My question is why do I need to dereference and reference a pointer for the following code to work? Doesn't ref/deref cancel each other? I would really appreciate if anyone could explain it like I'm five :)
Code:
template <typename T>
class binNode {
private:
T key;
public:
binNode * left;
binNode * right;
binNode * parent;
binNode() {
this->left = NULL;
this->right = NULL;
this->parent = NULL;
}
// arg constructor:
binNode (T key) {
this->key = key;
this->left = NULL;
this->right = NULL;
this->parent = NULL;
}
T getKey() {
return this->key;
}
void setKey(T key) {
this->key = key;
}
};
template<typename T> class Tree {
private:
binNode <T> *root;
public:
Tree() {
this->root = NULL;
}
Tree(binNode <T> * node) {
node->parent = NULL;
this->root = node;
}
/* THIS IS THE PART I DON'T GET */
void addNode(binNode<T> *&x, binNode<T> * node) { // what's up with the *&???
if (x == NULL) {
x = node;
return;
} else if (x->getKey() == node->getKey()) {
node->left = x;
node->parent = x->parent;
x->parent = node;
return;
}
if (node->getKey() < x->getKey()) {
addNode(x->left, node);
} else {
addNode(x->right, node);
}
}
void addNode(binNode<T> * node) {
addNode(this->root, node);
}
binNode<T> * treeSearch(binNode<T> * x, T key) {
if (x == NULL || key == x->getKey()) {
return x;
}
if (key < x->getKey()) {
return treeSearch(x->left, key);
} else {
return treeSearch(x->right, key);
}
}
void printOrdered() {
inorderTreeWalk(root);
cout << endl;
}
void inorderTreeWalk(binNode<T> * node) {
if (node != NULL) {
inorderTreeWalk(node->left);
cout << node->getKey() << '\t';
inorderTreeWalk(node->right);
}
}
};
Here is the main function ( #inlude
is not included)
int main() {
Tree<int> T (new binNode<int>(10));
// Tree<int> T = new binNode<int>(10);
T.addNode(new binNode<int> (11));
T.addNode(new binNode<int> (9));
T.addNode(new binNode<int> (8));
T.addNode(new binNode<int> (12));
T.printOrdered();
}
That's not a reference / dereference of a pointer , it's a reference to a pointer. It is necessary because...
void addNode(binNode<T> *&x, binNode<T> * node) {
if (x == NULL) {
x = node; // ...here...
return;
} else // ...
...you are assigning to the parameter x
.
If you hadn't passed the pointer x
by reference , you would assign to the local copy of the parameter:
void addNode(binNode<T> * x, binNode<T> * node) {
if (x == NULL) {
x = node; // this acts on the local copy only, and thus does nothing.
return;
} else // ...
Via the pointer (without the reference), you get a local copy of the address. Which means you can manipulate the value behind the pointer (in this case *x) which would change. But if you change the address itself, the address would behave like a local copy and you lose the address-changes after leaving the method.
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