What will be the output of the program under DOS?

#include<stdio.h>

int main()
{
    char huge *near *far *ptr1;
    char near *far *huge *ptr2;
    char far *huge *near *ptr3;
    printf("%d, %d, %d\n", sizeof(ptr1), sizeof(**ptr2), sizeof(ptr3));
    return 0;
}

Negative numbers are treated with 2's complement method.

1's complement: Inverting the bits ( all 1s to 0s and all 0s to 1s)
2's complement: Adding 1 to the result of 1's complement.

Binary of 1(2byte)     :  0000 0000 0000 0001
Representing -1:
1s complement of 1(2byte)    : 1111 1111 1111 1110
Adding 1 to 1's comp. result : 1111 1111 1111 1111
Right shift 1bit(-1>>1): 1111 1111 1111 1111 (carry out 1)
Hexadecimal            : f   f    f    f
(Filled with 1s in the left side in the above step)

Note:

1. Fill with 1s in the left side for right shift for negative numbers.
2. Fill with 0s in the right side for left shift for negative numbers.
3. Fill with 0s in the left side for right shift for positive numbers.
4. Fill with 0s in the right side for left shift for positive numbers.

enum takes the format like {0,1,2..) so pass=0, fail=1, atkt=2

stud1 = pass (value is 0)

stud2 = atkt (value is 2)

stud3 = fail (value is 1)

Hence it prints 0, 2, 1

Step 1: char ch; ch = 'A'; here variable ch is declared as an character type an initialized to 'A'.

Step 2: printf("The letter is"); It prints "The letter is".

Step 3: printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);

The ASCII value of 'A' is 65 and 'a' is 97.

Here

=> ('A' >= 'A' && 'A' <= 'Z') ? (A + 'a' - 'A'):('A')

=> (TRUE && TRUE) ? (65 + 97 - 65) : ('A')

=> (TRUE) ? (97): ('A')

In printf the format specifier is '%c'. Hence prints 97 as 'a'.

Step 4: printf("Now the letter is"); It prints "Now the letter is".

Step 5: printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');

Here => ('A' >= 'A' && 'A' <= 'Z') ? ('A') : (A + 'a' - 'A')

=> (TRUE && TRUE) ? ('A') :(65 + 97 - 65)

=> (TRUE) ? ('A') : (97)

It prints 'A'

Hence the output is

The letter is a
Now the letter is A

Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf("%d\n", num); It prints the value of variable num '30'
Step 3: Hence the output of the program is '30'

Yes. If a function contains two return statements successively, the compiler will generate "Unreachable code" warnings.

Example:

#include<stdio.h>
int mul(int, int); /* Function prototype */

int main()
{
    int a = 4, b = 3, c;
    c = mul(a, b);
    printf("c = %d\n", c);
    return 0;
}
int mul(int a, int b)
{
   return (a * b);
   return (a - b); /* Warning: Unreachable code */
}

Output:
c = 12