C Programming - Complicated Declarations

6.
What will be the output of the program?
#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;
}
4, 4, 4
2, 4, 4
4, 4, 2
2, 4, 8
Answer: Option
Explanation:
No answer description is available. Let's discuss.

7.
What will be the output of the program?
#include<stdio.h>
typedef unsigned long int uli;
typedef uli u;

int main()
{
    uli a;
    u b = -1;
    a = -1;
    printf("%lu, %lu", a, b);
    return 0;
}
4343445454, 4343445454
4545455434, 4545455434
4294967295, 4294967295
Garbage values
Answer: Option
Explanation:

The system will treat the negative numbers with 2's complement method.

For 'long int' system will occupy 4 bytes (32 bits).

Therefore,

Binary 1 : 00000000 00000000 00000000 00000001

To represent -1, system uses the 2's complement value of 1. Add 1 to the 1's complement result to obtain 2's complement of 1.

So, First take 1's complement of binary 1 (change all 0s to 1s and all 1s to 0s)

1's complement of Binary 1:

11111111 11111111 11111111 11111110

2's complement of Binary 1: (Add 1 with the above result)

11111111 11111111 11111111 11111111


In HexaDecimal

11111111 11111111 11111111 11111111 = FFFF FFFF FFFF FFFF

In Unsigned Integer

11111111 11111111 11111111 11111111 = 4294967295.


8.
What will be the output of the program in DOS (Compiler - Turbo C)?
#include<stdio.h>
double i;

int main()
{
    (int)(float)(char) i;
    printf("%d",sizeof(i));
    return 0;
}
4
8
16
22
Answer: Option
Explanation:
No answer description is available. Let's discuss.

9.
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;
}
4, 4, 4
4, 2, 2
2, 8, 4
2, 4, 8
Answer: Option
Explanation:
No answer description is available. Let's discuss.

10.
What will be the output of the program?
#include<stdio.h>

int main()
{
    struct s1
    {
        char *z;
    int i;
    struct s1 *p;
    };
    static struct s1 a[] = {{"Nagpur", 1, a+1} , {"Chennai", 2, a+2} , 
                            {"Bangalore", 3, a} };

    struct s1 *ptr = a;
    printf("%s,", ++(ptr->z));
    printf(" %s,", a[(++ptr)->i].z);
    printf(" %s", a[--(ptr->p->i)].z);
    return 0;
}
Nagpur, Chennai, Bangalore
agpur, hennai, angalore
agpur, Chennai, angalore
agpur, Bangalore, Bangalore
Answer: Option
Explanation:
No answer description is available. Let's discuss.