“没有呼叫错误的匹配函数”与g ++ [英] "No matching function for call error" with g++

问题描述

我有一个类A

  template< typename T& 
 class A：public std :: auto_ptr< T> 
 {
 typedef std :: auto_ptr< T>超; 
 public：
 A（）：Super（）{} 
 A（T * t）：Super（t）{} 
 A ：Super（o）{} 
 ... 
}; 
  

和IConstEnumerator

  template< typename T> 
 struct IConstEnumerator 
 {
 ... 
 virtual IEnumerator< T> * GetEnumerator（）= 0; 
 virtual IConstEnumerator< T> * GetEnumerator（）const = 0; 
}; 
  

当我运行此代码

  AP < IConstEnumerator< IPort> > pe = node.GetPorts（）。GetEnumerator（）; ; 
  

我找不到与g ++编译器匹配的错误。

 错误：没有匹配函数调用'AIR :: A< AIR :: IConstEnumerator< AIR :: IPort& > :: AP（AIR :: A< AIR :: IConstEnumerator< AIR :: IPort>）'
注意：候选项为：AIR :: A& T>&）[其中T = AIR :: IConstEnumerator< AIR :: IPort>] 
 note：AIR :: A  :: A（T *）[T = AIR :: IConstEnumerator   

A类有什么问题？

EDIT

使用复制构造函数似乎解决了这个问题。 A< IConstEnumerator< IPort> >

解决方案 > A 复制构造函数通过非const引用获取其参数：

  A（A< T> & o）：Super（o）{} 
 ^ not const 
  

尝试复制临时对象，非const引用不能绑定到临时对象。 Visual C ++有一个邪恶的扩展，允许这个工作;

如果您试图模仿 auto_ptr

，则必须小心避免依赖该扩展。 code>的复制构造函数，您还需要实现类似于 auto_ptr_ref 的东西，用作帮助器允许复制临时 auto_ptr s。我将介绍如何在接受的答案中实现这一点。 code> std :: auto_ptr 的复制构造函数？

对于值得的，从 std :: auto_ptr 有点奇怪;考虑使用组合而不是继承，如果你可以（在 std :: auto_ptr 中没有很多东西，你可以从中获益）。

I have a class A

template<typename T>
class A : public std::auto_ptr<T>
{
    typedef std::auto_ptr<T> Super;
public:
    A() : Super() { }
    A(T* t) : Super(t) { }
    A(A<T>& o) : Super(o) { }
    ...
};

and IConstEnumerator

template<typename T>
struct IConstEnumerator
{
    ...   
virtual IEnumerator<T>* GetEnumerator() = 0;
virtual IConstEnumerator<T>* GetEnumerator() const = 0;      
};

When I run this code

AP< IConstEnumerator<IPort> > pe = node.GetPorts().GetEnumerator(); ;

I got errors from not finding correct match with g++ compiler.

error: no matching function for call to ‘AIR::A<AIR::IConstEnumerator<AIR::IPort> >::AP(AIR::A<AIR::IConstEnumerator<AIR::IPort> >)’
note: candidates are: AIR::A<T>::A(AIR::A<T>&) [with T = AIR::IConstEnumerator<AIR::IPort>]
note:                 AIR::A<T>::A(T*) [with T = AIR::IConstEnumerator<AIR::IPort>]

What's wrong with the class A? It works well with MSVC.

EDIT

Using copy constructor explicitly seems to solve this issue. A< IConstEnumerator<IPort> > pe(node.GetPorts().GetEnumerator())

解决方案

Your A copy constructor takes its argument by non-const reference:

A(A<T>& o) : Super(o) { }
      ^ not const

Your example probably tries to copy a temporary object and the non-const reference can't bind to the temporary object. Visual C++ has an evil extension that allows this to work; you have to be careful to avoid relying on that extension if you want your code to be portable.

If you are trying to mimic auto_ptr's copy constructor, you need to also implement something similar to auto_ptr_ref, which is used as a helper to allow copying temporary auto_ptrs. I describe how this is accomplished in the accepted answer to How could one implement std::auto_ptr's copy constructor?

For what it's worth, deriving from std::auto_ptr is a bit odd; consider using composition instead of inheritance, if you can (there isn't a whole lot in std::auto_ptr that you'd benefit from by deriving from it anyway).

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