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

Defined in header `<cmath>`
	(1)
float erf ( float num ); double erf ( double num ); long double erf ( long double num );			(until C++23)
/floating-point-type/ erf ( /floating-point-type/ num );			(since C++23) (constexpr since C++26)
float erff( float num );		(2)	(since C++11) (constexpr since C++26)
long double erfl( long double num );		(3)	(since C++11) (constexpr since C++26)
SIMD overload (since C++26)
Defined in header `<simd>`
template< /math-floating-point/ V > constexpr /deduced-simd-t/<V> erf ( const V& v_num );		(S)	(since C++26)
Additional overloads (since C++11)
Defined in header `<cmath>`
template< class Integer > double erf ( Integer num );		(A)	(constexpr since C++26)

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)

S) The SIMD overload performs an element-wise std::erf on v_num.

(See math-floating-point and deduced-simd-t for their definitions.)

(since C++26)

A) Additional overloads are provided for all integer types, which are treated as double.

(since C++11)

Parameters

num

-

floating-point or integer value

Return value

If no errors occur, value of the error function of num, that is

2 \sqrt π \int num 0 e -t 2 d t

, is returned.
If a range error occurs due to underflow, the correct result (after rounding), that is

2*num \sqrt 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).

erf(x σ \sqrt 2)

is the probability that a measurement whose errors are subject to a normal distribution with standard deviation

σ

is less than

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

External links

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

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