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

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

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。 我们可以从头文件中使用std::chrono::high_resolution_clock 我们可以编写一个方法，以易于阅读的形式打印方法执行时间。

例如，要找到1到1亿之间的所有质数，大约需要1分40秒。 因此，执行时间打印为:

Execution Time: 1 Minutes, 40 Seconds, 715 MicroSeconds, 715000 NanoSeconds

代码在这里:

#include <iostream>
#include <chrono>

using namespace std;
using namespace std::chrono;

typedef high_resolution_clock Clock;
typedef Clock::time_point ClockTime;

void findPrime(long n, string file);
void printExecutionTime(ClockTime start_time, ClockTime end_time);

int main()
{
    long n = long(1E+8);  // N = 100 million

    ClockTime start_time = Clock::now();

    // Write all the prime numbers from 1 to N to the file "prime.txt"
    findPrime(n, "C:\\prime.txt"); 

    ClockTime end_time = Clock::now();

    printExecutionTime(start_time, end_time);
}

void printExecutionTime(ClockTime start_time, ClockTime end_time)
{
    auto execution_time_ns = duration_cast<nanoseconds>(end_time - start_time).count();
    auto execution_time_ms = duration_cast<microseconds>(end_time - start_time).count();
    auto execution_time_sec = duration_cast<seconds>(end_time - start_time).count();
    auto execution_time_min = duration_cast<minutes>(end_time - start_time).count();
    auto execution_time_hour = duration_cast<hours>(end_time - start_time).count();

    cout << "\nExecution Time: ";
    if(execution_time_hour > 0)
    cout << "" << execution_time_hour << " Hours, ";
    if(execution_time_min > 0)
    cout << "" << execution_time_min % 60 << " Minutes, ";
    if(execution_time_sec > 0)
    cout << "" << execution_time_sec % 60 << " Seconds, ";
    if(execution_time_ms > 0)
    cout << "" << execution_time_ms % long(1E+3) << " MicroSeconds, ";
    if(execution_time_ns > 0)
    cout << "" << execution_time_ns % long(1E+6) << " NanoSeconds, ";
}

简单程序查找函数执行时间。

#include <iostream>
#include <ctime> // time_t
#include <cstdio>

void function()
{
     for(long int i=0;i<1000000000;i++)
     {
        // do nothing
     }
}

int main()
{

time_t begin,end; // time_t is a datatype to store time values.

time (&begin); // note time before execution
function();
time (&end); // note time after execution

double difference = difftime (end,begin);
printf ("time taken for function() %.2lf seconds.\n", difference );

return 0;
}