std::imag(std::complex)
From cppreference.com
Defined in header <complex>
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(1) | ||
template< class T > T imag( const std::complex<T>& z ); |
(until C++14) | |
template< class T > constexpr T imag( const std::complex<T>& z ); |
(since C++14) | |
Additional overloads (since C++11) |
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Defined in header <complex>
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(A) | ||
float imag( float f ); double imag( double f ); |
(until C++14) | |
constexpr float imag( float f ); constexpr double imag( double f ); |
(since C++14) (until C++23) |
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template< class FloatingPoint > FloatingPoint imag( FloatingPoint f ); |
(since C++23) | |
(B) | ||
template< class Integer > double imag( Integer i ); |
(until C++14) | |
template< class Integer > constexpr double imag( Integer i ); |
(since C++14) | |
1) Returns the imaginary part of the complex number z, i.e. z.imag().
A,B) Additional overloads are provided for all integer and floating-point types, which are treated as complex numbers with zero imaginary part.
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(since C++11) |
Parameters
z | - | complex value |
f | - | floating-point value |
i | - | integer value |
Return value
1) The imaginary part of z.
A) decltype(f){} (zero).
B) 0.0.
Notes
The additional overloads are not required to be provided exactly as (A,B). They only need to be sufficient to ensure that for their argument num:
- If num has a standard (until C++23) floating-point type
T
, then std::imag(num) has the same effect as std::imag(std::complex<T>(num)). - Otherwise, if num has an integer type, then std::imag(num) has the same effect as std::imag(std::complex<double>(num)).
See also
accesses the imaginary part of the complex number (public member function) | |
returns the real part (function template) |