In the below example of AVR code, let's say the decimal value 24 is stored in the memory location 0x100 (also called result_location). What value is stored in memory location 0x100 after the following lines of code? Write your answer in base 10.

ldi r18, 14
ldi r19, 24
sts result_location, r18
add r18, r19
lds r18, result_location

Question

In the below example of AVR code, let's say the decimal value 24 is stored in the memory location 0x100 (also called result_location). What value is stored in memory location 0x100 after the following lines of code? Write your answer in base 10.

ldi r18, 14
ldi r19, 24
sts result_location, r18
add r18, r19
lds r18, result_location

BlackTom AI · Accepted Answer

The sts instruction writes the current register value into memory, so the memory at result_location becomes the value loaded into r18 before the add occurs.

1) Relevant concepts
- ldi Rx, K loads an immediate constant K into register Rx.
- sts addr, Rx stores the value of Rx into data memory at address addr.
- add Rx, Ry updates the register Rx with Rx + Ry (does not change memory).
- lds Rx, addr loads a value from data memory addr into register Rx.
- AVR register and memory operations are sequential; later register changes do not retroactively change memory unless a store is executed.

2) Step-by-step execution and register/memory state
Initial: memory[0x100] = 24 (given).
- ldi r18, 14
  - r18 ← 14
  - r19 unchanged
  - memory[0x100] = 24
- ldi r19, 24
  - r19 ← 24
  - r18 = 14
  - memory[0x100] = 24
- sts result_location, r18
  - memory[0x100] ← r18 = 14
  - Now memory[0x100] = 14
  - r18 = 14, r19 = 24
- add r18, r19
  - r18 ← 14 + 24 = 38 (fits in 8 bits; no wrap)
  - memory[0x100] remains 14 (no store executed)
- lds r18, result_location
  - r18 ← memory[0x100] = 14
  - memory[0x100] still = 14

3) Calculations shown
- After add: r18 = 14 + 24 = 38
- Memory after sts: 14; no further stores change it.

4) Why the correct answer is correct
The store (sts) occurred before the add, so the memory was overwritten with 14. The subsequent add changed only the register r18, not memory. The later lds simply reloads the stored value from memory. Therefore the value in memory remains 14.

5) Why other plausible answers are incorrect
- 24 would be incorrect because the initial 24 at 0x100 was overwritten by sts r18, which wrote 14.
- 38 would be incorrect because add changes r18 only; it does not write back to memory unless another store instruction is executed.

6) Conclusion
The value stored at memory location 0x100 after these instructions is 14, which matches the provided answer.

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