std::chrono::year_month_weekday::operator+=, std::chrono::year_month_weekday::operator-=
来自 cppreference.com
< cpp | chrono | year month weekday
constexpr std::chrono::year_month_weekday& operator+=( const std::chrono::years& dy ) const noexcept; |
(1) | (自 C++20 起) |
constexpr std::chrono::year_month_weekday& operator+=( const std::chrono::months& dm ) const noexcept; |
(2) | (自 C++20 起) |
constexpr std::chrono::year_month_weekday& operator-=( const std::chrono::years& dy ) const noexcept; |
(3) | (自 C++20 起) |
constexpr std::chrono::year_month_weekday& operator-=( const std::chrono::months& dm ) const noexcept; |
(4) | (自 C++20 起) |
修改时间点 *this 表示的持续时间 dy 或 dm.
1) 等效于 *this = *this + dy;.
2) 等效于 *this = *this + dm;.
3) 等效于 *this = *this - dy;.
4) 等效于 *this = *this - dm;.
对于可转换为 std::chrono::years 和 std::chrono::months 的持续时间,如果调用在其他情况下会导致歧义,则 years
重载 (1,3) 优先。
[编辑] 示例
运行此代码
#include <cassert> #include <chrono> #include <iostream> int main() { auto ymwi{1/std::chrono::Wednesday[2]/2021}; std::cout << ymwi << '\n'; ymwi += std::chrono::years(5); std::cout << ymwi << '\n'; assert(static_cast<std::chrono::year_month_day>(ymwi) == std::chrono::year(2026)/1/14); ymwi -= std::chrono::months(1); std::cout << ymwi << '\n'; assert(static_cast<std::chrono::year_month_day>(ymwi) == std::chrono::day(10)/12/2025); }
输出
2021/Jan/Wed[2] 2026/Jan/Wed[2] 2025/Dec/Wed[2]
[编辑] 另请参阅
(C++20) |
添加或减去 year_month_weekday 和一些年或月(函数) |