Test by Bikram | Programming | Test 2 | Question: 25

Question

Read the following program fragment:       

#include<stdio.h>
int main()
{
int p = 2, q = 5;
p = p^q;
q = q^p;
printf("%d%d",p,q);
return 0;
}

The output is:

• A. $5 \ 2$
• B. $2 \ 5$
• C. $7 \ 7$
• D. $7 \ 2$

Comments

I am not getting, why 2nd Printf is not giving 7

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That ^ is known as " Bitwise Exclusive OR Operator " .

The bitwise exclusive OR operator (^) compares each bit of its first operand to the corresponding bit of its second operand.

 If one bit is 0 and the other bit is 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.

p = 2 = 0010

q = 5 = 0101

p^q means = 0010 ^ 0101 = 0111 = 4+2+1 = 7

q^p means = 0101 ^ 0111 ( because now p is changed to 0111) =    0010 = 2

so output is 7,2 which is option D .

---

@aarati mishra

2nd Printf() is not giving 7 that's because p is no more 0010 it have changed to 0111 due to p^q in first assignment, now that changed p value we use in our 2nd assignment q^p .

Answer 1

^ is known as " Bitwise Exclusive OR Operator " .

The bitwise exclusive OR operator (^) compares each bit of its first operand to the corresponding bit of its second operand.

 If one bit is 0 and the other bit is 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.

p = 2 = 0010

q = 5 = 0101

p^q means = 0010 ^ 0101 = 0111 = 4+2+1 = 7

q^p means = 0101 ^ 0111 ( because now p is changed to 0111) =    0010 = 2

Hence  output is 7,2 which is option D .