In the below example of AVR code, let's say the decimal value 15 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.

lds r18, result_location
ldi r19, 2
add r18, r19
inc r19
lsl r18
sub r18, r19

Question

In the below example of AVR code, let's say the decimal value 15 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.

lds r18, result_location
ldi r19, 2
add r18, r19
inc r19
lsl r18
sub r18, r19

BlackTom AI · Accepted Answer

Since there is no store instruction (such as sts) in the code shown, the data memory at 0x100 is never written back to — it remains unchanged at 15.

Relevant concepts:
- lds r18, result_location reads the byte from data memory address 0x100 into register r18.
- ldi loads an immediate into a register; add/inc/lsl/sub operate only on registers and flags.
- No instruction in the sequence writes any register value back to memory, so the memory content stays the same.

Step-by-step register execution (all values in decimal, with hex where useful):
1. lds r18, result_location
   - r18 := 15 (memory at 0x100 is read but not modified)
2. ldi r19, 2
   - r19 := 2
3. add r18, r19
   - r18 := 15 + 2 = 17
4. inc r19
   - r19 := 2 + 1 = 3
5. lsl r18
   - Logical shift left: r18 := 17 << 1 = 34 (equivalent to multiply by 2)
   - In hex: 0x11 -> 0x22
6. sub r18, r19
   - r18 := 34 - 3 = 31

All modifications above affect only registers r18 and r19. Because there is no sts (store direct to data space) or similar memory write instruction after these operations, the original memory byte at address 0x100 remains 15.

Why the common alternative (31) is incorrect:
- A reader might report the final register r18 value (31) as the memory value, but registers and memory are distinct; without an explicit store, memory is not updated.

Conclusion: Memory location 0x100 still contains 15, which matches the provided answer.

类似问题

The following two pieces of code do the same thing. a = 7; b = 23; c = b - a; Suppose r18 is for storing a, r19 is for storing b, and r20 is for storing c. What is the missing line? ldi r18, 7 ; r18 = 7 (a) ldi r19, 23 ; r19 = 23 (b) mov r20, r19 ; r20 = r19 [MISSING LINE] ; This line does c = b - a

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