both asterisk and ampersand in a parameter c++

问题

I am reading a book about Binary Search Tree and something weird came up.

class BST
{
public:
   void insert(const Comparable & item)

private:
   BinaryNode *root;
   struct BinaryNode
   {
       Comparable element;
       BinaryNode *left;
       BinaryNode *right;
       BinaryNode(const Comparable & theElement, BinaryNode *lt, BinaryNode *rt) : 
          element(theElement), left(lt), right(rt) {}
   }
   void insert(const Comparable & item, BinaryNode * & t) const;
};

The private insert function is helper function for public insert function, and private insert function looks for the right place to insert using recursion.

Part that I don't understand is BinaryNode * & t in the parameter. What does it mean? Pointer of the address of t?

回答1:

In your expression BinaryNode * & t)

            BinaryNode*                & t
           -------------              -----
            BinaryNode pointer        t is reference variable

so t is reference to pointer of BinaryNode class.

Pointer of the address of t?

You are confused ampersand & operator in c++. that give address of an variable. but syntax is different.

ampersand & in front of some of variable like below:

BinaryNode b;
BinaryNode* ptr = &b;

But following way is for reference variable (its simple not pointer):

BinaryNode b;
BinaryNode & t  = b;

and your is like below:

BinaryNode b;
BinaryNode* ptr = &b;
BinaryNode* &t  = ptr;

回答2:

It's reference to pointer, you can change pointer in this function and it will changed outside. Simple example http://liveworkspace.org/code/1EfD0Q$8

来源：https://stackoverflow.com/questions/14314640/both-asterisk-and-ampersand-in-a-parameter-c

标签

c++

pointers

reference

memory-address