Question

explain this code, step by step

(8) Question 2a. Assume initially R4-4, R5-5, R6-6, and R7-7 Consider the following assembly code PUSH IR5,R61 Top-) 5,6 Bottom R4 0 R5-3 ASR R4,R4,R7 EOR R7,R5,R6 R7 - 1012 1102-0112- 3 PUSH {R7,R4) Top-> 0,3,5,6 <-Bottom POP R7,R5, R4,R6 R4-0,R5-3,R6 5,R7-6 (stack empty) R6-5 R7-6 After executing these five instructions, what will be the final values for these four registers?

Answer 1

PUSH {R5, R6} is pushing value of register R5 above value of register R6 in a stack. Stack contents = {5,6} from top to bottom.

ASR R4, R4, R7            ; arithmetic shift right (ASR) the contents of register R4 are shifted right 7 bits which is equivalent to divide the contents of R4 to , and storing the resulting value in register R4. Contents of R4 becomes 0.

2' 128

EOR R7, R5, R6          ; taking Ex-Or logic of the contents of register R5 and R6 (in binary format) and storing the result in register R7. XOR(1,1) = XOR(0,0) = 0 and XOR(1,0) = XOR(0,1) = 1. New contents of R7 is 3.

PUSH {R7, R4} is pushing value of register R7 above value of register R4 in a stack. Stack contents = {3,0,5,6} from top to bottom.

POP {R7, R5, R4, R6} is popping out the values one by one from top of stack and assigning to R6,R5,R4,R7 respectively

Final values are R7 = 6, R5 = 5, R4 = 0 and R6 = 3.