Question: Here *len* is at A[10] and *i* is at A[11]. Is there a way to catch these errors??
I tried compiling with gcc -Wall -W but no warnings are displayed.

```
int main()
{
  int A[10];
  int i, len;
  len = sizeof(A) / sizeof(0[A]);
  printf("Len = %d\n",len);
  for(i = 0; i < len; ++i){
    A[i] = i*19%7;
  }
  A[i] = 5;
  A[i + 1] = 6;
  printf("Len = %d i = %d\n",len,i);
  return 0;
}

```

**Output :**
Len = 10
Len = 5 i = 6

Answer: 
The array allocation of A is adjacent in memory to i and len. Remember that when you address via an array, it's exactly like using pointers, and you're walking off the end of the array, bumping into the other stuff you put there.

C by default does not do bounds checking. You're expected to be careful as a programmer; in exchange you get speed and size benefits.

Usually external tools, like lint, will catch the problems via static code analysis. Some libraries, depending on compiler vendor, will add additional padding or memory protection to detect when you've walked off the end.

Lots of interesting, dangerous, and non-portable things reside in memory at "random spots." Most of the house keeping for heap memory allocations occur in memory locations before the one the compiler gives you.

The general rule is that if you didn't allocate or request it, don't mess with it.
The short answer is that local variables are al-located on stack, and indexing is just like \*(ptr + index). So it could happen that the room for int y[N] is adjacent to the room for another int x; e.g. x is located after the last y. So, y[N-1] is this last y, while y[N] is the int past the last y, and in this case, by accident, it happens you get x (or whatever in your practical example). But it is absolutely not a sure fact what you can get going past the bounds of an array and so you can't rely on that. Even though undetected, it's a "index out of bound error", and a source of bugs.