What will be the output of the following pseudocode?integer pp, qq, rrset pp= 1, qq= 2, rr= 4if ( pp> 0 && qq> 0 && rr> 0 )pp= 1pp = pp >> pppp = pp >> ppelserr= 0qq= 1end ifprint pp + qq + rr[Note - >> - Bitwise right shift operator, it takes two numbers, right-shift the bits of the first operand the second operand decides the number of places to shift&& : Logical AND - The logical AND operator (&& ) returns the boolean value True (or 1) if either or both operands is True and returns False( or 0)]
Question
What will be the output of the following pseudocode?integer pp, qq, rrset pp= 1, qq= 2, rr= 4if ( pp> 0 && qq> 0 && rr> 0 )pp= 1pp = pp >> pppp = pp >> ppelserr= 0qq= 1end ifprint pp + qq + rr[Note - >> - Bitwise right shift operator, it takes two numbers, right-shift the bits of the first operand the second operand decides the number of places to shift&& : Logical AND - The logical AND operator (&& ) returns the boolean value True (or 1) if either or both operands is True and returns False( or 0)]
Solution
The pseudocode starts by setting the variables pp, qq, and rr to 1, 2, and 4 respectively.
Then it checks if all of these variables are greater than 0. Since they all are, it enters the if statement.
Inside the if statement, it first sets pp to 1.
Then it performs a bitwise right shift on pp by pp places. Since pp is 1, this is equivalent to dividing pp by 2^1, which results in 0.5. However, since pp is an integer, the decimal part is discarded and pp becomes 0.
Next, it performs another bitwise right shift on pp by pp places. Since pp is now 0, this operation has no effect and pp remains 0.
If the condition in the if statement was not met, it would have set rr to 0 and qq to 1. However, since the condition was met, this part of the code is not executed.
Finally, it prints the sum of pp, qq, and rr, which is 0 + 2 + 4 = 6.
So, the output of the pseudocode will be 6.
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