运算符<<<（ostream& amp; X）对于嵌套在类模板中的类X [英] operator<<(ostream&, X) for class X nested in a class template

问题描述

这个编译和工作原理（非嵌套模板）：

  #include< iostream> 
 
 template< typename T> Z类; 
 
 template< typename T> 
 std :: ostream&运算符<< （std :: ostream& os，const Z< T>）{
 return（os<<Z）; 
} 
 
 template< typename T> class Z {
 friend std :: ostream&运算符<< <> （std :: ostream& os，const Z&）; 
}; 
 
 int main（）{
 Z< int> z; 
 std :: cout<< z < std :: endl; 
} 
  

这不能编译（gcc 4.4和gcc 4.6，和0x模式）：

  #include< iostream> 
 
 template< typename T> Z类; 
 
 template< typename T> 
 std :: ostream&运算符<< （std :: ostream& os，const typename Z T :: ZZ&）{
 return（os<ZZ！ 
} 
 
 template< typename T> class Z {
 public：
 class ZZ {
 friend std :: ostream&运算符<< <> （std :: ostream& os，const ZZ&）; 
}; 
}; 
 
 
 int main（）{
 Z< int> :: ZZ zz; 
 std :: cout<< zz < std :: endl; 
} 
  

错误消息如下所示：

 错误：template-id'operator< <>'for'std :: ostream& <<<<（std :: ostream& 
 const Z< int> :: ZZ&）'不匹配任何模板声明
 error： :: cout<< zz'
  

在0x模式下，第二个错误消息不同，但含义是一样的。 / p>

我可以做我想做的吗？

EDIT 这里有一个非推导上下文的实例，它解释了错误消息。然而，问题仍然是：我可以为嵌套在类模板中的类有一个有效的操作符<<吗？

解决方案

Matthieu很好地解释了这个问题，但是在这种情况下可以使用一个简单的解决方法。
可以使用friend声明在类中实现friend函数：

  #include< iostream> 
 
 template< typename T> class Z {
 public：
 class ZZ {
 friend std :: ostream&运算符<< （std :: ostream& os，const ZZ&）{
 return os< ZZ！; 
} 
}; 
}; 
 
 
 int main（）{
 Z< int> :: ZZ zz; 
 std :: cout<< zz < std :: endl; 
} 
  

This one compiles and works like it should (non-nested template):

#include <iostream>

template<typename T> class Z;

template <typename T> 
std::ostream& operator<< (std::ostream& os, const Z<T>&) {
    return (os << "Z");
}

template<typename T> class Z {
    friend std::ostream& operator<< <> (std::ostream& os, const Z&);
};

int main () {
    Z<int> z;
    std::cout << z << std::endl;
}

This one doesn't compile (gcc 4.4 and gcc 4.6, in both 03 and 0x mode):

#include <iostream>

template<typename T> class Z;

template<typename T> 
std::ostream& operator<< (std::ostream& os, const typename Z<T>::ZZ&) {
    return (os << "ZZ!");
}

template <typename T> class Z {
  public:
    class ZZ {
        friend std::ostream& operator<< <> (std::ostream& os, const ZZ&);
    };
};


int main () {
    Z<int>::ZZ zz;
    std::cout << zz << std::endl;
}

The error message looks like this:

error: template-id ‘operator<< <>’ for ‘std::ostream& operator<<(std::ostream&,
const Z<int>::ZZ&)’ does not match any template declaration
error: no match for ‘operator<<’ in ‘std::cout << zz’

In the 0x mode the second error message is different, but the meaning is the same.

Is it possible to do what I want to do?

EDIT Apparently, there's an instance of non-deduced context here, which explains the error messages. The question, however, still stands: can I have a working operator<< for a class nested in a class template?

解决方案

Matthieu explained the problem very well, but a simple work-around can be used in this case. You can implement the friend function inside the class with the friend declaration:

#include <iostream>

template <typename T> class Z {
public:
  class ZZ {
    friend std::ostream& operator<< (std::ostream& os, const ZZ&) {
      return os << "ZZ!";
    }
  };
};


int main () {
  Z<int>::ZZ zz;
  std::cout << zz << std::endl;
}

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