


1.
How will you free the allocated memory ?
2.
What is the similarity between a structure, union and enumeration?
3.
What will be the output of the program ?
#include int main() { union a { int i; char ch[2]; }; union a u; u.ch[0]=3; u.ch[1]=2; printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i); return 0; } 4.
What will be the output of the program ?
#include int main() { union var { int a, b; }; union var v; v.a=10; v.b=20; printf("%d\n", v.a); return 0; } 5.
What will be the output of the program ?
#include int main() { struct value { int bit1:1; int bit3:4; int bit4:4; } bit={1, 2, 13}; printf("%d, %d, %d\n", bit.bit1, bit.bit3, bit.bit4); return 0; } Answer & Solution
Answer: (C) 1, 2, 3
Solution: Note the below statement inside the struct: int bit1:1; > 'int' indicates that it is a SIGNED integer. For signed integers the leftmost bit will be taken for +/ sign. If you store 1 in 1bit field: The left most bit is 1, so the system will treat the value as negative number. The 2's complement method is used by the system to handle the negative values. Therefore, the data stored is 1. The 2's complement of 1 is also 1 (negative). Therefore 1 is printed. If you store 2 in 4bits field: Binary 2: 0010 (left most bit is 0, so system will treat it as positive value) 0010 is 2 Therefore 2 is printed. If you store 13 in 4bits field: Binary 13: 1101 (left most bit is 1, so system will treat it as negative value) Find 2's complement of 1101: 1's complement of 1101 : 0010 2's complement of 1101 : 0011 (Add 1 to the result of 1's complement) 0011 is 3 (but negative value) Therefore 3 is printed. . 