What does *& mean?

Question

What is the meaning of Node *&front the below linked list code extract in C++?

 FrontBackSpilit(head,a,b) 

 Node * FrontBackSpilit(Node * head, Node *&front, Node * &back)

@Raufjavid Because you are passing the pointer by value, so if you want to modify the pointer within the function (not the thing it is pointing at) it will not be reflected back outside the function, because only the copy of the pointer is modified within the function. — Cory Kramer, Jan 14 '15 at 12:27
Related: http://stackoverflow.com/questions/5789806/meaning-of-and-in-c — Troyseph, Aug 11 '16 at 10:45

score 6 · Accepted Answer · answered Jan 14 '15 at 12:25

6

Node *&front

this is a reference to pointer to instance of Node, so if you call FrontBackSpilit like:

Node* pNode1;
Node* pBack;
FrontBackSpilit(NULL, pNode1, pBack);

and FrontBackSpilit will do for example front = new Node, then pNode1 will be assigned this value.

If instead of Node *&front you would have Node *front, then assignment to front will not change pNode1.

answered Jan 14 '15 at 12:25

marcinj

It means by using this ( Node *&front ) we are passing reference not value..Am i Right. – Rauf javid Jan 14 '15 at 12:28
It means Node* is passed by reference not by value. – drescherjm Jan 14 '15 at 12:29
@Raufjavid Yes. The *pointer* is being passed by *reference*. You can think of non-`const` reference parameters as "output parameters". – Joseph Mansfield Jan 14 '15 at 12:29
reference to pointer, this is actually equivalent to pointer to pointer : `Node **front`, then in function `*front = new Node;`, but its easier to access variable using `*&` – marcinj Jan 14 '15 at 12:30
@marcinj@Joseph Mansfield@drescherjm K...I understand..Thank You so much for your Help. – Rauf javid Jan 14 '15 at 12:31

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