CODEWITHSUNDEEP
c++templates
deeb
deeb

Reputation: 1382

g++ is giving "unresolved overloaded function type" with template parameters

Reduced to basic form, I'm trying to do something like this with a binary search tree:

template <class Item, class Key>
class bst
{
public:
    Item val;
    Key key;
    bst *left;
    bst *right;
private:
};

template <class Item, class Key, class Process, class Param>
void preorder_processing1(bst<Item, Key> *root, Process f, Param g)
{
    if (root == NULL) return;
    f(root->val, g);
    preorder_processing1(root->left, f, g);
    preorder_processing1(root->right, f, g);
}

template <class Item, class Key>
void print_path(const bst<Item, Key>* root, const Key target)
{
    if (root->key != target)
    {
        cout << root->key << " " << root->val << endl;
        if (target < root->key)
            print_path(root->right, target);
        else
            print_path(root->left, target);
    }
    else
        cout << root->key << " " << root->val << endl;
}

template <class Item, class Key>
void print_if_leaf(const bst<Item, Key>* test_node, const bst<Item, Key>* root)
{
    if (test_node->right == NULL && test_node->left == NULL)
    {
        print_path(root, test_node->key);
        cout << endl;
    }
}

template <class Item, class Key>
void print_paths_bst(const bst<Item, Key>* root)
{
    preorder_processing1(root, print_if_leaf, root);
}

When I call print_paths_bst, I get this: error: no matching function for call to ‘preorder_processing1(const bst<int, int>*&, <unresolved overloaded function type>, const bst<int, int>*&)’

I've tried forcing a cast on the preorder_processing1 call like 

preorder_processing1(root, print_if_leaf<Item, Key>, root);

...but this also hasn't resolved the problem.

Does anyone see what's wrong with this?

Upvotes: 0

Views: 3217

Answers (1)

Erik
Erik

Reputation: 91270

template <class Item, class Key, class Process, class Param>
void preorder_processing1(bst<Item, Key> *root, Process f, Param g)

This needs to take a const bst<Item, Key> *

Additionally, your call to f(root->val, g); will have problems - root->val isn't a bst<Item, Key> const *. You likely mean f(root, g)

Upvotes: 4

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