Using a function pointer as a function argument

Hello all,

I am trying to learn C++ on my spare time and am following the book "c++ primer plus" right now.. In one of the exercises I want to be able to use a function as one of the arguments to an other function. However, the function inside of the other functions arguments is not being called.. here is a piece of the code...

The prototype of the two functions:

Code:

int average_score(int ar[], int limit);
void display_scores(const int ar[], int limit, int (*avg_score)(int*, int));

The function call:

Code:

display_scores(golf_scores, num_of_scores, average_score);

and the definitions:

Code:

int average_score (int ar[], int limit)
{
	int average = 0;
	int x;
	for (x = 0; x < limit; x++)
	{
		average = average + ar[x];
	}
	average = average / limit;
	cout << "This is function average_score() " << endl ;
	return average;
}	

void display_scores(const int ar[], int limit, int (*avg_score)(int*, int))
{
	int x;
	for (x = 0; x < limit; x++)
	{
		cout << ar[x] << " ";
	}
	cout << endl << "The average score is " << avg_score << endl;
}

I know that it is not necessary to do it like this, but I would like to know why it doesn't work. I will attach the entire piece of code.. Thanks to anyone who can help me.. I worked on this problem for several hours tonight and could not figure it out.

That's because you're not calling the function:

Code:

cout << endl << "The average score is " << avg_score << endl;

You don't call a function with the name alone. You need to supply the arguments (or an empty set of parenthesis if there are none)
Try something like this:

Code:

cout << endl << "The average score is " << avg_score(ar, limit) << endl;

Function pointers are cool in C, but in C++, I would recommend using function objects which overload operator().

If you wanted that to be clearer:

Code:

typedef int (*func)(int*, int);

void display_scores(const int ar[], int limit, func avg_score){

}

third parameter is typedef'd