what does "*&" mean together?

Question

For example:

private:
    Node* curr;
public:
    Node*& Iterator::getCurr() {
        return curr;
    }

    bool Iterator::operator==(const Iterator& other) {
        return curr == other.getCurr();
    }

I'm getting error in this code:

passing ‘const Iterator’ as ‘this’ argument of ‘Node*& Iterator::getCurr()’ discards qualifiers [-fpermissive]

How should I fix it?

You're returning a reference to a pointer. It's hard to tell from your question whether that's what you want. — Cody Gray - on strike, Apr 16 '13 at 04:35
I'm sorry that I changed my question a lot. But, I think it's clearer now — Bek, Apr 16 '13 at 04:41
You will fix it by first providing a description of the error you encounter. Otherwise we won't be able to help. — Nbr44, Apr 16 '13 at 04:42

score 2 · Answer 1 · answered Apr 16 '13 at 04:35

2

don't read them together

if you see something like this:

Foo& foo();

Do you know what does the & means?

It is a reference to Foo

Then

Foo* foo();

What about this? this is Pointer to Foo

Then

Foo*& foo();

is reference to "Pointer to Foo"

answered Apr 16 '13 at 04:35

Adrian Shum

38,812
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score 2 · Accepted Answer · answered Apr 16 '13 at 04:36

Node*& means “reference to pointer to Node”. You can return it as normal. Accessing it, however, can be done two ways: the ‘normal’ way, where the “reference to” part will just be dropped, and the way preserving the reference. The advantage of the latter way is you can change the underlying curr value:

Node *&curr = iterator.getCurr();
curr = new Node();  // or something like that
// iterator.curr has been changed

score 0 · Answer 3 · answered Apr 16 '13 at 04:34

0

It's a reference to a pointer to Node.

answered Apr 16 '13 at 04:34

Ed S.

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what does "*&" mean together?

3 Answers3