operator= 和 C++ 中未继承的函数?

2021-12-17 00:00:00 inheritance operator-overloading c++ crtp

在我刚刚进行的测试之前，我认为只有构造函数在 C++ 中没有被继承.但显然，赋值 operator= 不是太......

Until a test I've just made, I believed that only Constructors were not inherited in C++. But apparently, the assignment operator= is not too...

  1. 这是什么原因?
  2. 是否有任何解决方法可以继承赋值运算符?
  3. operator+=, operator-=, ... 也是这种情况吗?
  4. 是否继承了所有其他函数(除了构造函数/operator=)?
  1. What is the reason of that ?
  2. Is there any workaround to inherit the assignment operator ?
  3. Is it also the case for operator+=, operator-=, ... ?
  4. Are all other functions (apart from constructors/operator=) inherited ?

事实上，我在做一些CRTP时遇到了这个问题:

In fact, I encountered this problem as I was doing some CRTP :

template<class Crtp> class Base
{
    inline Crtp& operator=(const Base<Crtp>& rhs) {/*SOMETHING*/; return static_cast<Crtp&>(*this);}
};

class Derived1 : public Base<Derived1>
{
};

class Derived2 : public Base<Derived2>
{
};

有什么解决方案可以让它工作吗?

Is there any solution to get that working ?

好的，我已经隔离了问题.为什么以下不起作用?如何解决问题?

EDIT : OK, I have isolated the problem. Why the following isn't working ? How to solve the problem ?

#include <iostream>
#include <type_traits>

// Base class
template<template<typename, unsigned int> class CRTP, typename T, unsigned int N> class Base
{
    // Cast to base
    public:
        inline Base<CRTP, T, N>& operator()()
        {
            return *this;
        }

    // Operator =
    public:
        template<typename T0, class = typename std::enable_if<std::is_convertible<T0, T>::value>::type>
        inline CRTP<T, N>& operator=(const T0& rhs)
        {
            for (unsigned int i = 0; i < N; ++i) {
                _data[i] = rhs;
            }
            return static_cast<CRTP<T, N>&>(*this);
        }

    // Data members
    protected:
        T _data[N];
};

// Derived class
template<typename T, unsigned int N> class Derived : public Base<Derived, T, N>
{
};

// Main
int main()
{
    Derived<double, 3> x;
    x() = 3; // <- This is OK
    x = 3;   // <- error: no match for 'operator=' in ' x=3 '
    return 0;
}

推荐答案

赋值运算符在技术上是继承的；然而，它总是被派生类的显式或隐式定义的赋值运算符隐藏(参见下面的注释).

The assignment operator is technically inherited; however, it is always hidden by an explicitly or implicitly defined assignment operator for the derived class (see comments below).

(13.5.3 赋值) 赋值运算符应由只有一个参数的非静态成员函数.因为是副本赋值运算符 operator= 是为类隐式声明的，如果未由用户声明，基类赋值运算符始终是被派生类的复制赋值运算符隐藏.

(13.5.3 Assignment) An assignment operator shall be implemented by a non-static member function with exactly one parameter. Because a copy assignment operator operator= is implicitly declared for a a class if not declared by the user, a base class assignment operator is always hidden by the copy assignment operator of the derived class.

您可以实现一个虚拟赋值运算符，它只是将调用转发到基类 operator=，如下所示:

You can implement a dummy assignment operator which simply forwards the call to the base class operator=, like this:

// Derived class
template<typename T, unsigned int N> class Derived : public Base<Derived, T, N>
{
public:
    template<typename T0, class = typename std::enable_if<std::is_convertible<T0, T>::value>::type>
    inline Derived& operator=(const T0& rhs)
    {
        return Base<Derived, T, N>::operator=(rhs);
    }
};

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