Q1: 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);
}

ANS:A - 1

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; The variable a is declared as an multidimensional integer array with size of 3 rows 4 columns. Step 2: int *ptr; The *ptr is a integer pointer variable. Step 3: ptr = &a[0][0]; Here we are assigning the base address of the array a to the pointer variable *ptr. Step 4: fun(&ptr); Now, the &ptr contains the base address of array a. Step 4: Inside the function fun(&ptr); The printf('%d\n', **p); prints the value '1'. because the *p contains the base address or the first element memory address of the array a (ie. a[0]) **p contains the value of *p memory location (ie. a[0]=1). Hence the output of the program is '1'