c++ - 仅为类的特定模板实例化声明成员函数

Question

是否可以仅为类的特定模板实例声明成员函数？这就是我想要这样做的原因:

// Polynomial<N> is a polynomial of degree N
template<int N>
class Polynomial {
public:
    //... various shared methods e.g...
    double eval(double x) const;
    Polynomial<N-1> derivative() const;
    Polynomial<N+1> integralFrom(double x0) const;
    // ... various shared operators etc.

    double zero() const; // only want Polynomial<1> to support this
    // only want Polynomial<2> and Polynomial<1> to support the following
    //     because the solutions rapidly become too difficult to implement
    std::vector<double> zeros() const;
    std::vector<double> stationaryPoints() const { return derivative().zeros();}

private:
    std::array<double,2> coeffs;
}

我目前的解决方法是从 Polynomial<N>::zeros() 中抛出异常对于 N>2但如果能在编译时检测到问题就好了。

Answer 1

您还可以使用 std::enable_if 使 SFINAE 远离零函数。

template< int I >
class Poly {

public:
    template<int Ib = I, typename = std::enable_if_t<Ib == 1> > 
    double zero() 
    {
        return 42;
    }
};

int main()
{
    Poly< 10 > does_not_compile;
    does_not_compile.zero();

    //Poly< 1 >  compile;
    //compile.zero();
}