Implementing cat>fileName command in NASM

[saugata bose]

I try to implement `cat>filename` command in NASM in Ubuntu 11.04 using system calls. My program is compiled successfully and run successfully (seems so). But whenever I tried to fire `cat filename` command it shows "No such file or directory" yet I see the file residing in the directory. And if I try to open the file by double clicking it shows me "You do not have the permissions necessary to open the file." Can you please help me to find the errors in my code?

The code is following:

    section .data
      msg:       dd "%d",10,0   
      msg1:   db "cat>",0
      length:   equ $-msg1
    section .bss
      a   resb 100
           len1   equ $-a
      b    resd 1
      c    resb 100      
      len2   equ $-c
    section .txt
     global main
    main:
     mov eax,4   ;;it will print cat>
     mov ebx,1
     mov ecx,msg1
     mov edx,length
     int 80h
    start:
     mov eax,3   ;;it will take the file name as input
     mov ebx,0
     mov ecx,a
     mov edx,len1
     int 80h
   
      mov eax,5   ;;it will create the file by giving owner read/write/exec permission
      mov ebx,a
      mov ecx,0100   
      mov edx,1c0h
      int 80h
   
     cmp eax,0
     jge inputAndWrite
     jmp errorSegment

    inputAndWrite:
     mov ,eax   

     mov eax,3   ;;take the input lines
     mov ebx,0
     mov ecx,c
     mov edx,len2
     int 80h

     mov edx,eax   ;;write the input lines in the file
     mov eax,4
     mov ebx,
     mov ecx,c
     int 80h   
   
     jmp done   
    errorSegment:   
     jmp done
    done:   
     mov eax, 1
     xor ebx, ebx
     int 80h

[saugata bose]

I try to implement `cat>filename` command in NASM in Ubuntu 11.04 using system calls. My program is compiled successfully and run successfully (seems so). But whenever I tried to fire `cat filename` command it shows "No such file or directory" yet I see the file residing in the directory. And if I try to open the file by double clicking it shows me "You do not have the permissions necessary to open the file." Can you please help me to find the errors in my code?

The code is following:

Code: [Select]

    section .data
   msg:     dd "%d",10,0
   msg1: db "cat>",0
   length: equ $-msg1
    section .bss
   a resb 100
           len1 equ $-a
   b resd 1
   c resb 100    
   len2 equ $-c
    section .txt
  global main
    main:
  mov eax,4   ;;it will print cat>
  mov ebx,1
  mov ecx,msg1
  mov edx,length
  int 80h
    start:
  mov eax,3   ;;it will take the file name as input
  mov ebx,0
  mov ecx,a
  mov edx,len1
  int 80h

      mov eax,5   ;;it will create the file by giving owner read/write/exec permission
      mov ebx,a
      mov ecx,0100
      mov edx,1c0h
      int 80h

  cmp eax,0
  jge inputAndWrite
  jmp errorSegment

    inputAndWrite:
  mov [b],eax

  mov eax,3   ;;take the input lines
  mov ebx,0
  mov ecx,c
  mov edx,len2
  int 80h

  mov edx,eax   ;;write the input lines in the file
  mov eax,4
  mov ebx,[b]
  mov ecx,c
  int 80h

  jmp done
    errorSegment:
  jmp done
    done:
  mov eax, 1
  xor ebx, ebx
  int 80h
   

[Frank Kotler]

Hi Saugata - I changed your "quote" and "/quote" to "code" and "/code". I think it looks a litle better - and makes it easier to copy and paste it to try.

But I wanted to ask you how you link this? I ASSume you're using "-f elf" or "-f elf32" (they're exactly the same) to assemble it. This "file or directory not found" (when it's right there!) is familiar to me. It can be caused by ld looking for a "dynamic linker" or "interpreter" that's not there. If you're using gcc to link it (which "global main" suggests) then gcc should find the right interpreter, but if you're using ld directly, you may need to tell it "-I/lib/ld-linux.so.2" or whatever's right for your system. You may need to tell gcc "-32".

Let me try this and see what I can learn...

Best,
Frank

[saugata bose]

Dear Frank,

thank you for your kind reply.
After running this code (cat>file1), file1 is created successfully. But there is nothing inside it. I don't get why my write function is not working here.

eagerly waiting for your rely.
And many many thanx for editing. Actually I don't know what is your code writing format in this forum.

