std::chrono::year_month_day::operator+=, std::chrono::year_month_day::operator-=
来自 cppreference.cn
constexpr std::chrono::year_month_day& operator+=( const std::chrono::years& dy ) const noexcept; |
(1) | (自 C++20 起) |
constexpr std::chrono::year_month_day& operator+=( const std::chrono::months& dm ) const noexcept; |
(2) | (自 C++20 起) |
constexpr std::chrono::year_month_day& operator-=( const std::chrono::years& dy ) const noexcept; |
(3) | (自 C++20 起) |
constexpr std::chrono::year_month_day& operator-=( const std::chrono::months& dm ) const noexcept; |
(4) | (自 C++20 起) |
通过时长 dy 或 dm 修改 *this 代表的时间点。
1) 等价于 *this = *this + dy;。
2) 等价于 *this = *this + dm;。
3) 等价于 *this = *this - dy;。
4) 等价于 *this = *this - dm;。
对于可以转换为 std::chrono::years 和 std::chrono::months 的时长,如果调用会产生歧义,则优先选择 years
重载 (1,3)。
[编辑] 示例
运行此代码
#include <cassert> #include <chrono> #include <iostream> int main() { constexpr auto monthsInYear{12}; auto ymd{std::chrono::day(1)/std::chrono::July/2020}; std::cout << "#1 " << ymd << '\n'; ymd -= std::chrono::years(10); std::cout << "#2 " << ymd << '\n'; assert(ymd.month() == std::chrono::July); assert(ymd.year() == std::chrono::year(2010)); ymd += std::chrono::months(10 * monthsInYear + 11); std::cout << "#3 " << ymd << '\n'; assert(ymd.month() == std::chrono::month(6)); assert(ymd.year() == std::chrono::year(2021)); // Handling the ymd += months "overflow" case. ymd = std::chrono::May/31/2021; // ok std::cout << "#4 " << ymd << '\n'; assert(ymd.ok()); ymd += std::chrono::months{1}; // bad date: June has only 30 days std::cout << "#5 " << ymd << '\n'; assert(not ymd.ok()); assert(ymd == std::chrono::June/31/2021); // Snap to the last day of the month, June 30: const auto ymd1 = ymd.year()/ymd.month()/std::chrono::last; std::cout << "#6 " << ymd1 << '\n'; assert(ymd1.ok()); assert(ymd1 == std::chrono::June/30/2021); // Overflow into the next month, July 1 (via converting to/from sys_days): const std::chrono::year_month_day ymd2 = std::chrono::sys_days{ymd}; std::cout << "#7 " << ymd2 << '\n'; assert(ymd2.ok()); assert(ymd2 == std::chrono::July/1/2021); }
输出
#1 2020-07-01 #2 2010-07-01 #3 2021-06-01 #4 2021-05-31 #5 2021-06-31 is not a valid date #6 2021/Jun/last #7 2021-07-01
[编辑] 参见
(C++20) |
添加或减去 year_month_day 以及若干年或月(函数) |