c++ - C++ 中可忽略的计时测量？

Question

考虑以下代码:

#include <iostream>
#include <string>
#include <chrono>

using namespace std;

int main()
{
    int iter = 1000000;
    int loops = 10;
    while (loops)
    {
        int a=0, b=0, c=0, f = 0, m = 0, q = 0;
        auto begin = chrono::high_resolution_clock::now();
        auto end = chrono::high_resolution_clock::now();
        auto deltaT = end - begin;
        auto accumT = end - begin;
        accumT = accumT - accumT;
        auto controlT = accumT;
        srand(chrono::duration_cast<chrono::nanoseconds>(begin.time_since_epoch()).count());

        for (int i = 0; i < iter; i++) {
            begin = chrono::high_resolution_clock::now();

            // No arithmetic operation

            end = chrono::high_resolution_clock::now();
            deltaT = end - begin;
            accumT += deltaT;
        }

        controlT = accumT; // Control duration
        accumT = accumT - accumT; // Reset to zero

        for (int i = 0; i < iter; i++) {
            auto n1 = rand() % 100;
            auto n2 = rand() % 100;
            begin = chrono::high_resolution_clock::now();
            c += i*2*n1*n2; // Some arbitrary arithmetic operation
            end = chrono::high_resolution_clock::now();
            deltaT = end - begin;
            accumT += deltaT;
        }
        // Print the difference in time between loop with no arithmetic operation and loop with
        cout << " c = " << c << "\t\t" << " | ";
        cout << "difference between the 1st and 2nd loop: "
             << chrono::duration_cast<chrono::nanoseconds>(accumT - controlT).count()
             << endl;
        loops--;
    }
    return 0;
}

它试图隔离操作的时间测量。第一个循环是建立基线的控制，第二个循环是任意算术运算。

然后它输出到控制台。这是示例输出:

 c =  2116663282   | difference between 1st and 2nd loop: -8620916
 c =   112424882   | difference between 1st and 2nd loop: -1197927
 c = -1569775878   | difference between 1st and 2nd loop: -5226990
 c =  1670984684   | difference between 1st and 2nd loop:  4394706
 c = -1608171014   | difference between 1st and 2nd loop:   676683
 c = -1684897180   | difference between 1st and 2nd loop:  2868093
 c =   112418158   | difference between 1st and 2nd loop:  5846887
 c =  2019014070   | difference between 1st and 2nd loop:  -951609
 c =   656490372   | difference between 1st and 2nd loop:   997815
 c =   263579698   | difference between 1st and 2nd loop:  2371088

这是非常有趣的部分:有时带有算术运算的循环比没有算术运算的循环完成更快(负差异)。这意味着记录当前时间的运算比算术运算慢，因此不可忽略。

有解决办法吗？

PS:是的，我知道您可以在begin 和end 之间包装整个循环。

设置机器:Core i7体系结构、Windows 10 64 位和 Visual Studio 2015

Answer 1

您的问题是您测量的是时间，而不是处理的指令数。时间会受到许多并非您真正期望或希望衡量的事物的影响。

相反，您应该测量时钟周期数。可以在 Agner Fog 的网站上找到一个用于此的库。他有很多关于优化的有用信息:

http://www.agner.org/optimize/#manuals

即使使用时钟周期，您仍然可以体验到结果的特殊性。如果处理器使用 out-of-order execution 可能会发生这种情况这使处理器能够优化操作的执行顺序。

如果您使用调试符号编译代码，编译器可能会注入(inject)额外的代码，这可能会影响结果。执行此类测试时，您应该始终在不带调试信息的情况下进行编译。