8051 logical AND practice — accumulator result Compute the accumulator after executing these two instructions (showing real newlines): MOV A, #0BH ANL A, #2CH

Difficulty: Easy

Correct Answer: 00001000

Explanation:


Introduction / Context:
This exercise checks understanding of bitwise AND on hexadecimal values in 8051 assembly. Being fluent in hex-to-binary conversion and bitwise operations is crucial for low-level I/O and mask manipulation.


Given Data / Assumptions:

  • MOV A, #0BH loads A with 0x0B.
  • ANL A, #2CH computes A = A AND 0x2C.
  • Binary conversions are performed to visualize the result.


Concept / Approach:
0x0B = 0000 1011, 0x2C = 0010 1100. The AND operator keeps only positions where both operands have 1s. The outcome is 0000 1000 (0x08).


Step-by-Step Solution:

1) Convert 0x0B → 0000 1011.2) Convert 0x2C → 0010 1100.3) AND bitwise: result → 0000 1000.4) Express in binary as 00001000 (matches the option).


Verification / Alternative check:
Compute in hex: 0x0B AND 0x2C = 0x08, which equals binary 00001000.


Why Other Options Are Wrong:

  • 11010111, 11011010, 00101000: do not match the correct AND result of the given operands.


Common Pitfalls:
Misaligning bits during conversion or confusing AND with OR/XOR. Always write both operands in full 8-bit binary before applying the operator.


Final Answer:
00001000

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