CODEWITHSUNDEEP
iosobjective-cassemblyarmstack-memory
Lorenzo B
Lorenzo B

Reputation: 33428

Clarifications on iOS Assembly Language

I'm investigating a little bit on how Objective-C language is mapped into Assembl. I've started from a tutorial found at iOS Assembly Tutorial.

The code snippet under analysis is the following.

void fooFunction() {
    int add = addFunction(12, 34);
    printf("add = %i", add);
}

It is translated into

 _fooFunction:
@ 1:
    push    {r7, lr}
@ 2:
    movs    r0, #12
    movs    r1, #34
@ 3:
    mov r7, sp
@ 4:
    bl  _addFunction
@ 5:
    mov r1, r0
@ 6:
    movw    r0, :lower16:(L_.str-(LPC1_0+4))
    movt    r0, :upper16:(L_.str-(LPC1_0+4))
LPC1_0:
    add r0, pc
@ 7:
    blx _printf
@ 8:
    pop {r7, pc}

About the assembly code, I cannot understand the following two points

-> Comment @1

The author says that push decrements the stack by 8 byte since r7 and lr are of 4byte each. Ok. But he also says that the two values are stored with the one instruction. What does it mean?

-> Comment @6

movw    r0, :lower16:(L_.str-(LPC1_0+4))
movt    r0, :upper16:(L_.str-(LPC1_0+4))

The author says the that r0 will hold the address of the "add = %i" (that can be find in the data segment) but I don't really get how the memory layout looks like. Why does he represent the difference L_.str-(LPC1_0+4) with the dotted black line and not with red one (drawn by me).

enter image description here

Any clarifications will be appreciated.

Edit

I'm missing the concept of pushing r7 onto the stack. What does mean to push that value and what does it contain?

Upvotes: 0

Views: 298

Answers (2)

Jester
Jester

Reputation: 58762

But he also says that the two values are stored with the one instruction. What does it mean?

That the single push instruction will put both values onto the stack.

Why does he represent the difference L_.str-(LPC1_0+4)

Because the add r0, pc implicitly adds 4 bytes more. To quote the instruction set reference:

Add an immediate constant to the value from sp or pc, and place the result into a low register.
Syntax: ADD Rd, Rp, #expr
where:
Rd   is the destination register. Rd mustbe in the range r0-r7.
Rp   is either sp or pc.
expr is an expression that evaluates (at assembly time) to a multiple of 4 in the range 0-1020.

If Rp is the pc, the value used is: (the address of the current instruction + 4) AND &FFFFFFFC.

Upvotes: 1

stonesam92
stonesam92

Reputation: 4457

For comment 1: The two values pushed to the stack are the values store in r7 and lr.

Two 4 byte values equals 8 bytes.

For comment 6: The label LPC1_0 is followed by the instruction

add r0, pc

which adds another 4 bytes to the difference between the two addresses.

Upvotes: 0

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