ARM Assembly: write number located on stack

Question

I have an ARM assembly code that atm just prints out "Hello World" from a string variable:

  .global _start

_start:
    mov r0, #40
    mov r1, #25
    bl func1
    mov r7, #1
    swi 0

func1:
    push {lr}
    add r2,r0,r1
    mov r0,r2
    bl func2
    pop {pc}

func2:
    mov r7, #4
    mov r0, #1
    mov r2, #12
    ldr r1, =string

    swi 0
    bx lr

.data
    string:
    .ascii "Hello World\n"

Now however, I want to print the result of the addition happening in func1 (result = 65) by writing it onto the stack and then handing the write() syscall the memory address of the result on the stack. I have already tried pushing the result at the beginning of func2 onto the stack with

push {r0}

and then somehow loading the address of that in r1 (which stores the parameter to print for the write() syscall) by accessing the stack pointer and an offset of 4, where the value should be stored if Im correct:

ldr r1, [sp, #4]

But this doesnt seem to work, I only get segmentation faults with everything I try like this.

Can anyone help me on how I can get the memory of address on the stack of the result and then proceed to print it with the write() syscall? Im ok with it if it prints the result as a letter instead of a number (since it will probably interprete it as an ASCII value), its just about me understanding how to get the address of the value on the stack and print it. Thx a lot :)

Jester Jester · Accepted Answer · 2017-05-12T13:25:08

To get the address 4 bytes past sp, you'd do add r1, sp, #4. However remember that push uses the stack in full descending mode, meaning sp points to the last item pushed. As such you don't need to add 4, you can just do mov r1, sp.

What you did with ldr was actually loading the value from the stack and even if you used the correct offset that would still not have been good as you need the address not the value.

ARM Assembly: write number located on stack

1 Answers