Cannot understand why variable i is set to zero? - c

Question

I cannot understand why i is set to 0 right after array is initialized to zero.
The program is working fine because I have reinitialized value of k to i.
But I could not find out why i becomes 0.
And why memset() is clearing the array, or setting the array to 0?
#include <stdio.h>
#include <stdlib.h>
#include <windows.h>
int main()
{
long long int i = 123456789;
long long int j = 987654321;
long long int cnt = 0;
int array[9] ;
int xyz, k, x, rem, se;
xyz = 0;
// printf("I = %llf", i);
for (i; (i < j) && (cnt < 100000); i++)
{
k = i;
x = 0;
for (se = 0; se <= 9; se++)
{
array[se] = 0;
}
/*************************************************/
i = k; // Here i becomes zero. Why?
/************************************************/
//memset(array, 0, 9);
while(k != 0)
{
rem = k % 10;
for(se = 0; se <= 9; se++)
{
if(rem == array[se])
{
xyz = 1;
break;
}
}
if(rem == array[se])
{
xyz = 1;
break;
}
array[x++] = rem;
k = k / 10;
}
if (xyz != 1)
{
cnt++;
// printf("Cnt = %d ", cnt);
// printf("The value i is = %lld\n", i);
// Sleep(10);
}
xyz = 0;
// printf("The value i is = %lld\n", i);
// printf("Cnt = %d \n", cnt);
fflush(stdin);
}
printf("The value i is = %lld \n", i-1);
return 0;
}

Answer 1

You have a buffer overflow; since the buffer is on the stack, it might be regarded as a form of Stack Overflow†.
int array[9]; // Elements array[0] .. array[8]
...
for (se = 0; se <= 9; se++)
{
array[se] = 0;
}
It must be k that is being overwritten with the extra 0; i is assigned 0 because that's the value in k. You invoke 'undefined behaviour' when you write outside the boundaries of an array, as you did here. Undefined behaviour means that anything can happen and it is OK. Sometimes, it will seem to work; sometimes, there'll be unexpected side effects. Avoid 'undefined behaviour' at all costs.
The idiomatic for loop is:
for (se = 0; se < 9; se++)
array[se] = 0;
Note the < instead of <=.
† There are those who would disagree. See the comments.
You also ask about the (commented out) call to memset():
//memset(array, 0, 9);
The third parameter to memset() is the size in bytes of the area of memory to be set. You are setting 9 bytes out of a total of (probably) 36 in the array, which is unlikely to be what you wanted.
Here, the array is defined in the same function so it is safe and sensible to write:
memset(array, 0, sizeof(array));
If array was a parameter passed to a function, that would not work correctly; you would need a different size. For example:
void somefunc(int array[], int num)
{
...
memset(array, 0, num * sizeof(array[0]));
...
}

Answer 2

You are overwriting the end of the array.
int array[9] ;
for(se = 0; se <= 9; se++)
{
array[se] = 0;
}
In C, arrays are indexed from 0, so the loop goes one step to far. The idiomatic syntax is:
for(se = 0; se < 9; se++)
In other words, use < with the number of elements as the value. Often written to remove the scary numerical constant, like so:
for(se = 0; se < sizeof array / sizeof *array; se++)
This uses the sizeof operator to automatically (at compile-time) compute the proper number of elements, so that if you change the array definition the loop remains correct.

Answer 3

One way to check this is to check the pointer of x and array[9] by printing &x and &array[9].
If they are same, you are surely overwriting as mentioned.