Cumulative Normal Distribution Function in C/C++

Boost is as good as the standard :D here you go: boost maths/statistical.

Here's a stand-alone C++ implementation of the cumulative normal distribution in 14 lines of code.

http://www.johndcook.com/cpp_phi.html

#include <cmath>

double phi(double x)
{
    // constants
    double a1 =  0.254829592;
    double a2 = -0.284496736;
    double a3 =  1.421413741;
    double a4 = -1.453152027;
    double a5 =  1.061405429;
    double p  =  0.3275911;

    // Save the sign of x
    int sign = 1;
    if (x < 0)
        sign = -1;
    x = fabs(x)/sqrt(2.0);

    // A&S formula 7.1.26
    double t = 1.0/(1.0 + p*x);
    double y = 1.0 - (((((a5*t + a4)*t) + a3)*t + a2)*t + a1)*t*exp(-x*x);

    return 0.5*(1.0 + sign*y);
}

void testPhi()
{
    // Select a few input values
    double x[] = 
    {
        -3, 
        -1, 
        0.0, 
        0.5, 
        2.1 
    };

    // Output computed by Mathematica
    // y = Phi[x]
    double y[] = 
    { 
        0.00134989803163, 
        0.158655253931, 
        0.5, 
        0.691462461274, 
        0.982135579437 
    };

        int numTests = sizeof(x)/sizeof(double);

    double maxError = 0.0;
    for (int i = 0; i < numTests; ++i)
    {
        double error = fabs(y[i] - phi(x[i]));
        if (error > maxError)
            maxError = error;
    }

        std::cout << "Maximum error: " << maxError << "\n";
}

I figured out how to do it using gsl, at the suggestion of the folks who answered before me, but then found a non-library solution (hopefully this helps many people out there who are looking for it like I was):

#ifndef Pi 
#define Pi 3.141592653589793238462643 
#endif 

double cnd_manual(double x)
{
  double L, K, w ;
  /* constants */
  double const a1 = 0.31938153, a2 = -0.356563782, a3 = 1.781477937;
  double const a4 = -1.821255978, a5 = 1.330274429;

  L = fabs(x);
  K = 1.0 / (1.0 + 0.2316419 * L);
  w = 1.0 - 1.0 / sqrt(2 * Pi) * exp(-L *L / 2) * (a1 * K + a2 * K *K + a3 * pow(K,3) + a4 * pow(K,4) + a5 * pow(K,5));

  if (x < 0 ){
    w= 1.0 - w;
  }
  return w;
}

Theres is no straight function. But since the gaussian error function and its complementary function is related to the normal cumulative distribution function (see here, or here) we can use the implemented c-function erfc (complementary error function):

double normalCDF(double value)
{
   return 0.5 * erfc(-value * M_SQRT1_2);
}

Which considers the relation of erfc(x) = 1-erf(x) with M_SQRT1_2 = √0,5.

I use it for statistical calculations and it works great. No need for using coefficients.