std::erfc, std::erfcf, std::erfcl
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Defined in header
<cmath>
(1)
float erfc ( float num );
(until C++23)
double erfc ( double num );
/*floating-point-type*/
erfc ( /*floating-point-type*/ num );
(since C++23) erfc ( /*floating-point-type*/ num );
(constexpr since C++26)
float erfcf( float num );
(2)
(since C++11) (constexpr since C++26)
long double erfcl( 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 >
(S)
(since C++26)
constexpr /*deduced-simd-t*/<V>
Additional overloads (since C++11)
Defined in header
<cmath>
template< class Integer >
double erfc ( Integer num );
(A)
(constexpr since C++26)
double erfc ( Integer num );
1-3) Computes the complementary error function of num, that is 1.0 - std::erf (num), but without loss of precision for large num. The library provides overloads of
std::erfc 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::erfc on v_num.- (See math-floating-point and deduced-simd-t for their definitions.)
A) Additional overloads are provided for all integer types, which are treated as double.
(since C++11)[edit] Parameters
num
-
floating-point or integer value
[edit] Return value
If no errors occur, value of the complementary error function of num, that is \(\frac{2}{\sqrt{\pi} }\int_{num}^{\infty}{e^{-{t^2} }\mathsf{d}t}\) 2
√π
∫∞nume-t2
dt or \({\small 1-\operatorname{erf}(num)}\)1-erf(num), is returned.
If a range error occurs due to underflow, the correct result (after rounding) is returned.
[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 is returned.
- If the argument is -∞, 2 is returned.
- If the argument is NaN, NaN is returned.
[edit] Notes
For the IEEE-compatible type double, underflow is guaranteed if num > 26.55.
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::erfc(num) has the same effect as std::erfc(static_cast<double>(num)).
[edit] Example
Run this code
#include <cmath> #include <iomanip> #include <iostream> double normalCDF(double x) // Phi(-∞, x) aka N(x) { return std::erfc(-x / std::sqrt (2)) / 2; } int main() { std::cout << "normal cumulative distribution function:\n" << std::fixed << std::setprecision (2); for (double n = 0; n < 1; n += 0.1) std::cout << "normalCDF(" << n << ") = " << 100 * normalCDF(n) << "%\n"; std::cout << "special values:\n" << "erfc(-Inf) = " << std::erfc(-INFINITY ) << '\n' << "erfc(Inf) = " << std::erfc(INFINITY ) << '\n'; }
Output:
normal cumulative distribution function: normalCDF(0.00) = 50.00% normalCDF(0.10) = 53.98% normalCDF(0.20) = 57.93% normalCDF(0.30) = 61.79% normalCDF(0.40) = 65.54% normalCDF(0.50) = 69.15% normalCDF(0.60) = 72.57% normalCDF(0.70) = 75.80% normalCDF(0.80) = 78.81% normalCDF(0.90) = 81.59% normalCDF(1.00) = 84.13% special values: erfc(-Inf) = 2.00 erfc(Inf) = 0.00
[edit] See also
C documentation for erfc
[edit] External links
Weisstein, Eric W. "Erfc." From MathWorld — A Wolfram Web Resource.