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What will be the contents of register AL after the following program has been executed in a 8086 microprocessor?
MOV AL, 7E 
MOV BL, 8C 
ADD AL, BL
  • a)
    0A and carry flag is reset
  • b)
    6A and carry flag is set
  • c)
    0A and carry flag is set
  • d)
    6A and carry flag is reset
Correct answer is option 'C'. Can you explain this answer?
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Explanation:
The program consists of three instructions:

1. MOV AL, 7E - This instruction moves the hexadecimal value 7E into the AL register.

2. MOV BL, 8C - This instruction moves the hexadecimal value 8C into the BL register.

3. ADD AL, BL - This instruction adds the contents of the BL register to the contents of the AL register and stores the result in the AL register.

To solve the problem, we need to convert the hexadecimal values to decimal values:

AL = 7E (in hexadecimal) = 126 (in decimal)
BL = 8C (in hexadecimal) = 140 (in decimal)

Adding these two decimal values we get:

AL + BL = 126 + 140 = 266

Since 266 is greater than 8 bits, we need to discard the most significant bit and consider only the least significant 8 bits, which is 0x0A (in hexadecimal) or 10 (in decimal).

Therefore, the contents of the AL register after the execution of the program will be 0x0A (in hexadecimal) or 10 (in decimal).

Since the sum of the two values did not result in a carry, the carry flag is set to 0.

Hence, the correct answer is option 'c': 0A and carry flag is set.
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What will be the contents of register AL after the following program has been executed in a 8086 microprocessor?MOV AL, 7EMOV BL, 8CADD AL, BLa)0A and carry flag is resetb)6A and carry flag is setc)0A and carry flag is setd)6A and carry flag is resetCorrect answer is option 'C'. Can you explain this answer?
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