测量操作块的时间

Question

我正在使用 chrono 库。我的代码如下：

auto begin = chrono::high_resolution_clock::now();
// operations
auto end = chrono::high_resolution_clock::now();
auto elapsed = chrono::duration_cast<chrono::nanoseconds>(end - begin);
cout<<”time: „<<elapsed.count()<<endl;

它总是显示 0。我做错了什么？

Answer 1

要测量两次之间的差异，请根据您想要的精度将差异保存在适当的对象中：

std::chrono::duration<double, std::milli> fp_ms = end - start;//difference in milliseconds
std::chrono::duration<double, std::nano> fp_ns = end - start;//difference in nanoseconds

然后，您可以适当地将其转换为秒（或者更好的是，使用由user chris 在 cmets 中提出的默认秒数duration<double> 建议）：

Live on Compiler Explorer

#include <iostream>
#include <chrono>
int main() {
    auto V = std::chrono::high_resolution_clock::now();
    for(int i = 0; i < 2000; i++)
        printf("%d ", i);
    printf("\n");
    auto Y = std::chrono::high_resolution_clock::now();
    std::chrono::duration<double> fp_s = Y - V;
    std::chrono::duration<double, std::milli> fp_ms = Y - V;
    std::chrono::duration<double, std::nano> fp_ns = Y - V;
    auto Zm = fp_ms.count()/1000.;
    auto Zn = fp_ns.count()/1000000000.;
    std::cout<<"Time from seconds in seconds is " << fp_s.count()<<std::endl;
    std::cout<<"Time from milliseconds in seconds is "<< Zm <<std::endl;
    std::cout<<"Time from nanoseconds in seconds is "<< Zn << std::endl;
//OP's method:
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(Y - V);
    std::cout<<"time: "<<elapsed.count()<<std::endl;
}

顺便说一句，你的方法也有效。请参阅上面的编译器资源管理器链接。