What will be the output of the following pseudocode?Integer p, q, rset p= 1, q= 4, r= 2if ( p&q)r =p&qend ifprint r- p - q[ Note- & bitwise AND - The bitwise AND operator ( & ) compares each bit of the first operand to the corresponding bit of the second operand if both bits are 1 the corresponding result bit is set to 1. otherwise, the corresponding result bit is set to 0]
Question
What will be the output of the following pseudocode?Integer p, q, rset p= 1, q= 4, r= 2if ( p&q)r =p&qend ifprint r- p - q[ Note- & bitwise AND - The bitwise AND operator ( & ) compares each bit of the first operand to the corresponding bit of the second operand if both bits are 1 the corresponding result bit is set to 1. otherwise, the corresponding result bit is set to 0]
Solution
The pseudocode is using bitwise AND operation. Let's break it down:
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The variables p, q, and r are set to 1, 4, and 2 respectively.
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The bitwise AND operation is performed between p and q. The binary representation of p (1) is 0001 and of q (4) is 0100. The bitwise AND operation between p and q results in 0000 because there are no positions where both p and q have a bit set to 1. So, p&q results in 0.
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The if condition checks if the result of p&q is true. In this case, p&q is 0, which is considered as false in boolean context. Therefore, the code inside the if block is not executed and r remains 2.
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Finally, the pseudocode prints the value of r - p - q, which is 2 - 1 - 4 = -3.
So, the output of the pseudocode will be -3.
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