recursive array counter

[zwA4jE8b]

I don't need help with the assignment, I am happy with it.
I am just wondering if there is a way to start at A[0] instead of A[l](l being the max position) without using variables or more parameters?

//Michael Ervin - CS3 - Assignment 2 - Recursive occurence counter - 8/24/2011 - 8:50pm

#include <iostream>

int occurences(int A[], int l, int t)
{
	if(l >= 0)
		if(A[l] == t)
			return 1 + occurences(A, l - 1, t);
		else
			return 0 + occurences(A, l - 1, t);
	return 0;
}

int main()
{
	int A[10] = {2, 5, 2, 6, 7, 8, 12, 2, 2, 0};
	std::cout << "The number 2 occurs " << occurences(A, 9, 2) << " times";
	std::cin.get();
	return 0;
}

i just can't come up with a method

[shacktar]

A std::vector could do the trick if you're allowed to use it.

[helios]

Since you have the length of the input, it can be done with the code you already have. You just need to move the pointer you pass to the function.

[shacktar]

I thought of that as well, however I realized that you'd have to declare the size as a const somewhere and that would conflict with the requirement of "without using variables". Maybe if CreativeMFS isn't as strict in that regard?

[zwA4jE8b]

thank you helios. I often forget simple things (array variable is a pointer to the base)

int occurences(int A[], int l, int t)
{
	if(l >= 0)
		if(*A == t)
			return 1 + occurences(++A, l - 1, t);
		else
			return 0 + occurences(++A, l - 1, t);
	return 0;
}