std::literals::chrono_literals::operator""h
From cppreference.com
Defined in header <chrono>
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constexpr std::chrono::hours operator ""h( unsigned long long hrs ); |
(1) | (since C++14) |
constexpr std::chrono::duration</*unspecified*/, std::ratio<3600,1>> operator ""h( long double hrs ); |
(2) | (since C++14) |
Forms a std::chrono::duration literal representing hours.
1) integer literal, returns exactly std::chrono::hours(hrs)
2) floating-point literal, returns a floating-point duration equivalent to std::chrono::hours
Parameters
hrs | - | the number of hours |
Return value
The std::chrono::duration literal.
Possible implementation
constexpr std::chrono::hours operator ""h(unsigned long long h) { return std::chrono::hours(h); } constexpr std::chrono::duration<long double, ratio<3600,1>> operator ""h(long double h) { return std::chrono::duration<long double, std::ratio<3600,1>>(h); } |
Notes
This operator is declared in the namespace std::literals::chrono_literals, where both literals and chrono_literals are inline namespaces. Access to this operator can be gained with:
- using namespace std::literals,
- using namespace std::chrono_literals, or
- using namespace std::literals::chrono_literals.
In addition, within the namespace std::chrono, the directive using namespace literals::chrono_literals; is provided by the standard library, so that if a programmer uses using namespace std::chrono; to gain access to the classes in the chrono library, the corresponding literal operators become visible as well.
Example
Run this code
#include <iostream> #include <chrono> int main() { using namespace std::chrono_literals; auto day = 24h; auto halfhour = 0.5h; std::cout << "one day is " << day.count() << " hours\n" << "half an hour is " << halfhour.count() << " hours\n"; }
Output:
one day is 24 hours half an hour is 0.5 hours
See also
constructs new duration (public member function of std::chrono::duration<Rep,Period> ) |