It is safe to use i+1 in O(n) searching? - c++

Question

I'm asking myself if it is safe to use i+1 to check if the next number is the same as current... Example :
int search(int el, int* a, int n) {
if(a == NULL && n<0)
{
return NULL;
}
for(int i=0; i<n; i++)
{
if((a[i] == el )&& (a[i+1] != el)) // here I check next element
{
return i; // if I find it, return a position.
}
}
return -1; // else return -1
}
If we have an array with length 4, then a[i+1] will be passed an array length, but program will still work.

Answer 1

No, accessing elements out of bounds is Undefined Behavior.
Your program may "seem" to work, but you cannot expect anything out of it. It could work on your machine, and crash on another one. Or it could work on all machines now, but not tomorrow. You should check if the index is out-of-bounds before using the subscript operator.
You may consider changing your cycle so that it never accesses out-of-bounds elements:
for (int i = 0; i < n - 1; i++)
// ^^^^^
Then, you would have to take care of the last element in the array separately, as a special case.

Answer 2

That's not correct, when i reaches its last value (n-1) you check the value of a nonexistent element (the C standard allows you to have a pointer to one-after-last element, but not to dereference it).
You can fix your code like this:
if((a[i] == el ) && ((i == n-1) || (a[i+1] != el)))

Answer 3

If n is the last element in the array, then i + 1 is safe in this case. If n is the number of elements in the array, i + 1 may appear to work most of the time, but it is not safe.
In that case, you're accessing an element outside the bounds of the array, which may do anything from giving you incorrect results to crashing your program. Most of the time it will appear to work, though, making the problem very hard to debug when it happens.

Answer 4

It is not clear to me what you're asking. Even if array access is expensive, accessing both a[i] and a[i+1] remains of O(N) complexity. What you can't do is adding complexity that's based on i (for example an additional loop from 0 to i), or modifying i (for example decrementing it based on some condition).
The problem, as others have already pointed out, is that the last element is compared to a nonexistent datum: either you'll get an error and a possible crash, or the program will appear to be working - and actually work most of the time - until the time where that unknown and possibly random last-and-one item will trigger the comparison, and yield an unexpected result.
You should check that the array size is at least 1 (that ought to be a special case anyway, can't run a compare on a single element!) and then loop only up to n-1. Or you could save the previous value in a temporary variable; depending on the platform, this will be a (possibly much faster) register, or a (possibly much slower) stack location. In most cases I'd just state my intention of comparing with the next element, as you did, and let the compiler sort it out.

Answer 5

No.
int main(void)
{
char pumpkin[8];
int a[4];
int i, p;
a[0] = 3760;
a[1] = 100001;
a[2] = 595959;
a[3] = 1886221680;
pumpkin[0] = 'p';
pumpkin[1] = 'u';
pumpkin[2] = 'm';
pumpkin[3] = 'p';
for (i = 0; i < 4; i++) {
p = search(a[i], a, 4);
if (p >= 0)
printf("Found it at position %d: %d.\n", p, a[i]);
else
printf("Value not found.\n");
}
return 0;
}
In my machine:
$ gcc -std=c11 -o boundserror boundserror.c
$ ./boundserror
Found it at position 0: 3760.
Found it at position 1: 100001.
Found it at position 2: 595959.
Value not found.
What happened ? Compiler wrote value 1886221680 both in a[3] and where a[4] would be if it existed. About the program working on your machine: read about the works on my machine concept.
http://www.codinghorror.com/blog/2007/03/the-works-on-my-machine-certification-program.html