Match List I with List II List I (Assembly code) List II (Content of register A after the code)...

71) Match List I with List II

List I (Assembly code)	List II (Content of register A after the code)
A. MOV A, #37H CPL A	I. 02H
B. MOV A, #37H ANL A, #8AH	II. F7H
C. MOV A, #37H ORL A, #8AH	III. FBH
D. MOV A, #37H XRL A, #8AH	IV. C8H

Choose the correct answer from the options given below:

1. A - IV, B - I, C - III, D - II [Option ID = 3113]
2. A - I, B - II, C - III, D - IV [Option ID = 3114]
3. A - IV, B - I, C - II, D - III [Option ID = 3115]
4. A - III, B - I, C - IV, D - II [Option ID = 3116]

Answer: A

Solution

To solve this assembly code problem, we convert the hexadecimal values to 8-bit binary and apply bitwise logic operations.

Binary Conversions:
• Register A (37H) = 0011 0111
• Operand (0CAH) = 1100 1010

A. CPL A (Complement A)

Flips every bit (NOT operation) of the original 37H.

37H	0	0	1	1	0	1	1	1
Result	1	1	0	0	1	0	0	0

Binary 1100 1000 = C8H. Match: A → IV

B. ANL A, #0CAH (Bitwise AND)

Output is 1 only where both bits are 1.

37H	0	0	1	1	0	1	1	1
0CAH	1	1	0	0	1	0	1	0
AND	0	0	0	0	0	0	1	0

Binary 0000 0010 = 02H. Match: B → I

C. ORL A, #0CAH (Bitwise OR)

Output is 1 if at least one bit is 1.

37H	0	0	1	1	0	1	1	1
0CAH	1	1	0	0	1	0	1	0
OR	1	1	1	1	1	1	1	1

Binary 1111 1111 = FFH. Match: C → III

D. XRL A, #0CAH (Bitwise XOR)

Output is 1 only if the bits are different.

37H	0	0	1	1	0	1	1	1
0CAH	1	1	0	0	1	0	1	0
XOR	1	1	1	1	1	1	0	1

Binary 1111 1101 = FDH. Match: D → II

Final Sequence: A-IV, B-I, C-III, D-II — (Option 1)

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