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 ?
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.
- Initial Conditions:
- \(AL = 19_{BCD}\)
- \(BL = 36_{BCD}\)
- 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 \(AL\).
- Applying DAA (Decimal Adjust for Addition):
- The
DAAinstruction 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}\\)
- Add 6 to lower four bits: \(D\)
- The
- 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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