Project Euler problem 1 example

problem1.s

        .data
output_text:
        .asciz "%d\n" ;; So we can printf later
        .balign 4

        .text
        .global main
        .extern printf ; load external function

;; This program calculates the sum of all positive integers evenly divisible by 3 or 5
;; The python equivalent is:
;; print(sum([i for i in range(1000) if i%3==0 or i%5==0]))
main:
        mov r6, #0 ;; r6 will hold the sum of numbers we find
        mov r5, #1 ;; r5 will be the var we increment each time

mainloop:
        cmp r5, #1000 ;; compare to 1000
        beq print_results ;; if it's equal, jump to the print_results section
        mov r0, r5  ;; r0 and r1 are the first two arguments sent to functions
        mov r1, #3  ;; So with r0==r5, and r1==3, it's like calling mod(VALUE_IN_R5, 3)
        bl mod ;; call the mod function
        cmp r0, #0 ;; If the return value (r0) is zero, its evenly divisible by 3
        beq addval ;; jump to addval if the above is true
        
        ;; The next few lines are the same as above, but just check the numbers divisibility with 5
        mov r0, r5
        mov r1, #5
        bl mod
        cmp r0, #0
        beq addval
        
        ;; Increment r5 by 1, and then start our loop over
        add r5, r5, #1
        b mainloop


;; If we jump here, then add the current value of r5 to r6
addval:
        add r6, r6, r5

;; Increment r5 by 1, and then start our loop over
;; This is placed here so it will automatically be reached if we jump to incr
inc_and_loop:
        add r5, r5, #1
        b mainloop
        
print_results:
        ;; Move our final sum to r1 so it can be an argument to printf
        mov r1, r6
        
        ;; Set our first argument as the value of output_text
        ldr r0, =output_text
        
        ;; call the printf function
        bl printf
        
        ;; Set our exit value to 0
        mov r0, #0
        
        ;; Setup the syscall we want, in this case it's 7, for exit
        mov r7, #1
        
        ;; perform the syscall
        swi #0

mod:
        ;; First of all, compare the dividend and divisor
        cmp r0, r1
        ;; if they are equal, there's no remainder
        beq no_remainder
        ;; Compare again
        cmp r0, r1
        ;; If the dividend is less than the divisor, exit early and force a remainer of 1
        blt less_than
        
        ;; Otherwise, subtract the divisor from the dividend
        sub r0, r0, r1
        
        ;; Compare the values now
        cmp r0, r1
        
        ;; If the dividend is still greater than or equal to the divisor, start at the top and run again with the new dividend
        bge mod
        
        ;; Otherwise, r0 now contains the remainder, so return
        bx lr

no_remainder:
        mov r0, #0
        bx lr
less_than:
        mov r0, #1
        bx lr

.global printf