto& start = std::chrono::high_resolution_clock::now();
        // function invocation using perfect forwarding
        std::forward<decltype(func)>(func)(std::forward<decltype(params)>(params)...);
        // get time after function invocation
        const auto& stop = std::chrono::high_resolution_clock::now();
        return stop - start;
     };

问题是，您只能测量一次执行，因此结果可能非常不同。为了获得可靠的结果，您应该测量大量的执行。 根据Andrei Alexandrescu在code::dive 2015会议上的演讲-编写快速代码I:

测量时间:tm = t + tq + tn + to

地点:

Tm -测量(观察)时间

T -实际感兴趣的时间

Tq -由量化噪声增加的时间

Tn -由各种噪声源添加的时间

To -开销时间(测量、循环、调用函数)

根据他在后面的演讲中所说的，你应该把大量执行中的最小值作为你的结果。 我鼓励你们去看他解释原因的那节课。

还有谷歌上的一个很好的库- https://github.com/google/benchmark。 这个库使用简单，功能强大。你可以在youtube上查看钱德勒·卡鲁斯的一些讲座，他在实践中使用了这个库。例如，2017年CppCon:钱德勒·卡鲁斯《无处可去》;

使用示例:

#include <iostream>
#include <chrono>
#include <vector>
auto timeFuncInvocation = 
    [](auto&& func, auto&&... params) {
        // get time before function invocation
        const auto& start = high_resolution_clock::now();
        // function invocation using perfect forwarding
        for(auto i = 0; i < 100000/*largeNumber*/; ++i) {
            std::forward<decltype(func)>(func)(std::forward<decltype(params)>(params)...);
        }
        // get time after function invocation
        const auto& stop = high_resolution_clock::now();
        return (stop - start)/100000/*largeNumber*/;
     };

void f(std::vector<int>& vec) {
    vec.push_back(1);
}

void f2(std::vector<int>& vec) {
    vec.emplace_back(1);
}
int main()
{
    std::vector<int> vec;
    std::vector<int> vec2;
    std::cout << timeFuncInvocation(f, vec).count() << std::endl;
    std::cout << timeFuncInvocation(f2, vec2).count() << std::endl;
    std::vector<int> vec3;
    vec3.reserve(100000);
    std::vector<int> vec4;
    vec4.reserve(100000);
    std::cout << timeFuncInvocation(f, vec3).count() << std::endl;
    std::cout << timeFuncInvocation(f2, vec4).count() << std::endl;
    return 0;
}

编辑: 当然，你总是需要记住，你的编译器可以优化或不优化某些东西。像perf这样的工具在这种情况下很有用。