8051 instruction semantics — does 'RL A' rotate the eight bits of the Accumulator left by one position? State whether the mnemonic RL A performs a one-bit left rotate confined to A (bit7 wraps to bit0, carry flag unchanged).
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ATrue
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BFalse
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COnly if CY = 1
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DIt rotates through the carry flag (RLC)
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EIt shifts left and fills 0 (not rotate)
Answer
Correct Answer: True
Explanation
Introduction / Context:The 8051 family provides two left-rotate instructions for the Accumulator: RL A and RLC A. Understanding the difference is essential for bit-level operations, look-up indexing, and simple arithmetic without disturbing the carry flag.
Given Data / Assumptions:
- Target CPU: 8051/8052 architecture
- Mnemonic under discussion: RL A
- Accumulator A is 8 bits wide (A7…A0)
- Carry flag CY exists in PSW but may or may not be affected depending on the instruction
Concept / Approach:RL A performs a pure 8-bit rotate left within A. Bit7 moves into bit0, and all other bits shift left by one. The carry flag is not used and not updated. In contrast, RLC A rotates through the carry: bit7 goes to CY and CY enters bit0.
Step-by-Step Solution:
Let A = b7 b6 b5 b4 b3 b2 b1 b0RL A result = b6 b5 b4 b3 b2 b1 b0 b7 (CY unchanged)RLC A result = b6 b5 b4 b3 b2 b1 b0 CY ; CY_new = b7Verification / Alternative check:Examine the flags after RL A; CY remains whatever it was. After RLC A, CY reflects previous b7. This confirms that RL does not use the carry path.
Why Other Options Are Wrong:
- False: contradicts the ISA definition.
- Only if CY = 1: CY is irrelevant for RL.
- Rotates through carry: that describes RLC, not RL.
- Shifts left filling 0: that would be a logical shift, not a rotate.
Common Pitfalls:
- Confusing RL with RLC and assuming CY always participates.
- Expecting CY to capture bit7 on RL; it does not.
Final Answer:True