Stack demonstration code from PASS1 workshop

stacks.s

; Move the stack pointer to point at the stack
ADRL       SP, stack

; Define the strings we're going to print
s_div      DEFB    "Dividing ", 0
ALIGN
s_by       DEFB    " by ", 0
ALIGN
s_equals   DEFB    " equals ", 0
ALIGN
s_rem      DEFB    " remainder ", 0
ALIGN


start

    ; Print out 'dividing'
    ADR     R0, s_div;
    SVC     3;

    ; Print 512
    MOV     R0, #512
    SVC     4;
    
    ; Print out 'by 10'
    ADR     R0, s_by;
    SVC     3;
    MOV     R0, #10
    SVC     4;

    ; Print out 'equals'
    ADR     R0, s_equals;
    SVC     3;
    
    ; Now do the division
    MOV     R0, #512;
    BL      div10;
    SVC     4;

    ; Print out 'remainder'
    ADR     R0, s_rem;
    SVC     3;
    MOV     R0, R1
    SVC     4;

    ; Stop
    SVC     2;



; N.b. div10 is an example method that divides an integer by ten.
; How it works is beyond the scope of COMP15111.
; However, you can see that the function pushes R2 on the stack, since
; otherwise it would be corrupted. R1 and R0 do not need to be pushed
; since div10 returns its result in those registers.

; takes argument in R0
; returns quotient in R0, remainder in R1
div10
    ; We're going to use these two registers in the function
    PUSH  {R2}; Or we could do 'STMFD  SP!, {R2}'
    ; Magic division thing!
    SUB    R1, R0, #10                       ; keep (x-10) for later
    SUB    R0, R0, R0, LSR #2
    ADD    R0, R0, R0, LSR #4
    ADD    R0, R0, R0, LSR #8
    ADD    R0, R0, R0, LSR #16
    MOV    R0, R0, LSR #3
    ADD    R2, R0, R0, ASL #2
    SUBS   R1, R1, R2, ASL #1                ; calc (x-10) - (x/10)*10
    ADDPL  R0, R0, #1                        ; fix-up quotient
    ADDMI  R1, R1, #10                       ; fix-up remainder
    ; We now get back the value of the registers we popped
    POP    {R2}; Or we could do 'LDMFD  SP!, {R2}'
    ; Branch back to where we came from
    MOV    PC, LR

; Setup a stack area (we can point the SP here).
            DEFS    1024
stack