Study the following code:

.section .data
n: .byte 0
k: .byte 1
result: .space 1

.section .text
.global asm_function

asm_function:
lds r16, n
lds r17, k

ldi r18, 1
cp r18, r16
brne else

ldi r18, 1
cp r17, r18
brne else

if:
ldi r18, 13
jmp after_if

else:
ldi r18, 18

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 0
k: .byte 1
result: .space 1

.section .text
.global asm_function

asm_function:
lds r16, n
lds r17, k

ldi r18, 1
cp r18, r16
brne else

ldi r18, 1
cp r17, r18
brne else

if:
ldi r18, 13
jmp after_if

else:
ldi r18, 18

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 section to understand the initial values.
- n is defined as a byte with value 0.
- k is defined as a byte with value 1.
- result is reserved with space for 1 byte, so it can store a single value.

Next, we step through the asm_function sequence to see how the registers are loaded and how the flow progresses.
1) lds r16, n loads the value at label n into r16. Since n is 0, after this instruction r16 = 0.
2) lds r17, k loads the value at label k into r17. Since k is 1, after this instruction r17 = 1.
3) ldi r18, 1 loads the immediate value 1 into r18. So, r18 = 1.
4) cp r18, r16 performs a compare between r18 and r16 as if doing a subtraction r18 - r16 for flag setting. Here r18 = 1 and r16 = 0, so the result is 1. The S, Z, C, and V flags are set accordingly, and importantly, the result is not equal to 0.
5) brne else checks the zero flag from the previous comparison. Since r18 and r16 are not equal (1 != 0), the zero flag is clear, so brne will branch to the label else.

Upon taking the else branch:
6) else: sets r18 to 18 by executing ldi r18, 18. Now r18 = 18.
7) after_if: sts result, r18 stores the value of r18 into the memory location labeled result. Therefore, result is written with the value 18.
8) ret returns from the function.

Putting this together, the code path uses the initial values n = 0 and k = 1, immediately failing the comparison between r18 and r16 and branching to the else block, which ends up storing 18 into result. There is no path in this code where the value 8 would be written to result given these initial conditions and the branch chosen by brne.

If a student wonders why not 8: note that the instruction sequence only places 18 into result in the shown path. The 8 would require a different control flow or different register values prior to the sts instruction, which does not occur in this execution trace.

ELEC1601 (ND) ELEC9601 (ND)

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