Find Output Of Program

Question 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;
}``````

A). 4, 4, 4
B). 2, 4, 4
C). 4, 4, 2
D). 2, 4, 8
Answer : Option A

Question 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;
}``````

A). 4343445454, 4343445454
B). 4545455434, 4545455434
C). 4294967295, 4294967295
D). Garbage values
Answer : Option C

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
11111111 11111111 11111111 11111111 = FFFF FFFF FFFF FFFF
In Unsigned Integer
11111111 11111111 11111111 11111111 = 4294967295.

Question 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;
}``````

A). 4
B). 8
C). 16
D). 22
Answer : Option B

Question 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;
}``````

A). 4, 4, 4
B). 4, 2, 2
C). 2, 8, 4
D). 2, 4, 8
Answer : Option B

Question 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;
}``````

A). Nagpur, Chennai, Bangalore
B). agpur, hennai, angalore
C). agpur, Chennai, angalore
D). agpur, Bangalore, Bangalore
Answer : Option D