What will be the output of the program in 16-bit platform (under DOS)?

#include<stdio.h>

int main()
{
    struct node
    {
        int data;
        struct node *link;
    };
    struct node *p, *q;
    p = (struct node *) malloc(sizeof(struct node));
    q = (struct node *) malloc(sizeof(struct node));
    printf("%d, %d\n", sizeof(p), sizeof(q));
    return 0;
}

Step 1: int arr[1]={10}; The variable arr[1] is declared as an integer array with size '2' and it's first element is initialized to value '10'(means arr[0]=10)

Step 2: printf("%d\n", 0[arr]); It prints the first element value of the variable arr.

Hence the output of the program is 10.

Step 1: int a[5] = {5, 1, 15, 20, 25}; The variable arr is declared as an integer array with a size of 5 and it is initialized to

a[0] = 5, a[1] = 1, a[2] = 15, a[3] = 20, a[4] = 25 .

Step 2: int i, j, m; The variable i,j,m are declared as an integer type.

Step 3: i = ++a[1]; becomes i = ++1; Hence i = 2 and a[1] = 2

Step 4: j = a[1]++; becomes j = 2++; Hence j = 2 and a[1] = 3.

Step 5: m = a[i++]; becomes m = a[2]; Hence m = 15 and i is incremented by 1(i++ means 2++ so i=3)

Step 6: printf("%d, %d, %d", i, j, m); It prints the value of the variables i, j, m

Hence the output of the program is 3, 2, 15

Step 1: int arr[]={2, 3, 4, 1, 6}; The variable arr is declared as an integer array and initialized.

Step 2: printf("%u, %u, %u\n", arr, &arr[0], &arr); Here,

The base address of the array is 1200.

=> arr, &arr is pointing to the base address of the array arr.

=> &arr[0] is pointing to the address of the first element array arr. (ie. base address)

Hence the output of the program is 1200, 1200, 1200

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions.

Step 2: printf("%u, %u\n", a+1, &a+1);

The base address(also the address of the first element) of array is 65472.

For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480

Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496

Hence the output of the program is 65480, 65496

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'

Since C is a compiler dependent language, in Turbo C (DOS) the output will be 2, but in GCC (Linux) the output will be 4.

Because ++ or -- cannot be done on enum value.