Question

我想确定哪个是用于某些操作的更好的容器，因此我必须检查向量与列表的运行时间。

在线参考对于此实现而言非常模糊或过于复杂。

任何帮助将不胜感激

Answer 1

如果您有权访问C++11编译器，

#include <chrono>
#include <iostream>

int main()
{
    std::chrono::steady_clock::time_point _start(std::chrono::steady_clock::now());
    // code you want to time here
    std::chrono::steady_clock::time_point _end(std::chrono::steady_clock::now());

    std::cout << std::chrono::duration_cast<std::chrono::duration<double>>(
                    _end - _start).count(); // in seconds, read more about std::chrono
}

Answer 2

这就是我使用它的方式：

#include <iostream>
#include <chrono>

    auto start = std::chrono::high_resolution_clock::now();
    ... code to measure here ... 
    auto diff = std::chrono::high_resolution_clock::now() - start;
    auto t1 = std::chrono::duration_cast<std::chrono::nanoseconds>(diff);
    std::cout << "Loop1: " << t1.count() << std::endl;

您需要使用g++ -std=c++11编译它，并且只要您有一个可用的C ++ 11编译器和库，它就应该可以工作。

Answer 3

这是我目前的代码：

#include <iostream>
#include <list>
#include <vector>
#include <chrono>
#include <stdlib.h>
#include <ctime>

using namespace std;

template <typename Container>
void fill_container(Container& c, int n)
{
int v;

for(int i =0; i<n; i++)
{
    v= rand() %12;
    c.push_back(v);
}

}

template<typename Container>
void print(Container& c, ostream& out =cout)
{
typename Container::iterator itr = c.begin();
cout << "now" ;
while(itr!=c.end())
{
    out << *itr << endl;
    itr++;
}

}

template <typename Container>
void removeOdd(Container& c)
{
 typename Container::iterator itr2 = c.begin();
 while(itr2!= c.end())
 {
        cout << *itr2 << endl;
         if(*itr2 %2 != 0 )
        {
            itr2 = c.erase(itr2);
        }
        itr2++;
 }

 }

int main()
{





//Timer ends
std::chrono::steady_clock::time_point _start(std::chrono::steady_clock::now());
// code you want to time here
vector<int> damn;
fill_container(damn, 20);
removeOdd(damn);
print(damn);
std::chrono::steady_clock::time_point _end(std::chrono::steady_clock::now());

std:: cout << std::chrono::duration_cast<std::chrono::duration<double>>( _end - _start).count();



return 0;
}

如何使用chrono来确定运行时

3 个答案: