問題描述

我遇到了重載 << 流運算符的問題，但我沒有找到解決方案:

I have a problem to overload the << stream operator and I don't find the solution :

template<class T, unsigned int TN>
class NVector
{
    inline friend std::ostream& operator<< (
        std::ostream &lhs, const NVector<T, TN> &rhs);
};

template<class T, unsigned int TN>
inline std::ostream& NVector<T, TN>::operator<<(
    std::ostream &lhs, const NVector<T, TN> &rhs)
{
    /* SOMETHING */
    return lhs;
};

它產生以下錯誤消息:

警告:朋友聲明‘std::ostream&operator<<(std::ostream&, const NVector&)' 聲明了一個非模板函數[-Wnon-template-friend]

warning : friend declaration ‘std::ostream& operator<<(std::ostream&, const NVector&)’ declares a non-template function [-Wnon-template-friend]

錯誤:'std::ostream&NVector::operator<<(std::ostream&, const NVector&)' 必須只取一個參數

error: ‘std::ostream& NVector::operator<<(std::ostream&, const NVector&)’ must take exactly one argument

如何解決這個問題?

非常感謝.

推薦答案

在你的代碼中有兩個不同的問題，第一個是 friend 聲明(正如警告明確說的，也許不是這樣清晰易懂)將單個非模板化函數聲明為友元.也就是說，當您實例化模板 NVector 時，它聲明了一個非模板化函數 std::ostream&operator<<(std::ostream&,NVector) 作為朋友.請注意，這與聲明您作為朋友提供的模板函數不同.

There are two different issues in your code, the first is that the friend declaration (as the warning clearly says, maybe not so clear to understand) declares a single non-templated function as a friend. That is, when you instantiate the template NVector<int,5> it declares a non-templated function std::ostream& operator<<(std::ostream&,NVector<int,5>) as a friend. Note that this is different from declaring the template function that you provided as a friend.

我建議您在類定義中定義友元函數.您可以在此答案中閱讀更多相關信息.

I would recommend that you define the friend function inside the class definition. You can read more on this in this answer.

template <typename T, unsigned int TN>
class NVector {
   friend std::ostream& operator<<( std::ostream& o, NVector const & v ) {
      // code goes here
      return o;
   }
};

或者，您可以選擇其他選項:

Alternatively you can opt for other options:

1. 將 operator<< 模板聲明為朋友(將授予對模板的任何和所有實例的訪問權限)，
2. 將該模板的特定實例聲明為朋友(編寫起來更麻煩)或
3. 完全避免友誼提供公共print(std::ostream&)成員函數并從非友元模板operator<<調用它.我仍然會選擇與非模板函數交朋友，并在模板化類中提供定義.

1. declare the operator<< template as a friend (will grant access to any and all instantiations of the template),
2. declare a particular instantiation of that template as a friend (more cumbersome to write) or
3. avoid friendship altogether providing a public print( std::ostream& ) member function and calling it from a non-friend templated operator<<. I would still opt to befriend the non-template function an provide the definition inside the templated class.

第二個問題是，當您想在左側參數的類之外定義一個運算符時，該運算符是一個自由函數(未綁定到類)，因此它應該不合格:

The second issue is that when you want to define an operator outside of the class of the left hand side argument, the operator is a free function (not bound to a class) and thus it should not be qualified:

template<class T, unsigned int TN>
inline std::ostream& operator<<(std::ostream &lhs, const NVector<T, TN> &rhs)
{
    /* SOMETHING */
    return lhs;
};

這篇關于C++:友元聲明‘聲明一個非模板函數的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！