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Question

In an 8086, AL = 19 BCD & BL = 36 BCD prior to execution of following example

ADD AL, BL

DAA

What will be content of AL after execution of the example ?

This question was previously asked in
UGC NET 2023 Paper 1 Question Paper (22-Jun-2023) (Shift 2)
The correct answer is

5 5

To solve this problem of adding two BCD numbers using the 8086 microprocessor, we first need to understand what happens during each of the given assembly instructions: ADD AL, BL and DAA.

  1. Initial Conditions:
    • \(AL = 19_{BCD}\)
    • \(BL = 36_{BCD}\)
  2. Performing ADD AL, BL:
    • This operation will add the contents of \(BL\) to \(AL\).
    • Adding \(19_{BCD} + 36_{BCD}\) as simple binary without considering BCD:
      • \(19_{BCD} = 25_{base\ 10}\\)
      • \(36_{BCD} = 54_{base\ 10}\\)
      • Binary sum:
        • \(11001_{binary} (25)\)
        • +
        • \(110010_{binary} (54)\)
        • =
        • \(101111_{binary} (79)\)
    • The result is stored in \(AL\).
  3. Applying DAA (Decimal Adjust for Addition):
    • The DAA instruction adjusts the result in \(AL\) to form a valid BCD.
    • Because the binary result \(79\) exceeds the maximum of \(9\) in lower 4 bits:
    • \(79_{binary} = 7D_{BCDfe}\)
    • The DAA will adjust it by:
      • Add 6 to lower four bits: \(D\)
        • So, \(85_{base\ 10} = 50_{BCD}\\)
  4. This adjustment gives the final correct BCD result in \(AL\) as 55.

Conclusion:

After executing these instructions, the content of AL becomes 55, which corresponds to option 2. Thus, the correct answer is 5 5.

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Important Questions from Microprocessors

  1. An 8086 address bus is a/an _____ -bit bus

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