Saugata

[Frank Kotler]

Well... I learned that I don't remember much about the "O_xxxx" flags to sys_open... or the permissions. I fiddled a little with your flags - surely "0100" is supposed to be "0100q" (octal)... but I "or"ed it with 2 for O_RDWR. I think I just butchered your permissions.

But the main thing that I think made a difference was that I zero-terminated your filename. sys_read does not return a zero-terminated string, it's likely to be terminated with a linefeed. I think I was creating a file named "fred\n" (fred was the filename I was using... or trying to use). After I zero-terminated it...

Code: [Select]

; get filename with sys_read
mov byte [ecx + eax - 1], 0
... things seemed to go better. I can see the "hello fred" I typed in the file "fred". The "permissions" seemed screwed up, but It "kinda works" at this point... mmm... I put the permissions back the way you had 'em... But "O_" flags as 102q... I think it was the zero-termination that fixed it.

I was getting weird results, before. The sys_open seemed to succeed, but when it came time to write the text to the file, the sys_write failed with 0xFFFFFFF7 or -9 - bad file number... ?

The code-writing syntax here is just the word "code" in square brackets - like a Nasm memory reference, not angle brackets like at SO - and  "/code" in square brackets at the end. No need for blank lines and leading spaces like at SO. I like this style better... probably because I'm used to it...

Best,
Frank

P.S. Maybe I shouldn't mention it, but I don't think this is how "cat" behaves, exactly...

[saugata bose]

Dear Frank,
 I am trying to solve the problem from last night. As I am a newbie in NASM, some concepts are lacking within me. I asked my classmates about it. But they also seem baffled. Thank you, very much.
Yet, I don't understand how to deal with "mov byte[ecx+eax -1],0". For your convenience I am inserting my MODIFIED version of the code. Can you please review the code?

Code: [Select]

section .data
msg: dd "%d",10,0
msg1: db "cat>",0
length: equ $-msg1

section .bss
a resb 100
len1 equ $-a
b resd 1
c resb 100
len2 equ $-c
section .txt
global main
main:
mov eax,4
mov ebx,1
mov ecx,msg1
mov edx,length
int 80h
start:
mov eax,3
mov ebx,0
mov ecx,a
mov edx,len1
int 80h
mov byte[ecx+eax -1],0          ;;I put your code here after taking the file name as input

mov eax,5
mov ebx,a
mov ecx,0100q
mov edx,1c0h
int 80h

cmp eax,0
jge inputAndWrite
jmp errorSegment

inputAndWrite:
mov [b],eax

mov eax,3
mov ebx,0
mov ecx,c
mov edx,len2
int 80h

mov edx,eax
mov eax,4
mov ebx,[b]
mov ecx,c
int 80h
jmp done
errorSegment:
jmp done
done:
mov eax, 1
xor ebx, ebx
int 80h
this also shows the same problem. File is created but the input lines are nor shown inside the file. I think ,I misplaced the "mov byte[ecx+eax -1],0".

[Frank Kotler]

Your zero-termination is in the right place. I've got to refresh my memory on the open flags, but I think you may be opening it for read-only. Try 0101q or 0102q in ecx for the sys_open.

Best,
Frank

[saugata bose]

Frank,
it works finally.
It works for 0101q

Thank you for all of your support. Actually I try to implement all cat operations individually. After learning from your lesson I try to implement "cat file1"
And in this case if I try to get now "cat file1" operations, here an interesting event occurs. file1 has not opened up in this case. The code is following:

Code: [Select]

ection .data
msg: dd "%d",10,0
msg1: db "cat ",0
length: equ $-msg1
section .bss
a resb 100
len1: equ $-a
b: resb 100
len2: equ $-b
section .text
global main
extern printf
main:
push ebp
mov ebp,esp

mov eax,4
mov ebx,1
mov ecx,msg1
mov edx,length
int 80h
start:
mov eax,3
mov ebx,0
mov ecx,a
mov edx,len1
int 80h
mov byte[ecx+eax -1],0
open:
mov eax,5
mov ebx,a
mov ecx,0000q
mov edx,1c0h
int 80h

cmp eax,0
jge read_write

pusha      ;;here for debugging purpose I tried to see the value of eax if file is not opened. And everytime eax remains negative,; means no file is opened up.
push msg1 ;;
call printf  ;;
add esp,4  ;;
popa         ;;

jmp exit
read_write:
pusha             ;;it is used to print the positive value of eax for succesfull opening of file
push eax
push msg
call printf
add esp,8
popa
exit:
mov ebp,esp
pop ebp
ret

Thank you again for your guidance, Frank.

Saugata