# std::chrono::year_month_day::operator sys_days, std::chrono::year_month_day::operator local_days

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< cpp‎ | chrono‎ | year month day

constexpr operator std::chrono::sys_days () const noexcept; | (1) | (since C++20)
explicit constexpr operator std::chrono::local_days () const noexcept; | (2) | (since C++20)

Converts `*this` to a std::chrono::time_point representing the same date as this `year_month_day`.

1) If ok() is true, the return value holds a count of days from the std::chrono::system_clock epoch (1970-01-01) to `*this`. The result is negative if `*this` represent a date prior to it.
Otherwise, if the stored year and month are valid (year().ok() && month().ok() is true), then the returned value is sys_days(year()/month()/1d) + (day() - 1d).
Otherwise (if year().ok() && month().ok() is false), the return value is unspecified.
A `sys_days` in the range [ std::chrono::days {-12687428}, std::chrono::days {11248737}], when converted to `year_month_day` and back, yields the same value.
2) Same as (1) but returns `local_days` instead. Equivalent to return local_days(sys_days(*this).time_since_epoch());.

### Notes

Converting to `sys_days` and back can be used to normalize a `year_month_day` that contains an invalid day but a valid year and month:

    using namespace std::chrono;
    auto ymd = 2017y/January/0;
    ymd = sys_days{ymd};
    // ymd is now 2016y/December/31

Normalizing the year and month can be done by adding (or subtracting) zero std::chrono::months:

    using namespace std::chrono;
    constexpr year_month_day normalize(year_month_day ymd){
        ymd += months{0}; // normalizes year and month
        return sys_days{ymd}; // normalizes day
    }
    static_assert(normalize(2017y/33/59) == 2019y/10/29);

### Example

| This section is incomplete Reason: no example

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