在C++中使用带有模板参数的标准数学函数

Question

我写了代码:

const double PI = 3.141592653589793;

namespace 
{
    const int N = 8;
    const double points[8] = {-0.9602898564975363, -0.7966664774136267, -0.5255324099163290, -0.1834346424956498,
        0.1834346424956498, 0.5255324099163290, 0.7966664774136267, 0.9602898564975363};
    const double weights[8] = {0.1012285362903706, 0.2223810344533744, 0.3137066458778874, 0.3626837833783621, 0.3626837833783621,
        0.3137066458778874, 0.2223810344533744, 0.1012285362903706};
    const double error = 1e-10;

    template <class TFunction, class TNumber>
    class ChangeOfVariables
    {
        TFunction f;
    public:
        ChangeOfVariables(TFunction f) : f(f){}
        TNumber operator() (TNumber x) { TNumber c=std::cos(x); return f(std::tan(x))/(c*c); } 
    };
}


class Quadrature
{
public:
    Quadrature(void) {};
    ~Quadrature(void) {};

    template <class TFunction, class TNumber>
    TNumber integrate(TFunction f, TNumber a, TNumber b) 
    {
        TNumber result = 0.0;
        for(int i=0; i<N; i++) 
        {
            result+= weights[i]*f(0.5*((b-a)*points[i]+(a+b)));
        }
        return 0.5*(b-a)*result;
    }

    template <class TFunction, class TNumber>
    TNumber integrateToInfty(TFunction f, TNumber a)
    {
        using std::atan; 
        return integrate<ChangeOfVariables<TFunction,TNumber>, TNumber>(ChangeOfVariables<TFunction,TNumber>(f), atan(a), PI/2); 
    }
};

在这里使用:

double dampendedExp(double x) {return std::exp(-2.0*x);}
int main()
{
    Quadrature quadrature;
    cout<<"e^(-2x)="<<quadrature.integrateToInfty(dampendedExp, 0)<<endl;
}

但编译器抱怨:

Error   1   error C2668: 'atan' : ambiguous call to overloaded function c:\users\ga1009\documents\dev\fouriertransform\fouriertransform\quadrature.h    48

当时的想法是,使之成为不同数量的类型,例如工作double和complex<double>,其中反正切定义.我该如何解决？

Answer 1

规范的解决方案是using std::atan; return integrate(atan(a), PI/2);. 这为您提供了在 的命名空间中进行参数依赖查找TFunction。