What will happen if in a C program you assign a value to an array element whose subscript exceeds the size of array?

What does the following declaration mean?
int (*ptr)[10];

In C, if you pass an array as an argument to a function, what actually gets passed?

What will be the output of the program ?

#include<stdio.h>

int main()
{
    int a[5] = {5, 1, 15, 20, 25};
    int i, j, m;
    i = ++a[1];
    j = a[1]++;
    m = a[i++];
    printf("%d, %d, %d", i, j, m);
    return 0;
}

What will be the output of the program ?

#include<stdio.h>

int main()
{
    static int a[2][2] = {1, 2, 3, 4};
    int i, j;
    static int *p[] = {(int*)a, (int*)a+1, (int*)a+2};
    for(i=0; i<2; i++)
    {
        for(j=0; j<2; j++)
        {
            printf("%d, %d, %d, %d\n", *(*(p+i)+j), *(*(j+p)+i), 
                                    *(*(i+p)+j), *(*(p+j)+i));
        }
    }
    return 0;
}

What will be the output of the program ?

#include<stdio.h>

int main()
{
    void fun(int, int[]);
    int arr[] = {1, 2, 3, 4};
    int i;
    fun(4, arr);
    for(i=0; i<4; i++)
        printf("%d,", arr[i]);
    return 0;
}
void fun(int n, int arr[])
{
    int *p=0;
    int i=0;
    while(i++ < n)
        p = &arr[i];
    *p=0;
}

What will be the output of the program ?

#include<stdio.h>
void fun(int **p);

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0};
    int *ptr;
    ptr = &a[0][0];
    fun(&ptr);
    return 0;
}
void fun(int **p)
{
    printf("%d\n", **p);
}

What will be the output of the program ?

#include<stdio.h>

int main()
{
    static int arr[] = {0, 1, 2, 3, 4};
    int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
    int **ptr=p;
    ptr++;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    *ptr++;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    *++ptr;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    ++*ptr;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    return 0;
}

What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

#include<stdio.h>

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
    printf("%u, %u\n", a+1, &a+1);
    return 0;
}

What will be the output of the program in Turb C (under DOS)?

#include<stdio.h>

int main()
{
    int arr[5], i=0;
    while(i<5)
        arr[i]=++i;

    for(i=0; i<5; i++)
        printf("%d, ", arr[i]);

    return 0;
}