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What will be the output of the following pseudo code?1. Integer p, q, r2. Set p=4, q=2, r=63. if((q-p-r) < (r - q))4. r=(q+p) & q5. End if6. r =3+q7. r =3+q8. r =3+q9. print p+q+r

Question

What will be the output of the following pseudo code?1. Integer p, q, r2. Set p=4, q=2, r=63. if((q-p-r) < (r - q))4. r=(q+p) & q5. End if6. r =3+q7. r =3+q8. r =3+q9. print p+q+r

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Solution

The output of the pseudo code will be 10. Here's the step by step explanation:

  1. The variables p, q, and r are set to 4, 2, and 6 respectively.
  2. The condition in the if statement, (q-p-r) < (r - q), translates to (2-4-6) < (6 - 2), which is -8 < 4. This condition is true.
  3. Since the condition is true, the code inside the if statement is executed. r is set to the result of (q+p) & q, which is (2+4) & 2. The & operator performs a bitwise AND operation. The binary representation of 6 (which is 2+4) is 110 and of 2 is 10. The result of the bitwise AND operation is 10, which is 2 in decimal.
  4. After the if statement, r is set to 3+q three times. Since q is 2, r is set to 5 each time.
  5. Finally, p+q+r is printed, which is 4+2+5 = 11.

This problem has been solved

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