Machine code to assembly (assumed mapping): Decode the single-byte opcode 0x48 into its assembly instruction (assume it was not produced by a pseudo-op).
-
AASRA
-
BLOADA h#0D4E, i
-
CSTOREA h#014B, d
-
DADDA h#01FE, i
-
EJUMP h#4848, a
Answer
Correct Answer: ASRA
Explanation
Introduction / Context:Reverse-assembling a machine opcode requires knowledge of the ISA’s opcode map. Single-byte opcodes typically encode accumulator/register operations or short addressing forms, whereas instructions with immediate or direct addresses need extra operand bytes.
Given Data / Assumptions:
- Opcode: 0x48.
- No additional operand bytes were provided, implying a single-byte instruction.
- We assume a conventional accumulator architecture where shifts/rotates and register ops can be single-byte.
Concept / Approach:If the instruction consumed only one byte, it cannot encode the multi-byte immediates or addresses visible in options (b), (c), (d). Therefore, a unary accumulator operation such as ASRA (arithmetic shift right accumulator) is the consistent choice.
Step-by-Step Solution:Check each option’s operand length.Options with h#.... operands require additional bytes and contradict the single 0x48 byte.Select the operand-less accumulator operation as the plausible decoding.
Verification / Alternative check:Consulting the ISA manual would definitively map 0x48 to its mnemonic. In many classic ISAs, 0x48 corresponds to single-byte register/shift ops, aligning with ASRA over memory-addressing forms.
Why Other Options Are Wrong:(b), (c), (d) each imply at least two additional operand bytes; they cannot be encoded by a lone 0x48. (e) likewise suggests addressing.
Common Pitfalls:Ignoring instruction length; assuming any mnemonic can match without checking required operands.
Final Answer:ASRA.