Please I need help having this compiled

[shaniwaketech]

Return the digit at the user specified index of an integer.  If the integer has n digits and the index is NOT in the range 0 <=index  <n  return -1   Start the digit numbering at 0.  Example,  if user input  is  4 (index) and the integer equals 123456790  the return value for the function is 5  (start index at 0) ;    if user input  is  40 (index) and the integer equals 123456790  the return value for the function is -1  



#include <iostream>
#include <istream>
#include <cstdlib>
#include <cassert> 
#include <string>
using namespace std;

int getIndex(int, int);
int main()
{
	int index = 0;
	int i = 0;
	cout << "Enter integer ";
	cin >> i;    	// 123466  sample user input 
	cout << "Enter index ";
	cin >> index;    	// 4 sample user input 
	
	cout << " The digit at index " << index << " is " << getIndex(index, i) << endl;

	system("pause");
	return 0;
} // main function finishes.

int getIndex(int index, int number )
{
	int returning = 0;
	int counter = 0;
	while (number)
	{
		number = number / 10;
		counter++;
	}
	cout << counter << endl;
	for (unsigned int i = 0; i < counter; counter--)
	{
if (number.at(i) == index)  // This is where I’m basically stuck  .at() didn’t compile 
	returning = counter;
	else
	returning = -1;
	}
	return returning;

}

[Ispil]

Well, let me make your life much, much easier:

Take the user input as a string. Yes, it is a number. However, for a problem like this, you would be much better off dealing with strings as character arrays- that way, you could just say something like i[index] for the digit rather than having to do all of this other stuff.

Also, .at(i) won't work here. What you want to do is modulus division by 10- if it isn't equal, divide the whole thing by 10.

[shaniwaketech]

Yeah unfortunately, the labs requests a pass by value to an integer. Otherwise, if it was a string as you said, It would have been easier.

I've done the modulus division ( N = number%10 ) but there the count will start from the opposite side(integer leftover), thus It will not give me the exact index. What you think?

[Ispil]

You will have to work backwards for this. There's no way to get a value starting at the largest digit- not unless you do some horrible modulus-division tree thing that would just be abysmal. You'll have to...

Wait. You could create a second number that is the exact reverse of the first, then use the index that way.

[shaniwaketech]

Oh yes. many many thanks. reversing it and then comparing the left from integers.
thank you for you help :)