结构取消引用运算符（operator->） [英] Structure dereference operator (operator->)

问题描述

我为迭代器编写一个简单的模板包装器，当通过结构dereference运算符时遇到一个绊脚石，主要是因为指针没有一个：

I am writing a thin template wrapper for iterators, and hit a stumbling block when passing through the structure dereference operator, mainly because pointers don't have one:

#include <vector>

struct mystruct {
    int member;
};

template<class iterator>
struct wrap {
   typedef typename std::iterator_traits<iterator>::pointer pointer;
   iterator internal;
   pointer operator->() {return internal.operator->();} //MARK1
};

int main() {
    wrap<std::vector<mystruct>::iterator> a;
    a->member;
    wrap<mystruct*> b;
    b->member;
    return 0;
}

http://ideone.com/XdvEz

prog.cpp: In member function ‘typename std::iterator_traits<_Iter>::pointer wrap<iterator>::operator->() [with iterator = mystruct*]’:
prog.cpp:18:   instantiated from here
prog.cpp:11: error: request for member ‘operator->’ in ‘((wrap<mystruct*>*)this)->wrap<mystruct*>::internal’, which is of non-class type ‘mystruct*’

以下方法有效，但我不认为它可以工作。也就是说，如果迭代器有一个奇怪的指针类型与指向 value_type 的指针不一样。 / p>

This following method works, but I don't think it's guaranteed to work. Namely, if an iterator has a strange pointer type that isn't the same as a pointer to a value_type.

   pointer operator->() {return &*internal;} //MARK3

推荐答案

标准间接说明重载 operator-> 必须返回一个指针，一个可转换为指针的对象，或一个重载 operator-> 的对象。最好的办法是返回 internal 。

The standard indirectly says that an overloaded operator-> has to either return a pointer, an object that is convertible to a pointer, or an object that has overloaded operator->. Your best bet is to just return internal.

§13.5.6[ ref] p1

表达式 x-> m 解释为（x.operator->（）） - > m

（以上适用于递归。）

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