&(*similarObject) 和similarObject 之间的区别?他们不一样吗?

问题描述

谁能给我解释一下

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

做一个解引用指针的地址有什么意义?指针本身不就是它的地址吗?

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

推荐答案

可能是 similarObject 的类型重载了 operator* 所以它返回了地址您正在传递给 dynamic_cast.

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) 和 x 可能不总是一样的.例如，考虑迭代器:

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

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

那么 it 和 &(*it) 是两个不同的东西:

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

• it的类型是std::map::iterator
• &(*it) 的类型是 std::pair*

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

他们完全不同.您的代码片段也可能发生类似的事情.

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

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