Question

Hey there I’m using a binary search method to find a number in an array, and I’m getting a StackOverflow error when looking for a number other than the one in the middle.

Here is my code:

public static  <T extends Comparable< ? super T>>
    int find(T [] a, T x, int low, int high){
    if(low>high)
        throw new IllegalArgumentException();
    int tmp = (high-low)/2;

    if(a[tmp].compareTo(x)==0)
        return tmp;
    else if(a[tmp].compareTo(x)>0)
        find(a,x,tmp,high);
    else if(a[tmp].compareTo(x)<0)
        find(a,x,low,tmp);
    return -1;
}

Also, if I try to look for a number under tmp, it returns -1.
I feel like I’m missing something but can’t figure out what.
Thanks in advanced!

Answer 1

This is the problem:

if(a[tmp].compareTo(x)>0)
    find(a,x,tmp,high);
else if(a[tmp].compareTo(x)<0)
    find(a,x,low,tmp);

You should be using tmp + 1 in the first case and tmp - 1 in the second. Otherwise if you’ve got (say) low = 0, high = 1 then you’ll potentially end up perpetually calling with the same arguments; tmp will end up being 0, and if x is more than a[0] you’ll just call find(a, x, 0, 1) again.

That’s my gut feeling, anyway. You should really log what happens in terms of the low/high values being used – I’m sure you’ll see some sort of repetition before it croaks.

EDIT: You’ve also got the comparison round the wrong way. a[tmp].compareTo(x) will return a value less than 0 if a[tmp] is less than x – i.e. you ought to look later in the array, not earlier.

EDIT: Currently your “exit with -1” code is broken – you’ll only ever return -1 if a[tmp].compareTo(x) returns a non-zero value which is neither above zero nor below zero. I challenge you to find such an integer 🙂 (It would also do it if compareTo were unstable, but that’s a separate issue…)

One option is to detect if high == low – at that point, if you haven’t hit the right value, you can return -1:

int comparison = a[tmp].compareTo(x); // Let's just compare once...
if (comparison == 0) {
   return tmp;
}
// This was our last chance!
if (high == low) {
   return -1;
}
// If a[tmp] was lower than x, look later in the array. If it was higher than
// x, look earlier in the array.
return comparison < 0 ? find(a, x, tmp + 1, high) : find(a, x, low, tmp - 1);