Dealing with segmentation fault in recursive binary search

Question

How can I prevent a segmentation fault with a recursive binary search that does not find the number it it is looking for in the array.

int recursive_binary_search(int A[], int beg, int end, int key)
{
    if(end < beg)
        return -1;
    int mid = beg + end / 2;
    if(A[mid] == key)
        return mid;
    else if(A[mid] > key)
        return recursive_binary_search(A, beg, mid - 1, key);
    else
        return recursive_binary_search(A, mid + 1, end, key);
}

your mid is not really a mid - missing parenthesis around beg + end — Aleksander Fular
– Aleksander Fular, Commented Sep 15, 2013 at 18:37
Does 'end' denote the last entry in the list or is it - as with stl - the entry after the end? If the latter, shouldn't you be writing if(end <= beg)? — kfsone
– kfsone, Commented Sep 15, 2013 at 18:38
e.g. if you have one element in A, would beg = 0 and end = 1? — kfsone
– kfsone, Commented Sep 15, 2013 at 18:39

Deepu · Accepted Answer · 2013-09-15 18:36:49Z

1

The following code might be sufficient,

int recursive_binary_search(int A[], int beg, int end, int key)
{
    if(end < beg)
        cout<<"\nKey not Found";
    int mid = beg + end / 2;
    if(A[mid] == key)
        return mid;
    else if(A[mid] > key)
        return recursive_binary_search(A, beg, mid - 1, key);
    else
        return recursive_binary_search(A, mid + 1, end, key);
}

answered Sep 15, 2013 at 18:36

Deepu

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Comments

Mars · Accepted Answer · 2013-09-15 18:41:21Z

0

Turns out I forgot the () around the definition for mid.

Correcting the definition to

int mid = (beg + end) / 2;

fixed the problem.

answered Sep 15, 2013 at 18:41

Mars

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