Explanation of output of this C code - c

Question

Here is my code:
#include <stdio.h>
#include <stdlib.h>
int main()
{
void check(int n, int arr[]);
int arr[] = {1, 2, 3, 4};
int i;
check(4, arr);
for (i = 0; i < 4; i++) {
printf("%d\n", arr[i]);
}
return 0;
}
void check(int n, int arr[]) {
int i = 0;
int *p = 0;
while (i++ < n)
p = &arr[i];
*p = 0;
}
I need an explanation for the output.
The original question I was asked, and the expected multiple-choice answers, are:

Answer 1

Please post your actual code, not what you intended to type. Actually copy-paste your real code.
Because you typed it in wrong.
You either put extra {} in here:
while(i++ < n) {
p = &arr[i];
*p = 0;
}
or you used a comma instead of a semicolon:
while(i++ < n)
p = &arr[i],
*p = 0;
and so the assignment to zero ran every time.
Edit to add: Yep, you put extra {} which the original question didn't have. So in your code, the "*p = 0" executes every time round the while loop, whereas the original question the "*p = 0" only executes once and clobbers some random data that is one past the end of the array.
(By the way, the answer to the original question is actually "it is undefined behaviour; the program doesn't necessarily print anything. Valid behaviours include printing 1 2 3 4, printing 42 42 42 42, crashing, and formatting your hard drive.")

Answer 2

I'm not sure what choices you're talking about, however the cause of your output is here:
int i=0;
int *p=0
while(i++<n)
{
p=&arr[i];
*p=0;
}
Variable 'i' starts at 0 but you increment it before entering the loop so the first index of the array is ignored. You then set a pointer to array index 'i' and then immediately dereference the pointer and set the value to 0;
Because of this any array you pass will always retain its first value whilst every other value will be zero.
if you want to include the first index of the array you'd be much better off doing:
for (int i = 0; i < n; ++i)
{
// stuff
}
With this, 'i' is not incremented until after the code between the braces has been executed.

Answer 3

In check(), i gets incremented after the comparison, but before the first statement inside. So the zero (first) element of the array is never set to 0, like the rest are. arr's 1 stays 1, and 2, 3, & 4 each become 0.
EDIT:
The OP code has changed since the version I discussed. It's a whole new problem now.

Answer 4

Some things you should know.
First:
you should declare your functions if you declare/define them after main.
Second:
when you declare an Array, the array starts from 0 to n and not from 1 to n.
So, if you declare int arr = {1,2,3,4} then you have arr[0],1,2,3 and not arr[1],2,3,4.
Third:
you should avoid code like:
while (i++ < n) {
p = &arr[i];
*p = 0;
}
And use:
while (i < n) {
p = &arr[i];
*p = 0;
i++;
}
Fourth:
What exactly did you expected from this:
int *p = 0;
Anyway, you just try to access a memory location that not belong to you.