std::norm(std::complex)
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std::complex 
  
 
 
 
 
 
 
(until C++20)  
(C++26)
  (C++26)
(C++26)
Defined in header 
 
 
<complex> 
  
 (1)
 
template< class T > 
T norm( const std::complex <T>& z );
 
 (until C++20)
T norm( const std::complex <T>& z );
template< class T > 
constexpr T norm( const std::complex <T>& z );
 
 (since C++20) 
constexpr T norm( const std::complex <T>& z );
Additional overloads (since C++11)
 
 
Defined in header 
 
 
<complex> 
  
 (A)
 
float       norm( float f );
 
 (until C++20)
double      norm( double f );
constexpr float       norm( float f );
 
 (since C++20) constexpr double      norm( double f );
(until C++23)
template< class FloatingPoint >
constexpr FloatingPoint norm( FloatingPoint f );
 
 (since C++23) 
constexpr FloatingPoint norm( FloatingPoint f );
 
 (B)
 
template< class Integer > 
double norm( Integer i );
 
 (until C++20)
double norm( Integer i );
template< class Integer > 
constexpr double norm( Integer i );
 
 (since C++20) 
constexpr double norm( Integer i );
1) Returns the squared magnitude of the complex number z.
A,B) Additional overloads are provided for all integer and floating-point types, which are treated as complex numbers with zero imaginary component.
 (since C++11)Contents
[edit] Parameters
 z
 -
 complex value
 f
 -
 floating-point value
 i
 -
 integer value
[edit] Return value
1) The squared magnitude of z.
A) The square of f.
B) The square of i.
[edit] Notes
The norm calculated by this function is also known as field norm or absolute square.
The Euclidean norm of a complex number is provided by std::abs, which is more costly to compute. In some situations, it may be replaced by std::norm, for example, if abs(z1) > abs(z2) then norm(z1) > norm(z2).
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::norm(num) has the same effect as std::norm(std::complex <T>(num)).
- Otherwise, if num has an integer type, then std::norm(num) has the same effect as std::norm(std::complex <double>(num)).
[edit] Example
Run this code
#include <cassert> #include <complex> #include <iostream> int main() { constexpr std::complex <double> z {3.0, 4.0}; static_assert(std::norm(z) == (z.real() * z.real() + z.imag() * z.imag())); static_assert(std::norm(z) == (z * std::conj (z))); assert (std::norm(z) == (std::abs(z) * std::abs(z))); std::cout << "std::norm(" << z << ") = " << std::norm(z) << '\n'; }
Output:
std::norm((3,4)) = 25