What is wrong with this code?

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Kristian_

Joined: 13 Nov 2004
Posts: 38

Kristian_ 19 Nov 2004, 13:19

Hi!
I'm having trouble with this simple code. It doesn't print anything out, but it should!

Code:

org 256
mov ah, 9        ; for text output
mov ax, 4     
push ax          ; push ax into the stack
pop ax           ; get the last push'ed value from the stack

mov bx, 4
cmp ax, bx       
je true              ; if(ax == bx){ goto true; } else {
jmp false         ; goto false; }

   true:
     dx, msg
     int 21h
     int 20h
   false:
    int 20h

msg DB "TRUE$"

And another thing. AX and BX registers are like normal 16 bit variables, but the difference are that they are in CPU? And CX is used for loops and DX is used for math (multiplication/devision) is it bad to use CX and DX the same way as AX, BX like normal variables? And what else registers are there, without any purpose like AX and BX? Thank you!

19 Nov 2004, 13:19

Matrix

Joined: 04 Sep 2004
Posts: 1166
Location: Overflow

Matrix 19 Nov 2004, 13:26

Kristian_ wrote:

Hi!
I'm having trouble with this simple code. It doesn't print anything out, but it should!
Code:
org 256
mov ah, 9        ; for text output
mov ax, 4     
push ax          ; push ax into the stack
pop ax           ; get the last push'ed value from the stack

mov bx, 4
cmp ax, bx        
je true              ; if(ax == bx){ goto true; } else {
jmp false         ; goto false; }

   true:
     dx, msg
     int 21h
     int 20h
   false:
    int 20h

msg DB "TRUE$"    
And another thing. AX and BX registers are like normal 16 bit variables, but the difference are that they are in CPU? And CX is used for loops and DX is used for math (multiplication/devision) is it bad to use CX and DX the same way as AX, BX like normal variables? And what else registers are there, without any purpose like AX and BX? Thank you!

that is because al is lower 8 bits of ax, and ax is lower 16 bits of eax,
so,
if you put 1 in ax, then ah will be zeroed out
btw.: ax has ah - high part and al - low part
you can use the registers for whatever except segment registers, and sp
if you modify them , you should then restore, or don't use stack for example if you're using ss,sp

you got this wrong there

Code:

mov ah, 9        ; for text output
mov ax, 4

ah is high part of ax
to see what you're doing:

Code:

mov ah,$09        ; for text output
mov ax,$0004    ; high 2 nibbles, then low 2 nibbles

19 Nov 2004, 13:26

crc

Joined: 21 Jun 2003
Posts: 637
Location: Penndel, PA [USA]

crc 19 Nov 2004, 13:34

Code:

org 256

mov ax, 4     
mov bx, 4
cmp ax, bx      
je true          ; if(ax == bx){ goto true; } else {
jmp false        ; goto false; }

        true:
          mov dx, msg    ; You forgot the 'mov'
          mov  ah, 9       ; Setting ax above cleared this, so we move it here
          int 21h
          int 20h
        false:
          int 20h

msg DB "TRUE$"

1) Setting ax will clear ah and al, so set ah later in the function.
2) No need to do push ax followed immediately by pop ax
3) This could be done more cleanly...

19 Nov 2004, 13:34

crc

Joined: 21 Jun 2003
Posts: 637
Location: Penndel, PA [USA]

crc 19 Nov 2004, 13:36

A cleaner version:

Code:

org 256

mov ax, 4     
mov bx, 4
cmp ax, bx      

jne false          ; skip the following if ax <> bx
                      ; True: do this
          mov dx, msg
          mov  ah, 9
          int 21h
false:              ; True condition will reach this and exit
          int 20h


msg DB "TRUE$"

19 Nov 2004, 13:36

Kristian_

Joined: 13 Nov 2004
Posts: 38

Kristian_ 19 Nov 2004, 17:55

Thank you! Very Happy

Quote:

ax has ah - high part and al - low part

What does it mean? For example AX register is 16bit:

[ AH 8bit]+[ ?? 8bit] = AX

AX has like two registers in it right? For example if I store this value into AX register 'ab', then AH will contain 'a' and other will contain b?

19 Nov 2004, 17:55

bubach

Joined: 17 Sep 2004
Posts: 341
Location: Trollhättan, Sweden

bubach 19 Nov 2004, 18:20

[ AH 8bit]+[ AL 8bit] = AX ...

example:

Code:

mov ah, 0x03
mov al, 0x7E     ; AX is now: 0x037E

19 Nov 2004, 18:20

Kristian_

Joined: 13 Nov 2004
Posts: 38

Kristian_ 19 Nov 2004, 19:00

Thank you!
So if AH = 1 and AL = 2 then AX = 12 NOT 3 right?
And is this the same for BX, CX and DX?

19 Nov 2004, 19:00

Matrix

Joined: 04 Sep 2004
Posts: 1166
Location: Overflow

Matrix 19 Nov 2004, 21:13

0 1 = BIT Boolean Algebra is 0 or 1
AL = 8 bits = byte
AL = 10101010 ; 8 bits
AH = 01010101 ; 8 bits
AX = 16 Bits = word
now AX is AH then AL
AH = 01010101 ; 8 bits + AL = 10101010 ; 8 bits =
AX = 01010101+10101010 = 0101010110101010
and if AX = 0101010110101010 then EAX low 16 bits are AX so eax is now
EAX = 16 times 0 then 16 BITS AX=0101010110101010
EAX = 0000000000000000 0101010110101010

all registers are the same EAX EBX ECX EDX

ESI (32 bit register) 's low 16 bits are accessible via SI
32 bit one > 16 bit part
EDI > DI
ESP > SP
EBP > BP
before you ask they do not have a byte part seperately

segments are 16 bits only
ds es fs gs ss

19 Nov 2004, 21:13

Kristian_

Joined: 13 Nov 2004
Posts: 38

Kristian_ 19 Nov 2004, 21:26

Thank you! But for what are segments used for?

19 Nov 2004, 21:26

vid
Verbosity in development

Joined: 05 Sep 2003
Posts: 7105
Location: Slovakia

vid 20 Nov 2004, 17:10

http://www.decard.net/?body=tajga&chapter=chap03
it's chapter 3.3. Not very detailed but may give you a pricture.

20 Nov 2004, 17:10

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