8051 assembly control flow Will the following code execute only once or loop repeatedly? STAT: MOV A, #01H JNZ STAT
Digital Electronics
The 8051 Microcontroller
Difficulty: Easy
Choose an option
-
ATrue
-
BFalse
-
CIt depends on the initial PSW flags
-
DOnly if interrupts are disabled
Answer
Correct Answer: False
Explanation
Introduction / Context:This question checks understanding of the 8051 conditional jump instruction JNZ (jump if A != 0) and how immediate moves affect the accumulator A.
Given Data / Assumptions:
- MOV A, #01H loads accumulator with 0x01 (decimal 1).
- JNZ LABEL branches when accumulator is nonzero.
- No instructions modify A between the move and the jump.
Concept / Approach:Because A is explicitly loaded with 1 before each JNZ, the zero condition is never met. Therefore, the branch is always taken, forming an infinite loop at STAT.
Step-by-Step Solution:
1) Execute MOV A, #01H → A = 1.2) Execute JNZ STAT → since A != 0, jump to STAT.3) Repeat the same two instructions forever because A is reloaded with 1 each iteration.4) No path clears A to zero, so the loop never exits.Verification / Alternative check:Single-step in a simulator: the program counter toggles between the two instructions indefinitely.
Why Other Options Are Wrong:
- True: Incorrect—code does not execute only once.
- It depends on the initial PSW flags: JNZ depends on A, not PSW flags.
- Only if interrupts are disabled: Interrupts are unrelated to JNZ behavior here.
Common Pitfalls:Confusing JNZ with conditional flags; on 8051, JNZ checks accumulator zero directly.
Final Answer:False — the sequence loops forever.