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ed_crumbpacker



Joined: 30 Jan 2015
Posts: 5
ed_crumbpacker
I give up. Could someone show me how to display the contents of a register (lets say eax) for debugging purposes.

If your wondering why I don't write this code myself, it is because the program I am trying to debug is "display the contents of register eax"...so if i can't see whats in it, I can't fix it...get the catch 22?

thanks
Post 30 Jan 2015, 18:06
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JohnFound



Joined: 16 Jun 2003
Posts: 3502
Location: Bulgaria
JohnFound
You need to convert the register to string and then to write the string to the console using sys_write system call (#4), with file descriptor = 1 (STDOUT) or 2 (STDERR).
Post 30 Jan 2015, 20:40
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ed_crumbpacker



Joined: 30 Jan 2015
Posts: 5
ed_crumbpacker
that's exactly what I'm trying to do. I have puppy linux so fasm is the only compiler (no c stuff). Every example I find uses printf or some other stuff. I tried 5 to 6 different methods but they all blow up or put out nothing and without a debugger I can't tell whats happening.
Post 30 Jan 2015, 22:02
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RIxRIpt



Joined: 18 Apr 2013
Posts: 50
RIxRIpt
Code:
    ;Prints RAX to STDOUT
    display_reg:
        lea rsi, [str_reg + 16]
        .loop:
            mov cl, al
            shr rax, 4
            and cl, 0x0F
            dec rsi
            add cl, '0'
            cmp cl, '9'
            jbe @f
                add cl, 'a' - '9' - 1
            @@:
            mov [rsi], cl
            test rax, rax
            jnz .loop
        mov edx, 17
        lea rsi, [str_reg]
        mov edi, STDOUT
        mov eax, sys_write
        syscall
        ret
    
Post 30 Jan 2015, 22:55
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ed_crumbpacker



Joined: 30 Jan 2015
Posts: 5
ed_crumbpacker
OK I had to change r to e because I'm 32-bit and created a str_reg as db ? (not sure if thats right). Is mov edi,STDOUT correct? Anyway I get a segmentation fault

Code:
format ELF executable
entry display_reg

str_reg   db   ?

;Prints RAX to STDOUT
display_reg:

     mov   eax,0x423f

  
        lea esi, [str_reg + 16]
        .loop:
                mov cl, al
                shr eax, 4
                and cl, 0x0F
                dec esi
                add cl, '0'
                cmp cl, '9'
                jbe @f
                        add cl, 'a' - '9' - 1
                @@:
                mov [esi], cl
                test eax, eax
                jnz .loop
        mov edx, 17
        lea esi, [str_reg]
        mov ebx, 1
        mov eax, 4
        int 0x80
        
_eoj:                         ;exit to LINUX
   mov    eax,1
   xor    ebx,ebx
   int    0x80    


another question...how do I get my code into one of those cool white boxes?

Edit by revolution: You now have a "cool white box"
Post 30 Jan 2015, 23:30
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17662
Location: In your JS exploiting you and your system
revolution
You need to make your buffer area larger to store the string:
Code:
str_reg   rb 32 ;make space for the entire string    
Post 30 Jan 2015, 23:36
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ed_crumbpacker



Joined: 30 Jan 2015
Posts: 5
ed_crumbpacker
thanks for the help...so this is the final outcome. When I run it all the register come out with about to same numbers

eax 8048054 ebx 8048058 ecx 804805C edx 8048060

This seems suspicious to me...I'll keep playing with it

Code:
format ELF executable
entry _start

reg_eax   rd   1
reg_ebx   rd   1
reg_ecx   rd   1
reg_edx   rd   1

prt_str   db   'e'
reg_id    db   ' '
          db   'x:'
str_reg   rb   32
str_end   db   0xa
str_reg_size = $-prt_str


_start:
   mov    [reg_eax],eax       ;save off registers
   mov    [reg_ebx],ebx
   mov    [reg_ecx],ecx
   mov    [reg_edx],edx

   mov    eax,reg_eax
   mov    [reg_id],'a'        ;place the reg letter in label
   call   display_reg
   
   mov    eax,reg_ebx
   mov    [reg_id],'b' 
   call   display_reg

   mov    eax,reg_ecx
   mov    [reg_id],'c' 
   call   display_reg
   
   mov    eax,reg_edx
   mov    [reg_id],'d' 
   call   display_reg
   jmp    _eoj

display_reg:
   lea    esi, [str_reg + 16]   
.loop:
   mov    cl, al                ;mov byte 
   shr    eax, 4                ;remove small byte
   and    cl, 0x0F              ;
   dec    esi                   ;
   add    cl, '0'               ;ex. if cl = 2 + 48 = 50 (ascii '2')
   cmp    cl, '9'
   jbe    @f
   add    cl, 'a' - '9' - 1     ;ex. 97 - 57 - 1 = 39 + 58 = 97(a)
@@:
   mov    [esi], cl             ;move character to esi
   test   eax, eax
   jnz    .loop

   mov    eax, 4                ;print the register
   mov    ebx, 1
   mov    ecx, prt_str
   mov    edx, str_reg_size
   int    0x80
   ret
        
_eoj:                           ;exit to LINUX
   mov    eax,1
   xor    ebx,ebx
   int    0x80    
Post 31 Jan 2015, 05:10
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RIxRIpt



Joined: 18 Apr 2013
Posts: 50
RIxRIpt
Quote:

When I run it all the register come out with about to same numbers

Because you are loading address of register value into eax: `mov eax,reg_ebx` instead of loading register value into eax: `mov eax,[reg_ebx]`
And 8 bytes for str_reg must be enough (instead of 32):
Code:
str_reg 8 dup '0' ;duplicate '0' eight times    

And please note that old contents of str_reg are not cleared out:
Code:
test   eax, eax ;exists loop as soon as eax is is empty
jnz    .loop    

If you don't want a counter, here're some ugly(?) fixes:
Code:
;Forcing to loop thru all digits of register:
cmp esi, str_reg ;instead of `test eax, eax`

;Or clearing out str_reg before the loop:
mov [str_reg + 0], '0000'
mov [str_reg + 4], '0000'
    
Post 31 Jan 2015, 11:38
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ed_crumbpacker



Joined: 30 Jan 2015
Posts: 5
ed_crumbpacker
yeah...I saw that this morning...shouldn't write code and watch hockey at the same time
Post 31 Jan 2015, 15:20
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