如何实现operator->对于迭代器类型？ [英] How to implement operator-> for iterator type?

问题描述

有没有办法实现operator->，而不仅仅是operator *。要使以下代码正常工作：

Is there a way to implement operator->, not only operator*. To have following code working:

Iterator<value> it = ...
i = (*it).get();
i = it->get(); // also works

我们说值类型有方法get。
当Iterator实现如下：

Let we say that value type has method get. When Iterator is implemnted as below:

template<T> class Iterator {
    T operator*() { return ... }
    T operator->()  { return ... }
 }

这里......是获得正确的T对象的实现。

Here ... is an implementation of getting right T object.

不知怎的，它赢了当我以这种方式实现它时，我的工作。我想我误解了一些事情。

Somehow it won't work when I implement it in this way. I think I misunderstand something.

推荐答案

operator-> 应该返回指针：

T * operator->();
T const * operator->() const;

operator * 如果你返回引用想用它来修改：

operator* should return a reference if you want to use it for modification:

T & operator*();
T operator*() const; // OR T const & operator*() const;

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