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c++: pass function as parameter to another function

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);
like image 919
Patrick Oscity Avatar asked Nov 04 '09 11:11

Patrick Oscity


2 Answers

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());
}
like image 165
CB Bailey Avatar answered Sep 28 '22 14:09

CB Bailey


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
}
like image 37
Zoli Avatar answered Sep 28 '22 15:09

Zoli