重载操作符<&lt ;:不能将lvalue绑定到'std :: basic_ostream< char>&& [英] Overloading operator<<: cannot bind lvalue to ‘std::basic_ostream<char>&&’

问题描述

我有一个使用嵌套类的类，并且想要使用嵌套类 operator 定义运算符<< / code>在上层类。这里是我的代码看起来像：

  #include< memory& 
 #include< iostream> 
 
 template< typename T> 
 struct classA {
 struct classB 
 {
 template< typename U> 
 friend inline std :: ostream&运算符<< （std :: ostream& out，
 const typename classA< U> :: classB& b）; 
}; 
 
 classB root; 
 
 template< typename U> 
 friend std :: ostream&运算符<< （std :: ostream& out，
 const classA< U>& tree）; 
}; 
 
 template< typename T> 
 inline std :: ostream&运算符<< （std :: ostream& out，
 const classA< T>& tree）
 {
 out<树木
 return out; 
} 
 
 template< typename T> 
 inline std :: ostream&运算符<< （std :: ostream& out，
 const typename classA< T> :: classB& b）
 {
 return out; 
} 
 
 int main（）
 {
 classA< int>一个; 
 std :: cout<<一个; 
} 
  





• ++ 11，operator<<<对于内部类似乎没有被编译器找到：

    so.hpp：24：7：error：no match for 'operator<<'in'out<< tree.classA< int> :: root'
   so.hpp：24：7：note：candidates are：... 
    

  li>
  

• 在使用std = c ++ 0x：

  时编译GCC 4.6和4.7时，
  
  

    so.hpp：21：3：error：can not bind'std :: ostream {aka std :: basic_ostream< char>}'lvalue to'std :: basic_ostream< char>&包括从/usr/include/c++/4.7/iostream:40:0，
  从so.hpp：2：
   /usr/include/c++/4.7/ostream:600:5：错误：初始化'std :: basic_ostream< _CharT，_Traits>的参数1 std :: operator<<<（std :: basic_ostream< _CharT，_Traits>&&&& const _Tp&）[with _CharT = char; _Traits = std :: char_traits< char> ;; _Tp = classA< int> :: classB]'
    

有人可以告诉我为什么这个代码不合法​​，什么是最好的方式来做我想要的？

解决方案

（ T 不能在调用嵌套的运算符<< 。一个简单的解决方法，我推荐一般，不仅仅是在这里，不是一个模板，而是一个单一的自由函数，为此你需要 define 函数inline：

 模板< typename T> 
 struct classA {
 struct classB 
 {
 friend inline std :: ostream& operator<<（std :: ostream& out，
 const classB& b）{
 //定义在这里
} 
}; 
 
 classB root; 
 
 friend std :: ostream& operator<<<（std :: ostream& out，
 const classA< U> & tree）{
 // definition here here 
} 
}; 
  

这两种方法有几个区别。最重要的是，这种方法将使编译器为模板的每个实例化定义 operator 的非模板化重载，这是因为它不再是模板，不依赖于推导参数。另一个副作用是，该方法有点更紧密（你只是一个函数），而在你的初始方法中，你结合模板和所有可能的实例化（可以是最后，这样定义的函数只能通过ADL找到，所以对于 operator 的重载更少了编译器考虑当参数不是 ClassA< T> 或 ClassA< T> :: ClassB 。

---

如何获得使用方法

  namespace {
 struct intruder {
 ClassA& ref; 
 intruder（ClassA& r）：ref（r）{} 
}; 
 template<> 
 std :: ostream& operator<<< intruder>（std :: ostream& _，ClassA< intruder> const& i）{
 std :: cout<< i.ref.private_member<< std :: endl; 
 return _; 
} 
} 
  

替代 >

或者，您可以熟悉模板的特殊化。这将解决入侵者问题，因为它只对运算符到 ClassA< intruder> ，其影响要小得多。但这不会解决你的特定问题，因为类型仍然不可推断。

I have a class that uses a nested class, and want to use the nested class operator<< to define operator<< in the upper class. Here is how my code looks like:

#include <memory>
#include <iostream>

template<typename T>
struct classA {
  struct classB
  {
    template<typename U>
    friend inline std::ostream& operator<< (std::ostream &out,
                                            const typename classA<U>::classB &b);
  };

  classB root;

  template<typename U>
  friend std::ostream& operator<< (std::ostream &out,
                                   const classA<U> &tree);
};

template<typename T>
inline std::ostream& operator<< (std::ostream &out,
                                 const classA<T> &tree)
{
  out << tree.root;
  return out;
}

template<typename T>
inline std::ostream& operator<< (std::ostream &out,
                                 const typename classA<T>::classB &b)
{
  return out;
}

int main()
{
  classA<int> a;
  std::cout << a;
}

• When compiling without support for C++11, the definition of operator<< for the inner class seems not to be found by the compiler:

  so.hpp:24:7: error: no match for ‘operator<<’ in ‘out << tree.classA<int>::root’
  so.hpp:24:7: note: candidates are: ...
  

• With GCC 4.6 and 4.7 when compiling with std=c++0x:

  so.hpp:21:3: error: cannot bind ‘std::ostream {aka std::basic_ostream<char>}’ lvalue to ‘std::basic_ostream<char>&&’
  In file included from /usr/include/c++/4.7/iostream:40:0,
                   from so.hpp:2:
  /usr/include/c++/4.7/ostream:600:5: error:   initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = classA<int>::classB]’
  

Can someone tell me why this code is not legal, and what's the best way to do what I want?

解决方案

Bo provided the reason why this is happening (the type T is not deducible in the call to the nested operator<<. A simple workaround for this, and something that I recommend in general, not only here, is not befriending a template, but rather a single free function. For that you will need to define the function inline:

template<typename T>
struct classA {
  struct classB
  {
    friend inline std::ostream& operator<< (std::ostream &out,
                                            const classB &b) {
       // definition goes here
    }
  };

  classB root;

  friend std::ostream& operator<< (std::ostream &out,
                                   const classA<U> &tree) {
       // definition goes here
  }
};

There are a couple of differences among the two approaches. The most important one is that this approach will have the compiler define a non-templated overload for operator<< for each instantiation of the template, which because it is no longer a template, does not depend on deducing the arguments. Another side effects are that the approach is a little tighter (you are only befriending one function, while in your initial approach you befriended the template and all possible instantiations (which can be used as a loophole to gain access to your class internals). Finally the functions so defined will only be found through ADL, so there are less overloads of operator<< for the compiler to consider when the argument is not ClassA<T> or ClassA<T>::ClassB.

---

How access can be gained with your approach

namespace {
   struct intruder {
       ClassA & ref;
       intruder( ClassA& r ) : ref(r) {}
   };
   template <>
   std::ostream& operator<< <intruder>( std::ostream& _, ClassA<intruder> const& i ) {
       std::cout << i.ref.private_member << std::endl;
       return _;
   }
}

Alternative

Alternatively you can befriend a particular specialization of a template. That will solve the intruder problem, as it will only be open to operator<< to ClassA<intruder>, which has a much lesser impact. But this will not solve your particular issue, as the type would still not be deducible.

这篇关于重载操作符&lt;&lt ;:不能将lvalue绑定到'std :: basic_ostream&lt; char&gt;&amp;&amp;的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！