std::literals::chrono_literals::operator""us
From cppreference.com
| Defined in header <chrono>
|
||
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.
Parameters
| us | - | the number of microseconds |
Return value
The std::chrono::duration literal.
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);
}
|
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, orusing 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 <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
See also
| constructs new duration (public member function of std::chrono::duration<Rep,Period>)
|