为什么编译器说:'enable_if'不能用于禁用此声明 [英] why compiler said: 'enable_if' cannot be used to disable this declaration

问题描述

template <bool Cond, typename Type = void>
using Enable_if = typename std::enable_if<Cond, Type>::type;

class Degree;

template <typename T>
constexpr inline bool Is_Degree() {
    return std::is_base_of<Degree, T>::value;
}

class Degree {
public:
    std::size_t inDeg = 0;
};

template <typename Satellite = Degree>
class Vertex: public Satellite {
public:
    explicit Vertex(int num): n(num) {}
private:
    std::size_t n;
};

template <typename Satellite = Degree>
class Edge {
public:
    // i want have different constructor depending on 
    // whether Vertex is (directly or indirectly) derived from Degree
    Edge(Enable_if<Is_Degree<Satellite>(), Vertex<Satellite> &>fromVertex,
        Vertex<Satellite> &toVertex)
        : from(fromVertex), to(toVertex){ ++to.inDeg; }
    Edge(Enable_if<!Is_Degree<Satellite>(), Vertex<Satellite> &>fromVertex, 
        Vertex<Satellite> &toVertex)
        : from(fromVertex), to(toVertex){}
private:
    Vertex<Satellite> &from;
    Vertex<Satellite> &to;
};

编译器在第2行抱怨:

"在' std :: __ 1 :: enable_if< false，Vertex< Degree>&> ':' enable_if '不能用于禁用此声明."

"No type named 'type' in 'std::__1::enable_if<false, Vertex<Degree> &>': 'enable_if' cannot be used to disable this declaration."

如果删除Edge的第二个构造函数，则没有错误.我想知道原因，以及如何达到评论中所述的目的.

There is no error if I remove the second constructor of Edge. I want know why, and how to attain my purpose as described in the comment.

推荐答案

这是因为替换发生在即时上下文 . std :: enable_if 中涉及的类型模板参数应该直接来自模板，当上下文要求存在功能/专业化功能时，编译器会尝试实例化该模板，并且在此之前未知.否则，编译器可以自由拒绝您的代码.

This is because substitution takes place (and fails) outside of an immediate context. Type template parameters involved in std::enable_if should come directly from a template that a compiler attempts to instantiate when a function/specialization is required to exist by a context, and that are unknown prior to that point. Otherwise, the compiler is free to reject your code.

可能的解决方法是将构造函数转换为模板，并将其参数默认为包含类的模板参数的值:

A possible workaround is to turn the constructors into templates and default their parameters to the value of the template parameter of the enclosing class:

template <typename S = Satellite>
//                 ^-----v
Edge(Enable_if<Is_Degree<S>(), Vertex<Satellite> &>fromVertex,
    Vertex<Satellite> &toVertex)
    : from(fromVertex), to(toVertex){ ++to.inDeg; }

template <typename S = Satellite>
//                 ^------v
Edge(Enable_if<!Is_Degree<S>(), Vertex<Satellite> &>fromVertex, 
    Vertex<Satellite> &toVertex)
    : from(fromVertex), to(toVertex){}

演示

这篇关于为什么编译器说:'enable_if'不能用于禁用此声明的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！