Study the following code:

.section .data
n: .byte 10
k: .byte 1
result: .space 1

.section .text
.global asm_function

asm_function:
lds r16, n
lds r17, k

ldi r18, 7
cp r18, r16
brsh else

ldi r18, 4
cp r17, r18
brsh else

if:
ldi r18, 12
jmp after_if

else:
ldi r18, 17

after_if:
sts result, r18
ret

.end

What is the value stored in result after this code is run?

Question

Study the following code:

.section .data 
n: .byte 10
k: .byte 1
result: .space 1

.section .text
.global asm_function

asm_function:
lds r16, n
lds r17, k

ldi r18, 7
cp r18, r16
brsh else

ldi r18, 4
cp r17, r18
brsh else

if:
ldi r18, 12
jmp after_if

else:
ldi r18, 17

after_if:
sts result, r18
ret

.end

What is the value stored in result after this code is run?

BlackTom AI · Accepted Answer

We start by examining the data and the registers that will be used by the function.

- n contains the byte 10, so after lds r16, n, r16 will be 10.
- k contains the byte 1, so after lds r17, k, r17 will be 1.
- r18 is loaded with 7 via ldi r18, 7.

Step through the comparisons and branches in order:

1) cp r18, r16 compares r18 (7) with r16 (10). This performs an unsigned comparison: 7 vs 10. Since 7 is not higher or equal to 10, the condition for brsh else is false, so we do not take the branch to else. We proceed to the next instruction.

2) Next, ldi r18, 4 sets r18 to 4. Then cp r17, r18 compares r17 (1) with r18 (4). Again this is unsigned: 1 vs 4. Since 1 is not higher or equal to 4, the brsh else condition is false, so we do not branch to else here either. We continue to the next instruction.

3) The label if: the instruction at this point is ldi r18, 12. This loads 12 into r18, and the next instruction is jmp after_if, which unconditionally jumps to after_if. There is no execution of the code under else in this path.

4) at after_if: sts result, r18 stores the current r18 value (12) into the memory location labeled result. Therefore result gets the value 12.

5) Finally, ret returns from the function.

Summary: Both comparison branches were not taken because 7 is not >= 10 and 1 is not >= 4 in unsigned terms. The code then forces r18 to 12 and stores that into result. Hence, the value stored in result after this code runs is 12.

类似问题

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