i am currently implementing a binary tree in c++ and i want to traverse it with a function called in_order().
is there any way to pass a function as an argument, so that i can do things like below (without having to write the code to traverse the list more than once)?
struct tree_node; // and so on
class  tree;      // and so on
void print_node () {
  // some stuff here
}
// some other functions
tree mytree();
// insert some nodes
mytree.in_order(print_node);
mytree.in_order(push_node_to_stack);
mytree.in_order(something_else);
                Yes, you can do this in a number of ways. Here are two common possibilities.
Old-style function pointers
class mytree
{
    // typedef for a function pointer to act
    typedef void (*node_fn_ptr)(tree_node&);
    void in_order(node_fn_ptr)
    {
        tree_node* pNode;
        while (/* ... */)
        {
        // traverse...
        // ... lots of code
        // found node!
            (*fnptr)(*pNode);
            // equivalently: fnptr(*pNode)
        }
    }
};
void MyFunc(tree_node& tn)
{
    // ...
}
void sample(mytree& tree)
{
    // called with a default constructed function:
    tree.inorder(&MyFunc);
    // equivalently: tree.inorder(MyFunc);
}
Using functors
With a template member, works with function pointers
class mytree
{
    // typedef for a function pointer to act
    typedef void (*node_fn_ptr)(tree_node&);
    template<class F>
    void in_order(F f)
    {
        tree_node* pNode;
        while (/* ... */)
        {
        // traverse...
        // ... lots of code
        // found node!
            f(*pNode);
        }
    }
};
struct ExampleFunctor
{
    void operator()(tree_node& node)
    {
        // do something with node
    }
}
void sample(mytree& tree)
{
    // called with a default constructed function:
    tree.inorder(ExampleFunctor());
}
                        Yes, you can use a function pointer as a parameter to in_order. You may also need to overload it, in case the passed functions' signatures don't match. For functions like print_node, declare in_order like this (provided its return type is void as well):
void tree::in_order( void (*)() )
{
   //implementation
}
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