Simple QuickSort Algorithm giving Stack Overflow E

Question

My friend has a small problem and I'm at the end of my knowledge. He wrote a Simple (he got it in school) QuickSort Algorithm And it produces a StackOverflow Error. I know it means it calls itself recursive too many times somewhere, but I can't get the logical Error - please help me!

Here is the code (I'm leaving out some code as that is only to show it in 2 text areas):

int array [] = new int [10];
...
 public static void quicksort (int array[],int l,int r){
int i = l;
int j = r;
int mitte = array [(l+r)/2];

while (i<j) {
  while  (array[i]<mitte) {
    i++;
  } // end of if
  while  (mitte<array[i]) {
    j--;
  } // end of if
  if (i<=j) {
    int merke =array[i];
    array[i] = array [j];
    array [j] = merke ;
    i++;
    j--;
  } // end of if
  if (i<j) {
    quicksort(array,l,j);
  } // end of if
  if (l<r) {
    quicksort(array,l,r);
  } // end of if
} // end of while
}

It is called like this:

 quicksort(array,0,9);

But, if We call it and the 2 numbers are the same, it gives no Overflow.

If more code is needed, here is the full Code on pastebin: http://pastebin.com/cwvbwXEu

Answer 1

First, you have an infinite loop here:

while  (mitte<array[i]) {
    j--;
  } // end of if

It needs to be array[j]

Secondly (and leading to the infinite recursion), in the second call to quicksort

if (l<r) {
  quicksort(array,l,r);
} // end of if

In recursion, you always need to shorten the range that you call yourself with, or else it'll be infinite. I haven't worked out exactly what you're doing, but I think you mean:

if (i<r) {
   quicksort(array,i,r);
 } // end of if

Answer 2

This call:

if (l<r) {
  quicksort(array,l,r);
}

recursively calls quicksort with the same arguments that were passed in, rather than calling with a smaller subproblem to solve. It will therefore infinitely recurse.

Answer 3

if (l<r) 
quicksort(array,l,r);

Isnt l always less than r? this will cause an infinite recursion which is why you dont get an overflow if both values are the same.