Computer Science – 4.2 Assembly Language | e-Consult
4.2 Assembly Language (1 questions)
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Execution Trace:
Initial State: R0 = 20, R1 = 0, R2 = 3
LOAD R1, R0: The value of R0 (20) is loaded into R1.
- R0 = 20
- R1 = 20
- R2 = 3
ADD R1, R2: The value of R2 (3) is added to the value in R1 (20). The result (23) is stored back in R1.
- R0 = 20
- R1 = 23
- R2 = 3
Final State: R0 = 20, R1 = 23, R2 = 3
Answer: The value stored in R1 after execution is 23.