CODEWITHSUNDEEP
c++
CodeDoctorJL
CodeDoctorJL

Reputation: 1254

what does the void(*f) mean in -> "methodName( void (*f)(const Datatype&));"?

So, I'm looking at one of the codes my professor gave me, but I don't know what the void(*f) means, could anyone clarify?

template<class T>
void BinaryTree<T>::inorder( void (*f)(const T&), BTNode<T> *node ) const //<-- right here
{
  if (!node)
    return;
  inorder(f, node->left); 
  f(node->elem);
  inorder(f, node->right);
}

Upvotes: 0

Views: 423

Answers (3)

Timothy Shields
Timothy Shields

Reputation: 79491

In C++ a pointer F to a function that maps type X to type Y is denoted as Y (*F)(X).

Upvotes: 0

vlad_tepesch
vlad_tepesch

Reputation: 6883

this is afunction pointer. the function gets a unction pointer as parameter.

but since this is already template c++ i would try to avoid this. It would be better to use functors instead. (but may be its a function pointer for a reason)

Upvotes: 0

juanchopanza
juanchopanza

Reputation: 227438

It is a pointer to a function returning void, and taking const T& parameter. The name of the pointer if f.

void foo(const T&);   // a function declaration

void (*f)(const T&);  // function pointer

f = &foo;             // Can assign &foo to f, return type and signature match

Upvotes: 5

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