SIMEAE禁用构造函数,如果转换从“双”到“T”存在

下面是我能想到的最有意义的程序来重现我在这个问题上的困境。由于的构造函数之间的冲突,程序无法编译 LinearForm<double> . 为了解决这场冲突,我想 LinearForm<V>::LinearForm(double) 当且仅当不存在转换时 double 到 V . 我该怎么做?(它会解决构造器之间的冲突吗?)

#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;
}