The example I showed above will run,but it will not exit.

For a program to be a program,it must be complete.ie it must exit.

we must tell dos we want to exit :It is done simply by using the operator “int” which calls to interrupt.

(0..0FFH) and instruct him what to do, so lets discuss  some int’s :

009h – the keyboard interrupt

010h – the screen interrupt

013h – the sectors, (format, read, write) interrupt

016h – the keyboard interrupt

017h – the printer interrupt

021h – the main dos command

each interrupt got many command and each one needs difrrent parameters

(if you want to know get the complete int list from your local BBS)

INT 21 – TERMINATE

ah – 4ch

al – errorlevel

so in order to initiate this interrupt we will do :

mov ah,4ch

mov al,0 ; if you want it can every value

int 21h ; and then the program will exit to dos

but how will the compiler know where to start, well we have to tell

him, the compiler will know to start excatly in the first, label –

label is a name which you can jump to him, the label will be like :

Name : – “start :”, “now :”, “rt :” all of those are labels, so this is

a basic (working) assembly program :

 

Sseg segment

db 10 dup (?) ; will be disscused late

ends

Dseg segment

ends

Cseg segment

assume cs:cseg,ds:dseg,ss:sseg

start : ; the compiler will start here

mov ah,4ch

mov al,0

int 21h ; terminate

ends ; the segment must finish before the label

end start ; close the label

end ; tell compiler script ends here

 

That was easy.I hope.If you could not get it fair and square, I have some eBooks as links in the home page. Just check out them and come back.

 

It is the main assembly struc, but it can be much more complex.

now in the sseg (stack) there was “db”. what is that?

well if (most of the times) assembly registers are not enough so you (like pascal,c) can define variables, how ?

 

well you have the operators – db,dw,dd,dt :

 

DB – define byte, the var will be 1 byte (8 bit)

DW – define word, the var will be 2 bytes (word, 16 bit)

DD – define double, the var will be 4 bytes (double word,32 bit)

DT – define ten, then var will be 10 bytes (not used)

the decleration goes like this :

name type [number] value.

examples : try db 1 – it will define “try” as 1 byte with starting value 1

tru dw 4 – it will define “tru” as 2 byte with starting value 4

ytu dw ? – it will define “ytu” as 2 byte with unknown starting

value

now lets say that you want to define an array (like pascal : a:array[1..10]of

like c : int c[100])

you do as follow : name type number dup(value).

lets say : gvr db 10 dup(2) – it will define “gvr” as 10 byte which each one

has value of 2.

but what if I want to define a messege or multiple values then what ??

then it should be like :

messege – mes db “this is a messege”,10,13,”$”

now the numbers after the second “”” are optional, it’s needed only

for some interrupt.

multiple values (vectors, etc.) then :

vec db 23,54,76,34,234,14,23,123,245

db 124,34,245,23,52,34,52,43,13

db 213,213,123,123

the db can be dw,dd or dt, and you can do it for how much you want

but you must not pass the 64k limit.

but what if you want to write a messege then ???

you will search the int which do that :

int 21h – write string

ah – 9

ds:dx points to string (in the end “$” for termination)

so how will we do it ???

 

 

Sseg segment

db 10 dup (?)

ends

Dseg segment

msg db “dr. encryption”,10,13,”$” ; for termination

ends

Cseg segment

assume cs:cseg,ds:dseg,ss:sseg

start : ; the compiler will start here

mov dx,offset msg ; ds is already pointing to dseg

mov ah,9

int 21h ; whop there he is

mov ah,4ch

mov al,0

int 21h ; terminate

ends ; the segment must finish before the label

end start ; close the label

end ; tell compiler script ends here

 

 

now what offset is, well offset will return the offset of the argument

so if you write :

mov dx,offset msg

then dx will be the offset of the msg

(offset is 16bit and so is segment ,so “mov al,offset …” will be wrong)

the operator seg will return the segment of the argument, but if you change

ds and not return his original value the system will hang.

so then the stack comes to buisness.

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