std::erf, std::erff, std::erfl
From cppreference.com
Defined in header <cmath> | ||
(1) | ||
float erf (float num ); double erf (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) |
longdouble erfl(longdouble 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> | (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.
| (since C++26) |
A) Additional overloads are provided for all integer types, which are treated as double. | (since C++11) |
Contents |
[edit]Parameters
num | - | floating-point or integer value |
[edit]Return value
If no errors occur, value of the error function of num, that is2 |
√π |
0e-t2
dt, is returned.
If a range error occurs due to underflow, the correct result (after rounding), that is
2*num |
√π |
[edit]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.
[edit]Notes
Underflow is guaranteed if |num|<DBL_MIN*(std::sqrt(π)/2).
erf(x |
σ√2 |
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)).
[edit]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
[edit]See also
(C++11)(C++11)(C++11) | complementary error function (function) |
C documentation for erf |
[edit]External links
Weisstein, Eric W. "Erf." From MathWorld — A Wolfram Web Resource. |