Question

因此，在Chrono图书馆的c++11中提供了duration_cast：

计算是在最广泛的类型中完成的，并且只有在完成时才会转换为结果类型，就好像通过static_cast一样转换

c++17的floor：

返回ToDuration中可表示的最大持续时间t，该持续时间小于或等于d

因此，对于所有x，这两个调用的结果将是相等的：

chrono::duration_cast<chrono::seconds>(x)
chrono::floor<chrono::seconds>(x)

Answer 1

据我所知，与static_cast和std::floor之间的差异相同：负数被舍入而不是截断为零。

#include <iostream>
#include <chrono>
using namespace std::chrono_literals;

int main() {
    std::cout << "duration_cast:" << std::endl;
    std::cout << "1.4s: " << std::chrono::duration_cast<std::chrono::seconds>(1400ms).count() << std::endl;
    std::cout << "1.5s: " << std::chrono::duration_cast<std::chrono::seconds>(1500ms).count() << std::endl;
    std::cout << "1.6s: " << std::chrono::duration_cast<std::chrono::seconds>(1600ms).count() << std::endl;
    std::cout << "-1.4s: " << std::chrono::duration_cast<std::chrono::seconds>(-1400ms).count() << std::endl;
    std::cout << "-1.5s: " << std::chrono::duration_cast<std::chrono::seconds>(-1500ms).count() << std::endl;
    std::cout << "-1.6s: " << std::chrono::duration_cast<std::chrono::seconds>(-1600ms).count() << std::endl;

    std::cout << "floor:" << std::endl;
    std::cout << "1.4s: " << std::chrono::floor<std::chrono::seconds>(1400ms).count() << std::endl;
    std::cout << "1.5s: " << std::chrono::floor<std::chrono::seconds>(1500ms).count() << std::endl;
    std::cout << "1.6s: " << std::chrono::floor<std::chrono::seconds>(1600ms).count() << std::endl;
    std::cout << "-1.4s: " << std::chrono::floor<std::chrono::seconds>(-1400ms).count() << std::endl;
    std::cout << "-1.5s: " << std::chrono::floor<std::chrono::seconds>(-1500ms).count() << std::endl;
    std::cout << "-1.6s: " << std::chrono::floor<std::chrono::seconds>(-1600ms).count() << std::endl;
    return 0;
}

duration_cast:
1.4s: 1
1.5s: 1
1.6s: 1
-1.4s: -1
-1.5s: -1
-1.6s: -1
floor:
1.4s: 1
1.5s: 1
1.6s: 1
-1.4s: -2
-1.5s: -2
-1.6s: -2

https://wandbox.org/permlink/SsmpRz6RkvbL6Sru

floor和duration_cast之间有什么区别？

1 个答案: