Re: confused asm newbie

[Frederic Marmond <fmarmond@eprocess.fr>]

Jason Roberts wrote:

> I have a few questions concerning how the stack is used and how memory 
> is accessed
> and what it looks like.I've read books, which only serve to confuse 
> me, or the author tries
> to mystify you into believing assembly is dark magic only for the elite. 

? I though it was really dark magic only for the elite !
;)

>
> Thanks in advance for any help and/or suggestions--
>
>   in C  If I code char *p = "hello" then *p will be 'h' and *(p+1) 
> will be 'e', and likewise
>   p[4] will derefrence byte 5 ,which is 'o'.If I  ++p  the value at p  
> will be whatever is at the address        of the next byte,etc
>
>   now, how does this relate to asm registers??? 

First thing:
I think you are not really a newbie, so, you problably already know 
that, but keep in mind that C is a typed langage, and ASM is not.
That's to say, a register can not know what type is the data it holds.
mov al,[foobar]
=> get 1 byte from @ foobar, and put it to the 1 byte al register
move edi,[foobar]
=> get the 4 bytes at @ msg, and put them into the 4 bytes edi register.

Ok?
The bytes you take as input are a memory region, pointed by the foobar 
variable. It may be a part of a pixels image, a part of a sound, of a 
text, of binary code, ...

Some assembleurs allows a pre-processing to check for data types, but it 
is not always the case.

When you use:
mov al,[foobar]
foobar is an offset in the memory, and the [] says to the compilator to 
get the data at this address (pointer)
but a
mov al,foobar
will (fail if the compilator check for data types => 32 bits fit into a 
8 bit register) or get the 8 lower bits from the foobar value (the 
address) and store them into al

It's the same as the '*' in C
foobar value;foobar *ptr;ptr=&value
ptr => the pointer (foobar)
*ptr=>the value pointed ([foobar])

Now, your questions:

>
>   based on code below  'msg' (line 5) is a pointer or 'label' which 
> represents an address which             holds 3              bytes 
> plus LF,CF,and NULL.
>   based on my current understanding--
>    msg = address of first byte
>   [msg] = value at base,'a'
>   add [msg],4 adds ascii value 4 to 'a' so it's now 'e'
>   add msg,3   adds 3 to pointer, so msg is now base+3, so  now [msg] 
> should be 'c'
>   am I right on this???
>
>    also, when I used edi to store string in _start (line 61)  I mov'ed 
> 'string' without the brackets
>    for some reason the use of brackets is neccessary in the case of
>    msg (line 84) ...why so?
>    also, the stack issue:
>    based on my knowledge the stack grows downward but reads upward, 
> i.e. if I push
>    edi then I have pushed 4 bytes onto stack- and so sp decrements 4 
> times and points
>    at last item pushed.  Sp only points to top of stack and knows 
> nothing about memory below
>    unless we tell it too by explicityly moving it down,which is 
> allocating space basically,malloc()
>    for the C gang. Am I getting it?

[FYI]
On x86, you have several modes in which your CPU can operate (to keep 
simple, Real mode (16 bits, 64ko segments), and protected mode(32bits, 
up to 4go 'segments')).
Linux runs in protected mode (the most powerfull mode).
In that mode, you can tell the stack to grow up or to grow down. The 
most often, it keep the default, growing downward.
[/FYI]

>
>    My only concern is this:
>    what does edi look like after line 84???
>    we have 6 bytes going into a 4-byte register...
>    my guess is:
>    6162630a with the CR and NULL being ignored. 

Yes, it's true: only 4 bytes can stand into a 4 bytes register... (or 
sign a software pattent, you have created a ziped register! ;))

>
>    If I'm right then what does the stack look like after the push? 

it will contain the "abc\10" string

>
>    From what I see the values in the registers are pushed from the 
> low-byte up to high so
>    that 61 is on top of stack, or worded differently, esp holds the 
> address of where 61 is.

it depends on which system you are (little/big endian).
I don't remember on little endian (x86) if bytes or words are swaped..

>
>    If I got all this right say Kudos! ,otherwise please point me narrow--
>    Thankyou for your time and patience. 

have fun!

>
>     Paul
>
>
> line 1:  section .data
> line 2:
> line 3:  string     db "Hello insanity!",10,13,0
> line 4:  error_msg  db "Null Pointer!",10,13,0
> line 5:  msg        db "abc",10,13,0
> line 6:
> line 7:
> line 8:
> line 9:  section .text
> line 10:
> line 11:  global _start
> line 12:
> line 13:  jmp _start
> line 14:
> line 15:  ;routine which puts length of string at es:edi into ecx and 
> returns
> line 16:  get_len:
> line 17:           jmp chk_edi
> line 18:
> line 19:  _inc:
> line 20:           inc ecx
> line 21:   jmp scan
> line 22:
> line 23:   ;check that edi is not null or scasb will fail
> line 24:  chk_edi:
> line 25:           or edi,edi
> line 26:           cmovz eax,edi
> line 27:   dec eax
> line 28:           ret
> line 29:
> line 30:
> line 31:  scan:
> line 32:
> line 33:   scasb
> line 34:           jne _inc
> line 35:           ret ; but what if first byte is null? well when ret 
> we will check now wont we!
> line 36:
> line 37:
> line 38:
> line 39:  error:
> line 40:
> line 41:
> line 42:   mov  edx,15     ;length of buffer
> line 43:   mov  eax,4  ;sys_write
> line 44:   mov  ebx,1       ;file descriptor
> line 45:   mov  ecx,error_msg  ;buffer
> line 46:
> line 47:   int  0x80
> line 48:
> line 49:
> line 50:           mov ebx,-1
> line 51:   mov eax,1
> line 52:   int 0x80
> line 53:
> line 54:
> line 55:
> line 56:
> line 57:  _start:
> line 58:
> line 59:   mov eax,0 ;used for scasb
> line 60:   mov ecx,0 ;will be the length value of string,init to  0
> line 61:   mov edi,string    ;if  given null will produce an 
> exception-- which I handle
> line 62:   cld
> line 63:   call get_len
> line 64:   cmp eax,0
> line 65:           js error
> line 66:
> line 67:
> line 68:
> line 69:  write:
> line 70:
> line 71:           or ecx,ecx       ;check if string was null
> line 72:   jz exit          ; could implement exception handler
> line 73:   mov  edx,ecx     ;length of buffer
> line 74:   mov  eax,4  ;sys_write
> line 75:   mov  ebx,1       ;file descriptor
> line 76:   mov  ecx,string  ;buffer
> line 77:
> line 78:   int  0x80
> line 79:
> line 80:
> line 81:
> line 82:
> line 83:  _test:
> line 84:           mov edi,[msg]
> line 85:           push edi
> line 86:
> line 87:           mov  edx,4         ;length of buffer
> line 88:   mov  eax,4    ;sys_write
> line 89:   mov  ebx,1         ;file descriptor
> line 90:   mov  ecx,esp       ;buffer
> line 91:   add  dword [ecx],4 ;ecx should be 'e'
> line 92:           int  0x80
> line 93:
> line 94:   ;will print 'ebc' on stdout
> line 95:
> line 96:
> line 97:  exit:
> line 98:
> line 99:   mov ebx,eax
> line 100:   mov eax,1
> line 101:   int 0x80
>
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