The exercise is to take a word from keyboard, search the DOS environment and if that word is there, display the line with this word from DOS environment.
Here is my code:
format binary
org 100h
start:
mov es,[ds:02ch]
xor si,si
;*****************************
;*****************************
mov ah,9
mov dx,string
int 21h
mov ah,10
mov dx,word
int 21h
mov ah,0
int 16h
;*****************************
;*****************************
xor di,di
xor bx,bx
start_while:
mov di,$-word
jge equal
mov dl,[es:si]
mov bl,dl
mov dl,[bx+di]
cmp dl,[string+di]
jne next_line
add di,1
jmp start_while
next_line:
inc si
cmp dl,0
jnz notexist
equal:
mov ah,2
int 21h
jmp end
notexist:
mov ah,9
mov dx,d_exist
int 21h
jmp end
end:
mov ah,08h
int 21h
ret
;**************************************
;**************************************
string db 'THE WORD:',10,13,"$"
word db 6
db 0
times 22 db "$"
d_exist db 'variable does not exist',10,13,"$'
The compiler says: mov dl,[dl+di] error. I am a begginer, how to fix the code? I have no idea.
Aside from the issue previously mentioned with x86 base plus index addressing, here are some additional questions and problems, some of which are leading to your undesired results:
start:
mov es,[ds:02ch]
xor si,si
mov ah,9 ; Writes out THE WORD:
mov dx,string
int 21h
mov ah,10 ; Accepts user input into "word"
mov dx,word
int 21h ; reads buffered input for "word"
; reads keyboard key
; -->**** WHY? Input was just read above with INT 21 / AH = 10
;
mov ah,0
int 16h ; AH = scan code, AX = ASCII char
xor di,di
xor bx,bx
start_while:
mov di,$-word ; Moves address of current location
; minus address of "word" to di -->***WHY?
; The last operation to affect flags was xor bx,bx, which results
; in a zero. So this jump will ALWAYS be taken
jge equal
... All the code here up until the 'equal' label is skipped
equal:
mov ah,2 ; write the character in dl to stdout
; -->*** BUT the contents of dl are unknown
int 21h
jmp end ; exit the program
The above comments should explain why you aren't seeing any output or why your terminal switches off (perhaps some control character in dl).
The expression $-word in mov di,$-word is not the length of the word at word. The $ symbol represents the address of the current location in which the $ appears, so in the above, it's actually the address of that mov instruction. Therefore, you're getting some odd negative number in di than you wanted. If you want the length of the word, you should set a new data item after word, a common pattern is to make it an item after the definition of word:
word db 6 ; This defines one byte, value = 6
db 0
times 22 db "$"
word_len db $-word ; Word length is current address - addr of "word"
; I'm assuming all 24 bytes above are for "word"