Question

Note: All assembly code should be well commented-similar level of detail to samples. 7. Write ARM code that decodes the immediate value in an instruction-takes the last 12 bits and turns it into the value it represents a. Start by loading r1 with the value E3A01CFA (machine code for a move immediate into register) b. Isolate bits 0-7 (the rightmost 8 bits) the 8-bit binary number - into r2 c. Isolate bits 8-11 d. Multiply r3 by 2 to get the actual rotation amount e. Rotate r2 to the right by that many bits. Place your answer in r4 (the four bits left of 0-7)-4 bits indicating the rotation- into r3

Answer 1

        ;       since in ARM immediate operand can only be a 32 bit number such that it can be
       ;       formed by shifting an 8 bit number
       ;       hence the number should be in the form:
       ;       from 0x00000000 to 0x000000FF
       ;       or from 0x00000100 to 0x0000FF00 etc.
       ;       means only 1 byte or 8 bits of the immediate operand can be non zero
       ;       hence to move 0xE3A01CFA we need to first move the first right most byte into register
       ;       and then add another bytes into register
       MOV       R1, #0x000000FA          ; move first right most byte into register r1
       ADD       R1, R1, #0x00001C00      ; adding second right most byte
       ADD       R1, R1, #0x00A00000      ; adding third right most byte
       ADD       R1, R1, #0xE3000000      ; adding fourth right most byte
       ;       Now register r1 contains 0xE3A01CFA
       MOV       R2, #0x00000000          ; setting r2 to 0
       MOV       R3, #0x00000000          ; setting r3 to 0
       MOV       R4, #0x00000000          ; setting r4 to 0
       ;       isolating right most byte of r1 into register r2 by doing and operation
       AND       R2, R1, #0x000000FF
       ;       isolating bits 8-11 into register r3
       AND       R3, R1, #0x00000F00
       ;       rotate to bring the number to the right most
       LSR       R3, R3, #8
       ;       multiplying r3 with 2 by just rotating r3 1 bit left
       LSL       R3, R3, #1
       ;       rotating r2 right as many bits as the value in r3 is. and storing the result into
       ;       register r4
       ROR       R4, R2, R3

At right in the image, you can see the value of registers.

the code 0xE3A01CFA

is a code for a mov instruction into a register with value #FA00. So at the end in our program, the register r4 contains the value #FA00.

The tool that i have used to visualize is "Visual" arm emulator.