重載運算符'='和'+'

Question

我正在學習如何使用模板以及如何重載運算符。我設法超載operator[]，但我遇到了超載operator+和operator=的問題。這裏是我的代碼：

template <class T> 
class A 
{ 
public: 
    //... 
    friend A<T>& A<T>::operator+ (A<T>&, const A<T>&); 
    friend A<T>& A<T>::operator= (A<T>&, const A<T>&); 
}; 

template<class T> A<T>& A<T>::operator+ (A<T>& left, const A<T>& right) 
{ 
    //some functions 
return left; 
} 

template<class T> A<T>& A<T>::operator= (A<T>& left, const A<T>& right) 
{ 
    //some functions 
    return left; 
} 

Whenver我嘗試編譯，我得到這些錯誤：

'+': is not a member of 'A<T>'

'=': is not a member of 'A<T>'

'operator =' must be a non-static member

我在做什麼錯？

---

編輯：

我已經成功地更新代碼：

template <class T> 
class A 
{ 
public: 
    //... 
    A<T> operator+ (A<T>); 
    A<T> operator= (A<T>, const A<T>); 
}; 

template<class T> A<T> A<T>::operator+ (A<T> right) 
{ 
    //some functions 
    return *this; 
} 

template<class T> A<T> operator= (A<T> right) 
{ 
    //some functions 
    return *this; 
} 

貌似operator+作品現在很好，但是編譯器會發出此錯誤：

'operator=' must be a non static member

爲什麼它是一個靜態成員，我該如何解決它？

Answer 1

對於初學者賦值運算符必須爲非靜態成員函數

從C++標準（13.5.3分配）

1 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 (12.8), a base class assignment operator is always hidden by the copy assignment operator of the derived class.

其次（11.3好友）

1 A friend of a class is a function or class that is given permission to use the private and protected member names from the class. A class specifies its friends, if any, by way of friend declarations. Such declarations give special access rights to the friends, but they do not make the nominated friends members of the befriending class.

因此例如這個定義

template<class T> A<T>& A<T>::operator+ (A<T>& left, const A<T>& right) 
         ^^^^^ 
{ 
//some functions 
return left; 
} 

不正確。至少您應該刪除A<T>::，因爲操作員不是班級的成員。

Answer 2

作爲非靜態成員實現的運算符只能接受1個輸入參數 - 右側操作數。左邊的操作數是運算符被調用的對象this。

作爲靜態成員或非成員實現的運算符必須接受2個輸入參數（左側和右側操作數）。

您的operator=被聲明爲具有2個輸入參數的非靜態成員，這是錯誤的。

此外，operator+是爲了返回一個新的對象，它是兩個輸入對象的副本添加在一起。不要返回對調用操作符的對象的引用。而operator=是爲了返回對被分配的對象的引用。

試試這個：

template <class T> 
class A 
{ 
public: 
    //... 
    A<T> operator+(const A<T>&) const; 
    A<T>& operator=(const A<T>&); 
}; 

template<class T> A<T> A<T>::operator+(const A<T>& right) const 
{ 
    A<T> result(*this); 
    //some functions to add right to result as needed... 
    return result; 
} 

template<class T> A<T>& A<T>::operator=(const A<T>& right) 
{ 
    // some functions to copy right into this... 
    return *this; 
}