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。

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

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

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

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