c++ - 如果存在从 `double`到 `T`的转换，SFINAE禁用构造函数

Question

以下是我可以提出的最有意义的最小程序，以重现该问题的困境。由于LinearForm<double>的构造函数之间的冲突，程序无法编译。要解决此冲突，我希望仅当不存在从LinearForm<V>::LinearForm(double)到double的转换时才启用V。我怎样才能做到这一点？ (并且它将解决构造函数之间的冲突吗？)

#include <type_traits>
#include <array>

template<int N>
struct Vector{
    std::array<double,N> coords;

    Vector(std::array<double,N> coords) : coords(coords) {}

    // implicit conversions between scalar and Vector<1>
    template<int Nd = N, std::enable_if_t<Nd==1>>
    Vector(double scalar) : coords(scalar) {}
    template<int Nd = N, std::enable_if_t<Nd==1>>
    operator double() const {return coords[0];}

    double dot(Vector<N> u) const {
        double acc = 0;
        for(int i=0; i<N; i++){
            acc += coords[i]*u.coords[i];
        }
        return acc;
    }

    static Vector<N> zero(){ return Vector<N>(std::array<double,N>{}); }
};

template<typename V> // V is domain element type i.e. LinearForm maps from V to double
struct LinearForm {
    V v;
    LinearForm(V v) : v(v) {}

    //template<typename Vd=V, typename = std::enable_if_t</* WHAT TO PUT IN HERE */>>
    LinearForm(double v) : LinearForm(V::zero())
    {
        if(v != 0){
            throw std::runtime_error("LinearForm cannot be non-zero constant.");
        }
    }
    double operator()(V u){return u.dot(v);}
};

int main()
{
    LinearForm<Vector<2>> lf(Vector<2>({1,2}));
    LinearForm<Vector<2>> zf = 0;

    LinearForm<double> slf = 0;

    auto u = Vector<2>({3,4});
    lf(u); // returns some value
    zf(u); // should return zero for any u

    return 0;
}

Answer 1

您可能要使用 std::is_convertible 。然后，这将完成工作:

template <typename Vd=V,
          typename std::enable_if_t<std::is_convertible<double, Vd>::value>::type* = nullptr>
LinearForm(double v) : LinearForm(V::zero())
{
    if(v != 0){
        throw std::runtime_error("LinearForm cannot be non-zero constant.");
    }
}

要编译您的代码，我需要再添加两点:

LinearForm(int value) : v(value) {}

并修改

template<int Nd = N, std::enable_if_t<Nd==1>>
Vector(double scalar) : coords(scalar) {}

至

template<int Nd = N>
Vector(double scalar) : coords({scalar}) {}

Live example