c++ - 继承模板赋值运算符

Question

在使用其父类的模板方法之一编写子类时，我经常遇到编译问题。例如，我写了这个，但我不知道为什么编译:

#include <iostream>
#include <type_traits>
#include <algorithm>

        /********************     **********     ********************/

template <class T, unsigned N>
struct A
{
    T data[N];

    template <class U>
    inline auto operator= ( const U& other ) -> decltype(*this)
        { this->assign(other); return *this; }

    template <class U>
    void assign( const U& other )
        { assign_impl( other, std::is_arithmetic<U>() ); }

    template <class U>
    void assign_impl( const U& val, std::true_type const )
        { std::fill( data, data+N, static_cast<T>(val) ); }
};

// ------------------------------------------------------------------------

template <class T, unsigned N>
struct B
    : public A<T,N>
{
    // Needed in order to compile
    using A<T,N>::operator=;

    void show()
    {
        for (unsigned i=0; i<N; ++i)
            std::cout<< this->data[i] << " ";
            std::cout<< std::endl;
    }
};

// ------------------------------------------------------------------------

int main()
{
    B<double,5> b;
    b = -5.1; 
    b.show();

    b.assign(3.14159);
    b.show();
}

包括声明 using A<T,N>::operator=;如果我想将此运算符与 B<T,N> 的实例一起使用，则这是必需的, 但我从未指定方法 assign应该是可见的。可见是因为operator=使用它？

Answer 1

方法assign在你的类里面可见B因为你正在使用 struct它具有公共(public)默认封装和 public继承。

至于operator= :

(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 class if not declared by the user, a base class assignment operator is always hidden by the copy assignment operator of the derived class.