Why does the iterator of List in STL use the (*node).data instead of node->data?

Question

I recently view the source code of SGI STL. I want to know whether I can use the "->" operator to replace the (*node).data to implement the operator*(), like this:

reference operator*() const {return (*node).data;}

replaced by:

reference operator*() const {return node->data;}

in addition:

node is a pointer which points to a struct object, like this:

template<class T>
struct __list_node {
    typedef void * void_pointer;
    void_pointer prev;
    void_pointer next;
    T data;
};

Probably, yes. What type is node and does it overload operator-> or unary operator*? — Joseph Mansfield
– Joseph Mansfield, Commented Dec 22, 2013 at 14:44
@herohuyongtao They usually should be, but nothing enforces that. Just like ++a and a++ could affect a in different ways – or just like c << n could have side-effects and mean something different from bit-wise shifting, to name the ever-popular early overload changing an operator meaning. — Christopher Creutzig
– Christopher Creutzig, Commented Dec 26, 2013 at 13:34

Joseph Mansfield · Accepted Answer · 2013-12-22 14:56:54Z

9

In most cases (for example, when node is a pointer), these will be equivalent. The x->y operator is defined as being equivalent to (*(x)).y. However, it's possible to overload operator* or operator->, in which case they may not behave as you would expect (but they should).

As you have said, in this case node is just a pointer, so they are equivalent.

edited Dec 22, 2013 at 14:56

answered Dec 22, 2013 at 14:50

Joseph Mansfield

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1 Comment

Benjamin Lindley Over a year ago

"in which case they may not behave as you would expect" -- But if they don't, please stop using that library. The author is a bad person.

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