Study the following code:

.section .data
n1: .byte 12
n2: .byte 13
result: .space 1

.section .text
.global asm_function

asm_function:
lds r16, n1
lds r17, n2

loop_check:
mov r18, r17
add r18, r16

cpi r18, 50
brsh after_loop

cpi r17, 8
brlo after_loop

loop:
dec r17
add r16, r17
jmp loop_check

after_loop: lds r16, n1
sts result, r16
ret

.end

What is the value stored in result after this code has finished?

Question

Study the following code:

.section .data 
n1: .byte 12
n2: .byte 13
result: .space 1

.section .text
.global asm_function

asm_function:
lds r16, n1
lds r17, n2

loop_check:
mov r18, r17
add r18, r16

cpi r18, 50
brsh after_loop

cpi r17, 8
brlo after_loop

loop:
dec r17
add r16, r17
jmp loop_check

after_loop: lds r16, n1
sts result, r16
ret

.end

What is the value stored in result after this code has finished?

BlackTom AI · Accepted Answer

We are given a small AVR-like assembly snippet and asked what value ends up in 'result' after execution. First, restate the key data: - n1 holds the byte 12 - n2 holds the byte 13 - result is a one-byte space for storage - The code loads n1 into r16 and n2 into r17 at the start of the function. Now walk through the loop step by step, keeping track of r16, r17, and the control flow: - Initialization: r16 = 12, r17 = 13 - Enter loop_check: - r18 = r17; r18 becomes 13 - r18 = r18 + r16 => 13 + 12 = 25 - Compare r18 with 50 (cpi r18, 50): 25 is not >= 50, so the brsh after_loop is not taken - Compare r17 with 8 (cpi r17, 8): 13 is not < 8, so brlo after_loop is not taken - Enter loop body: dec r17 => 12; add r16, r17 => r16 = 12 + 12 = 24 - Jump back to loop_check - Second iteration start: r16 = 24, r17 = 12 - r18 = r17; r18 = 12 - r18 = r18 + r16 => 12 + 24 = 36 - cpi r18, 50: 36 < 50, so brsh not taken - cpi r17, 8: r17 = 12, still not < 8, so brlo not taken - loop: dec r17 => 11; r16 = 24 + 11 = 35 - Jump to loop_check - Third iteration: r16 = 35, r17 = 11 - r18 = r17; r18 = 11 - r18 = r18 + r16 => 11 + 35 = 46 - cpi r18, 50: 46 < 50, brsh not taken - cpi r17, 8: 11 < 8? no; brlo not taken - loop: dec r17 => 10; r16 = 35 + 10 = 45 - Jump to loop_check - Fourth iteration: r16 = 45, r17 = 10 - r18 = r17; r18 = 10 - r18 = r18 + r16 => 10 + 45 = 55 - cpi r18, 50: 55 >= 50, so brsh after_loop is taken, exit the loop - After_loop label: - lds r16, n1 loads 12 back into r16 (reset to original n1 value) - sts result, r16 stores 12 into the memory location 'result' - ret exits the function Conclusion from the trace: the value stored in result is 12, not 45. The initial guess of 45 likely came from the intermediate sum in the loop, but the code resets r16 to n1 (12) right before storing. If you’re comparing with the provided answer 45, note that 45 was the value of r16 after the third iteration, but the final store uses the restored n1 value (12). The crucial step is the 'lds r16, n1' right before 'sts result, r16'.

Similar Questions

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

What is the three letter instruction in the AVR assembly language that copies the value of one register into another?

In the AVR assembly language, how many numbers are stored in the data memory after the following lines of code? .section .data .space 26, 8

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler