understanding pointers and casting in assembly

Question

I was given a function in assembly which basically converted uppercase letters to lowercase letters. Here is some of the assembly,

 Q1:
    pushq %rbp
    movq %rsp, %rbp 
    subq $24, %rsp
    movq %rdi, -24(%rbp)
    movl $0, -4(%rbp)
    movl $0. -8%(%rbp) 
    jmp .L2
L2:
    movl -4(%rbp) %edx
    movq -24(%rbp), %rax
    addq %rdx, %rax
    movzbl (%rax), %eax
    testb %al, %al
    jne .L4
    ...

Much of the rest is repetitive but L2 is what really is confusing me. This is my logic so far: We store param1 into -24(%rbp). We create local1 and local2, set them both to 0 and then jump to L2. I move local1 into %edx, param1 into %rax. Now this is where things get confusing for me, I was told the following line, addq ended up in local1 being a pointer to param1. I just reasoned add local1 + param1 and store them into %rax. How is that possible?

Next is, movzbl. From my understanding we dereference %rax so we get something like eax = (int) rax.

I was also told to think of it as converting a char to int. Which one is true, how do I know that I'm typecasting? What about if %rax didn't have parentheses around it? Is it an int because it's 4 bytes and %eax is a 32 bit register. Thank you in advance for your help, I'm kind of lost here....

Jester Jester · Accepted Answer · 2015-11-18T02:13:43

local1 is not a pointer, it's an index (a counter). That code is doing something like:

void toupper(char* text)
{
    int i = 0;  /* at rbp-4 */
    int j = 0;  /* unused, at rbp-8 */
    int ch;     /* in eax */
    while((ch = *(text + i)) != 0)
    {
        ...
    }
}

Note that in C pointer arithmetic *(text + i) is of course equivalent to text[i].

Yes, the movzbl is converting an unsigned char to an int you can see that from the instruction name itself: MOVe Zero extended Byte to Long.

The parentheses denote pointer dereferencing.

understanding pointers and casting in assembly

1 Answers