指针列表 C++
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Pointer list c++
提问by user1519221
The code below:
下面的代码：
#include <iostream>
#include <list>
class A
{
public:
void printHello(){std::cout << "hello";}
};
int main(int argc, char *argv)
{
std::list<A*> lista;
lista.push_back(new A());
for(std::list<A*>::iterator it=lista.begin();it!=lista.end();++it)
{
//how to get to printHello method?
//it doesn't work
it->printHello();
}
return 0;
}
This code doesn't work. My question is how to get to method 'printHello' by iterator it? Thanks.
此代码不起作用。我的问题是如何通过迭代器获取方法“printHello”呢？谢谢。
回答by Daniel Frey
You want
你要
(*it)->printHello();
as the *itreturns the stored pointer A*and only then you can apply ->.
因为*it返回存储的指针A*，只有这样你才能申请->。
回答by P0W
De-referencing itwill give you pointer to A, then you need to access the methods or data members.
取消引用it将为您提供指向 A 的指针，然后您需要访问方法或数据成员。
So use :
所以使用：
(*it)->printHello();
(*it)->printHello();
回答by TASagent
Let me expand on Daniel's answer.
让我扩展丹尼尔的回答。
When you stick an asterisk in front of a variable, it is called 'dereferencing'. Used this way, the Asterisk is a 'Dereference Operator'. To put it noob-ishly (I don't know what level of understanding you have offhand), *pMyPointeracts like it was the Object that the pMyPointer was pointing to. If it was a Pointer to a Pointer, then the result is just the Pointer.
当您在变量前添加星号时，它被称为“取消引用”。以这种方式使用，星号是一个“解引用运算符”。简单地说（我不知道你有什么水平的理解），*pMyPointer就像 pMyPointer 指向的对象一样。如果它是一个指向指针的指针，那么结果就是指针。
As an example, when you call a method on a pointer, you use the Into Operator ->.
例如，当您在指针上调用方法时，您使用 Into 运算符->。
These two oftendo the same thing:
这两个经常做同样的事情：
pMyPointer->MyFunction();
pMyPointer->MyFunction();
(*pMyPointer).MyFunction();
(*pMyPointer).MyFunction();
In the case of the C++ iterators, the Dereference Operator is overwritten to return the object stored in its position. In this case, what is stored in its position is a pointer, so you still have to use ->unless you stick another Dereference Operator in there.
在 C++ 迭代器的情况下，解引用运算符被覆盖以返回存储在其位置的对象。在这种情况下，它的位置存储的是一个指针，所以你仍然必须使用，->除非你在那里插入另一个解引用运算符。
回答by villekulla
Just change following line
只需更改以下行
it->printHello();
to
到
(*it)->printHello();
The operator*() gives access to the contained data of the container, which in your case is a pointer. When not using pointers in containers, just using operator->() would work, too.
operator*() 可以访问容器中包含的数据，在您的情况下是指针。当不在容器中使用指针时，只使用 operator->() 也可以。
