Scroll to navigation

std::erf,std::erff,std::erfl(3) C++ Standard Libary std::erf,std::erff,std::erfl(3)

NAME

std::erf,std::erff,std::erfl - std::erf,std::erff,std::erfl

Synopsis


Defined in header <cmath>
float erf ( float num );


double erf ( double num ); (until C++23)


long double erf ( long double num );
/* floating-point-type */ (since C++23)
erf ( /* floating-point-type */ num ); (constexpr since C++26)
float erff( float num ); (1) (2) (since C++11)
(constexpr since C++26)
long double erfl( long double num ); (3) (since C++11)
(constexpr since C++26)
Additional overloads (since C++11)
Defined in header <cmath>
template< class Integer > (A) (constexpr since C++26)
double erf ( Integer num );


1-3) Computes the error function of num.
The library provides overloads of std::erf for all cv-unqualified floating-point
types as the type of the parameter.
(since C++23)


A) Additional overloads are provided for all integer types, which are (since C++11)
treated as double.

Parameters


num - floating-point or integer value

Return value


If no errors occur, value of the error function of num, that is
\(\frac{2}{\sqrt{\pi} }\int_{0}^{num}{e^{-{t^2} }\mathsf{d}t}\)


2

π


∫num
0e^-t2
dt, is returned.
If a range error occurs due to underflow, the correct result (after rounding), that
is \(\frac{2\cdot num}{\sqrt{\pi} }\)


2*num

π


is returned.

Error handling


Errors are reported as specified in math_errhandling.


If the implementation supports IEEE floating-point arithmetic (IEC 60559),


* If the argument is ±0, ±0 is returned.
* If the argument is ±∞, ±1 is returned.
* If the argument is NaN, NaN is returned.

Notes


Underflow is guaranteed if |num| < DBL_MIN * (std::sqrt(π) / 2).


\(\operatorname{erf}(\frac{x}{\sigma \sqrt{2} })\)erf(


x
σ

2


) is the probability that a measurement whose errors are subject to a normal
distribution with standard deviation \(\sigma\)σ is less than \(x\)x away from the
mean value.


The additional overloads are not required to be provided exactly as (A). They only
need to be sufficient to ensure that for their argument num of integer type,
std::erf(num) has the same effect as std::erf(static_cast<double>(num)).

Example


The following example calculates the probability that a normal variate is on the
interval (x1, x2):

// Run this code


#include <cmath>
#include <iomanip>
#include <iostream>


double phi(double x1, double x2)
{
return (std::erf(x2 / std::sqrt(2)) - std::erf(x1 / std::sqrt(2))) / 2;
}


int main()
{
std::cout << "Normal variate probabilities:\n"
<< std::fixed << std::setprecision(2);
for (int n = -4; n < 4; ++n)
std::cout << '[' << std::setw(2) << n
<< ':' << std::setw(2) << n + 1 << "]: "
<< std::setw(5) << 100 * phi(n, n + 1) << "%\n";


std::cout << "Special values:\n"
<< "erf(-0) = " << std::erf(-0.0) << '\n'
<< "erf(Inf) = " << std::erf(INFINITY) << '\n';
}

Output:


Normal variate probabilities:
[-4:-3]: 0.13%
[-3:-2]: 2.14%
[-2:-1]: 13.59%
[-1: 0]: 34.13%
[ 0: 1]: 34.13%
[ 1: 2]: 13.59%
[ 2: 3]: 2.14%
[ 3: 4]: 0.13%
Special values:
erf(-0) = -0.00
erf(Inf) = 1.00

See also


erfc
erfcf
erfcl complementary error function
(C++11) (function)
(C++11)
(C++11)
C documentation for
erf

External links


Weisstein, Eric W. "Erf." From MathWorld — A Wolfram Web Resource.

2024.06.10 http://cppreference.com