what this function does..

[transgalactic2]

int what2(int *arr1,int *arr2,int i)
{
   if(*arr1==-1) return 0;
   if (i)
       return *arr1-what2(arr+1,arr2+1,1-i);
   else
       return *arr2-what2(arr+1,arr2+1,1-i);
}

i was told that both arrays end by -1
i tried arr1: 1,2,-1
arr2: 3,4,-1
i=2
so i get:
1-what2(2,4,-1)==>1-(4-what(-1,-1,2))=
it never ends??

[brewbuck]

If i == 0, then 1-i == 1. If i == 1, then 1-i == 0. I suspect that the only valid values for i are 0 and 1, and this is acting as a toggle to switch between two options at each level. The whole things ends when the end of the first array is reached.

[transgalactic2]

so its if it i=1
then we sum on the diagonal from arr1
arr1: 1 2 6 -1
arr2: 3 4 5 -1
if starts from i=1
1-4+6
if starts from i=0
3-2+5
correct?

what is the word definition for this diagonal thing??

[Jesdisciple]

For your original question, I suggest that you loop through the source arrays to see if -1 ever occurs. If it does, print something and stop looping. If you get a segfault while doing that, you've run out of the array and into some other memory you're not supposed to be looking at - i.e., -1 isn't in the array.

But your logic looks right to me, and the what2 function is recursive.