c++ - 5 秒内迭代向量 5 c

Question

如何在 5 秒内迭代向量 5？

例子：

for(int i = 0; i < count; i++) {
    cout << vector[count] << endl;
    //sleep 5 seconds
}

谢谢！

Answer 1

假设向量有 5 个元素：

#include <chrono>
#include <thread> /* introduced with c++11,
                     make sure your compiler is up to date
                     and is set to compile c++ with this version
                     of the STL and standard
                  */

for (int i = 0; i < count; ++i)
{
    cout << vector[count] << endl;
    std::this_thread::sleep_for(std::chrono::seconds(1));
}

您也可以使用Sleep(n)from<windows.h>替代。