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?
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*
it
is std::map<int, int>::iterator
&(*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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