Question

I’m learning C++ language and I’m trying to write BST, but something goes wrong.
I try to add element to empty tree, root is NULL, but after adding element root is still NULL despite of the fact that addiing was successful (I saw it in debug mode, node is set as tmp). I have no idea why it happens.

struct Node
{
    int data;
    Node* left;
    Node* right;
};

struct Tree
{
    Node* root;
};

Tree createTree()
{
    Tree tmp;
    tmp.root = NULL;
    return tmp;
}

void addToNode(Node* node, int value)
{
    Node* tmp = new Node;
    tmp->data = value;
    tmp->left = NULL;
    tmp->right = NULL;
    if(node == NULL)
        node = tmp;
    else if(value >= node->data)
        addToNode(node->right, value);
    else
        addToNode(node->left, value);
}

void add(Tree* tree, int value)
{
    addToNode(tree->root, value);
}

int _tmain(int argc, _TCHAR* argv[])
{
    Tree tree = createTree();
    add(&tree, 10);
    printf("%d", tree.root->data);
    scanf("%*s");
    return 0;
}

Answer 1

When you are passing your pointer into the function, you create a local version of the pointer. This local variable (node) does indeed point into the same memory that the outer pointer you were passing. However, any attempt to change this variable (not the memory it points to, but the pointer variable itself) will only change the local variable.

So your node points to the same memory location as your tree, but the node variable itself isn’t equal to the tree variable, so your changes are not visible from the outer function.

It sounds complicated, sorry, but it’s exacly the same thing as in this:

void foo( int a )
{
    a++;
}
int main()
{
    int var = 5;
    foo( var );
    std::cout << var;
}

Of course in this case the var will not change, it’s the a that is being changed inside the function.

To fix the issue, pass a reference to the pointer instead of the pointer itself:

void addToNode(Node*& node, int value)