std::literals::chrono_literals::operator""us
From cppreference.com
Defined in header <chrono>
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constexpr std::chrono::microseconds operator""us( unsigned long long us ); |
(1) | (since C++14) |
constexpr std::chrono::duration</*unspecified*/, std::micro> operator""us( long double us ); |
(2) | (since C++14) |
Forms a std::chrono::duration literal representing microseconds.
1) Integer literal, returns exactly std::chrono::microseconds(us).
2) Floating-point literal, returns a floating-point duration equivalent to std::chrono::microseconds.
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[edit] Parameters
us | - | the number of microseconds |
[edit] Return value
The std::chrono::duration literal.
[edit] Possible implementation
constexpr std::chrono::microseconds operator""us(unsigned long long us) { return std::chrono::microseconds(us); } constexpr std::chrono::duration<long double, std::micro> operator""us(long double us) { return std::chrono::duration<long double, std::micro>(us); } |
[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 d1 = 250us; std::chrono::microseconds d2 = 1ms; std::cout << d1 << " = " << d1.count() << " microseconds\n" << 1ms << " = " << d2.count() << " microseconds\n"; }
Output:
250us = 250 microseconds 1ms = 1000 microseconds
[edit] See also
constructs new duration (public member function of std::chrono::duration<Rep,Period> )
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