C ++模板元编程：重载运算符(C++ template meta programming: overloading operators)

 我目前正在玩模板元编程。 我正在尝试使用tmp创建一个有限状态机。 我知道网上有几个实现，但我想自己实现一个练习。 
 
 我正在尝试为基类内的模板化基类的模板化导数重载运算符。 假设我们有一个基类： 
 
template<typename Input>
class Base
{
public:
    virtual ~Base() = default;    
    virtual bool operator()(const Input& input) const = 0;

    template<typename Lhs, typename Rhs>
    constexpr Derivation1<Input, Lhs, Rhs> operator||(const Lhs& left, const Rhs& right) const;

    template<typename Lhs, typename Rhs>
    constexpr Derivation2<Input, Lhs, Rhs> operator&&(const Lhs& left, const Rhs& right) const;
};
 
 及其2个推导： 
 
template<typename Input, typename... TSpecialized>
class Derivation1 : public Base<Input>
{
public:
    bool operator()(const Input& input) const override
    {
        // ...
    }
};

template<typename Input, typename... TSpecialized>
class Derivation2 : public Base<Input>
{
public:
    bool operator()(const Input& input) const override
    {
        // ...
    }
};
 
 以及我们在基类中声明的运算符的定义： 
 
template <typename Input>
template <typename Lhs, typename Rhs>
constexpr Derivation1<Input, Lhs, Rhs> Base<Input>::operator||(const Lhs& left, const Rhs& right) const
{
    return Derivation1<Input, Lhs, Rhs>();
}

template <typename Input>
template <typename Lhs, typename Rhs>
constexpr Derivation2<Input, Lhs, Rhs> Base<Input>::operator&&(const Lhs& left, const Rhs& right) const
{
    return Derivation2<Input, Lhs, Rhs>();
}
 
 Rhs和Lhs类型也是基类的推导。 
 
 当我尝试使用运算符时： 
 
Derivation3<int, 10, 20> left;
Derivation4<int, 300, 290> right;

auto result = left || right;
 
 我收到一个错误，指出运算符的重载没有匹配参数。 两个派生都具有相同的基类型： Base<int> ，其中应声明重载。 然后变量result应该是Derivation1类型（就像我们在上面的代码中声明的那样）。 
 
 在这种情况下，如何正确地重载operatros？ 

I'm currently playing around with template metaprogramming. I'm trying to make a finite state machine by using tmp. I know that there are several implementations in the web but I want to implement one by myself as an exercise.
 
I'm trying to overload operators for templated derivatives of a templated base class inside the base class. Assume we have a base class:
 
template<typename Input>
class Base
{
public:
    virtual ~Base() = default;    
    virtual bool operator()(const Input& input) const = 0;

    template<typename Lhs, typename Rhs>
    constexpr Derivation1<Input, Lhs, Rhs> operator||(const Lhs& left, const Rhs& right) const;

    template<typename Lhs, typename Rhs>
    constexpr Derivation2<Input, Lhs, Rhs> operator&&(const Lhs& left, const Rhs& right) const;
};
 
and 2 of its derivations:
 
template<typename Input, typename... TSpecialized>
class Derivation1 : public Base<Input>
{
public:
    bool operator()(const Input& input) const override
    {
        // ...
    }
};

template<typename Input, typename... TSpecialized>
class Derivation2 : public Base<Input>
{
public:
    bool operator()(const Input& input) const override
    {
        // ...
    }
};
 
and the definition of the operators that we declared in the base class:
 
template <typename Input>
template <typename Lhs, typename Rhs>
constexpr Derivation1<Input, Lhs, Rhs> Base<Input>::operator||(const Lhs& left, const Rhs& right) const
{
    return Derivation1<Input, Lhs, Rhs>();
}

template <typename Input>
template <typename Lhs, typename Rhs>
constexpr Derivation2<Input, Lhs, Rhs> Base<Input>::operator&&(const Lhs& left, const Rhs& right) const
{
    return Derivation2<Input, Lhs, Rhs>();
}
 
The types Rhs and Lhs are derivations of the base class aswell.
 
When I try to use the operators like:
 
Derivation3<int, 10, 20> left;
Derivation4<int, 300, 290> right;

auto result = left || right;
 
I'm getting an error that says no overload of the operator matches the arguments. Both derivations have the same base type: Base<int> in which the overloads should be declared. The variable result should then be of type Derivation1 (like we declared it in the code above).
 
How do I overload the operatros properly in this case?

原文：https://stackoverflow.com/questions/41626239