Question

To get the command line arguments in x86_64 on Mac OS X, I can do the following:

_main:
  sub   rsp, 8         ; 16 bit stack alignment
  mov   rax, 0
  mov   rdi, format
  mov   rsi, [rsp + 32]
  call  _printf

Where format is “%s”. rsi gets set to argv[0].

So, from this, I drew out what (I think) the stack looks like initially:

 top of stack
               <- rsp after alignment
return address <- rsp at beginning (aligned rsp + 8)
  [something]  <- rsp + 16
    argc       <- rsp + 24
   argv[0]     <- rsp + 32
   argv[1]     <- rsp + 40
    ...            ...
bottom of stack

And so on. Sorry if that’s hard to read. I’m wondering what [something] is. After a few tests, I find that it is usually just 0. However, occasionally, it is some (seemingly) random number.

Also, could you tell me if the rest of my stack drawing is correct?

Answer 1

You have it close.

argv is an array pointer, not where the array is. In C it is written char **argv, so you have to do two levels of dereferencing to get to the strings.

 top of stack
               <- rsp after alignment
return address <- rsp at beginning (aligned rsp + 8)
  [something]  <- rsp + 16
    argc       <- rsp + 24
   argv        <- rsp + 32
   envp        <- rsp + 40  (in most Unix-compatible systems, the environment
    ...            ...       string array, char **envp)
bottom of stack
 ...
somewhere else:
   argv[0]     <- argv+0:   address of first parameter (program path or name)
   argv[1]     <- argv+8:   address of second parameter (first command line argument)
   argv[2]     <- argv+16:  address of third parameter (second command line argument)
    ...
   argv[argc]  <-  argv+argc*8:  NULL