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What will be the output of the following pseudocode?Integer a, b, cset a= 1, b= 5for (each c from 6 to 7)a = a & ca = a + cend forprint a + b[ 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 a, b, cset a= 1, b= 5for (each c from 6 to 7)a = a & ca = a + cend forprint a + b[ 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]

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Solution

The pseudocode provided is a simple loop that performs bitwise AND operation and addition. Here's how it works:

  1. Initialize a to 1 and b to 5.
  2. For each c from 6 to 7, do the following:
    • Perform a bitwise AND operation between a and c, and assign the result back to a.
    • Add c to a, and assign the result back to a.
  3. After the loop, add a and b, and print the result.

Now, let's break it down:

  • For c = 6, a = 1. The binary representation of 6 is 110, and of 1 is 001. The bitwise AND operation results in 000, or 0. So, a becomes 0. Then, a = a + c = 0 + 6 = 6.
  • For c = 7, a = 6. The binary representation of 7 is 111, and of 6 is 110. The bitwise AND operation results in 110, or 6. So, a remains 6. Then, a = a + c = 6 + 7 = 13.
  • Finally, a + b = 13 + 5 = 18.

So, the output of the pseudocode will be 18.

This problem has been solved

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