std::chrono::duration<Rep,Period>::operator=
来自 cppreference.com
duration& operator=( const duration &other ) = default; |
(自 C++11 起) | |
将一个 duration
的内容赋值给另一个。
[编辑] 参数
other | - | 要复制的 duration |
[编辑] 示例
运行此代码
#include <chrono> #include <iostream> int main() { using namespace std::chrono_literals; std::chrono::hours z_hours{}; std::chrono::seconds z_seconds{}; z_hours = 2h; // ok, no conversion needed z_seconds = z_hours; // First, the converting ctor is used to create a temporary object of `lhs`s type. // This ctor implicitly invokes the casting function // chrono::duration_cast<std::seconds>(z_hours). The resulting `rhs` rvalue // has the same type as `lhs`, and the `operator=` finally performs the assignment. std::cout << "hours: " << z_hours.count() << '\n'; std::cout << "seconds: " << z_seconds.count() << '\n'; z_seconds -= 42s; // z_hours = z_seconds; // compile-time error (which is good): incompatible types. // The library avoids the implicit cast to prevent a potential precision loss. z_hours = std::chrono::duration_cast<std::chrono::hours>(z_seconds); // ok z_hours = std::chrono::duration_cast<decltype(z_hours)>(z_seconds); // ditto std::cout << "hours: " << z_hours.count() << '\n'; std::cout << "seconds: " << z_seconds.count() << '\n'; std::chrono::duration<double, std::ratio<3600>> z2_hours{}; z2_hours = z_seconds; // ok, no truncation, implicit cast std::cout << "hours: " << z2_hours.count() << '\n'; }
输出
hours: 2 seconds: 7200 hours: 1 seconds: 7158 hours: 1.98833
[编辑] 另请参阅
构造新的 duration (公共成员函数) |