std::chrono::year_month_day_last::operator+=, std::chrono::year_month_day_last::operator-=
From cppreference.com
< cpp | chrono | year month day last
constexpr std::chrono::year_month_day_last& operator+=(const std::chrono::years& dy) const noexcept; |
(1) | (since C++20) |
constexpr std::chrono::year_month_day_last& operator+=(const std::chrono::months& dm) const noexcept; |
(2) | (since C++20) |
constexpr std::chrono::year_month_day_last& operator-=(const std::chrono::years& dy) const noexcept; |
(3) | (since C++20) |
constexpr std::chrono::year_month_day_last& operator-=(const std::chrono::months& dm) const noexcept; |
(4) | (since C++20) |
Modifies the time point *this
represents by the duration dy
or dm
.
1) Equivalent to *this = *this + dy;
2) Equivalent to *this = *this + dm;
3) Equivalent to *this = *this - dy;
4) Equivalent to *this = *this - dm;
Notes
Durations that are convertible to std::chrono::months, but not std::chrono::years, can be directly added to or subtracted from a year_month_day_last
. Durations convertible to std::chrono::years cannot because such durations are also convertible to std::chrono::months, resulting in an ambiguity:
using namespace std::chrono; using decades = duration<int, std::ratio_multiply<std::ratio<10>, years::period>>; using kilomonths = duration<int, std::ratio_multiply<std::kilo, months::period>>; auto ymdl = 2001y/April/last; ymdl += decades{1}; // error, ambiguous ymdl += kilomonths{1}; // OK
See also
adds or subtracts a year_month_day_last and some number of years or months (public member function) |