std::chrono::year_month_day::operator+=, std::chrono::year_month_day::operator-=

 

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C 样式日期和时间

 

std::chrono::year_month_day

成员函数

year_month_day::year_month_day

year_month_day::operator+=year_month_day::operator-=

year_month_day::yearyear_month_day::monthyear_month_day::day

year_month_day::operator sys_daysyear_month_day::operator local_days

year_month_day::ok

非成员函数

operator==operator<=>

operator+operator-

operator<<

from_stream

辅助类

formatter<std::chrono::year_month_day>

hash<std::chrono::year_month_day> (C++26)

 

constexpr std::chrono::year_month_day&     operator+=( const std::chrono::years& dy ) const noexcept;	(1)	(自 C++20)
constexpr std::chrono::year_month_day&     operator+=( const std::chrono::months& dm ) const noexcept;	(2)	(自 C++20)
constexpr std::chrono::year_month_day&     operator-=( const std::chrono::years& dy ) const noexcept;	(3)	(自 C++20)
constexpr std::chrono::year_month_day&     operator-=( const std::chrono::months& dm ) const noexcept;	(4)	(自 C++20)

通过持续时间 *this 表示的时间点 dy 或 dm 进行修改。

1) 等同于 *this = *this + dy;.

2) 等同于 *this = *this + dm;.

3) 等同于 *this = *this - dy;.

4) 等同于 *this = *this - dm;.

对于可以转换为 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，则首选 years 重载 (1,3)。

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

另请参阅

operator+operator- (C++20)

添加或减去 year_month_day 和一些年数或月数 (函数)