std::literals::chrono_literals::operator""ms
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std::chrono::duration
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operator""ms
(C++14)
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(C++20)
Helper classes
Defined in header
<chrono>
constexpr std::chrono::milliseconds
operator""ms( unsigned long long ms );
(1)
(since C++14)
operator""ms( unsigned long long ms );
constexpr std::chrono::duration </*unspecified*/, std::milli >
operator""ms( long double ms );
(2)
(since C++14)
operator""ms( long double ms );
Forms a std::chrono::duration literal representing milliseconds.
1) Integer literal, returns exactly std::chrono::milliseconds (ms).
2) Floating-point literal, returns a floating-point duration equivalent to std::chrono::milliseconds .
[edit] Parameters
ms
-
the number of milliseconds
[edit] Return value
The std::chrono::duration literal.
[edit] Possible implementation
constexpr std::chrono::milliseconds operator""ms(unsigned long long ms) { return std::chrono::milliseconds (ms); } constexpr std::chrono::duration <long double, std::milli > operator""ms(long double ms) { return std::chrono::duration <long double, std::milli >(ms); }
[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; constexpr auto d1{250ms}; constexpr std::chrono::milliseconds d2{1s}; std::cout << d1 << " = " << d1.count() << " milliseconds\n" << d2 << " = " << d2.count() << " milliseconds\n"; }
Output:
250ms = 250 milliseconds 1000ms = 1000 milliseconds