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std::imag(std::complex)

From cppreference.com
< cpp‎ | numeric‎ | complex
 
 
 
 
Defined in header <complex>
(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)
Defined in header <complex>
(A)
float       imag( float f );

double      imag( double f );

long double imag( long double f );
(until C++14)
constexpr float       imag( float f );

constexpr double      imag( double f );

constexpr long double imag( long double f );
(since C++14)
(until C++23)
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.
(since C++11)

Contents

[edit] Parameters

z - complex value
f - floating-point value
i - integer value

[edit] Return value

1) The imaginary part of z.
A) decltype(f){} (zero).
B) 0.0.

[edit] 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)).

[edit] See also

accesses the imaginary part of the complex number
(public member function) [edit]
returns the real part
(function template) [edit]
C documentation for cimag