&（* similarObject）和similarObject之间的区别？他们不一样吗？ [英] Difference between &(*similarObject) and similarObject? Are they not same?

问题描述

有人可以向我解释这个问题。

  dynamic_cast< SomeObject *>（&（* similarObject） 
  

执行取消引用指针的地址的意义是什么？

解决方案

这可能是 similarObject 已经重载运算符* ，因此它返回的东西的地址，你传递到 dynamic_cast 。

&（* x）和 x 可能不会总是相同的东西。例如，想到迭代器：

  std :: map< int，int> :: iterator it = v.begin ; 
  

然后 it 和是两个不同的东西：

• 它是 std :: map


• &（* it）是 std :: pair< int，int> *
/ ul>



它们不是 。类似的事情也可能发生在你的代码片段。




Can someone please explain this to me

dynamic_cast<SomeObject *>( &(*similarObject) );

What is the point of doing the address of a dereferenced pointer? Wouldn’t the pointer itself just be the address of it?

解决方案

It may be that the type of similarObject has overloaded operator* and so it returns something whose address you're passing to dynamic_cast.

&(*x) and x may not be always the same thing. For example, think of iterator:

std::map<int, int>::iterator it = v.begin();

Then it and &(*it) are two different thing:

• The type of it is std::map<int, int>::iterator
• The type of &(*it) is std::pair<int,int>*

They're not at all same. Similar thing may happen with your code-snippet as well.

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