std::literals::chrono_literals::operator""h
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std::chrono::duration
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operator""h
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Defined in header
<chrono>
constexpr std::chrono::hours
operator""h( unsigned long long hrs );
(1)
(since C++14)
operator""h( unsigned long long hrs );
constexpr std::chrono::duration </*unspecified*/, std::ratio <3600,1>>
operator""h( long double hrs );
(2)
(since C++14)
operator""h( long double hrs );
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 .
[edit] Parameters
hrs
-
the number of hours
[edit] Return value
The std::chrono::duration literal.
[edit] 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); }
[edit] 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.
[edit] Example
Run this code
#include <chrono> #include <iostream> int main() { using namespace std::chrono_literals; auto day = 24h; auto halfhour = 0.5h; std::cout << "one day is " << day.count() << " hours (" << day << ")\n" << "half an hour is " << halfhour.count() << " hours (" << halfhour << ")\n"; }
Output:
one day is 24 hours (24h) half an hour is 0.5 hours (0.5h)