I want to convert all the ascii numbers that i stored in the array 'VECTOR' in binary form because i want to make the sum of them, but i don't know how to count how many digits that a number has to perform the convert.. can someone modify my function ASCBIN2??
DOSSEG
.MODEL SMALL
.STACK 32
DOSSEG
.MODEL SMALL
.STACK 32
.DATA
VECTOR DB 5 DUP(20H,20H,20H) ;20 DE NR PE MAXIM 3 CIFRE
KBD DB 4,0,0,0,0,0
TEN_POWER DW 1000,100,10,1
NUMERE DB 0Dh,0Ah,'Nr=$'
MEDIE_ELEM_DIV3 DB 0
NR_ELEMENTE_DIV3 DB 0
SUMA DW 0
MSJ_SUMA DB 'SUMA=$'
MSJ_CONTOR DB 'NR_DIV3=$'
NUMAR DW 0 ; numar in cod binar
MESAJ_SUMA DB 0Dh,0Ah,'Suma = $'
.CODE
START:
MOV AX, @DATA
MOV DS, AX
CALL CITESTE
CALL CRLF
CALL AFISARE
CALL CRLF
CALL ASCBIN
MOV AH, 4CH
INT 21H
CITESTE:
MOV CX, 5
MOV DI,(OFFSET VECTOR) + 3
AGAIN: PUSH CX
MOV DX,OFFSET NUMERE
MOV AH,9
INT 21H ; afiseaza sir de interogare
MOV [KBD+1],0
MOV AH,0Ah
MOV DX,OFFSET KBD
INT 21H ; citeste numar cu 1 pana la 3 cifre
MOV CL,[KBD+1]
MOV CH,0
MOV SI,(OFFSET KBD)+2
PUSH DI
SUB DI,CX
NEXT:
MOV AL,[SI]
MOV [DI],AL
INC SI ; memoreaza numar
INC DI
LOOP NEXT
POP DI
ADD DI,3
POP CX
LOOP AGAIN
RET
AFISARE:
MOV CX, 5
MOV SI,OFFSET VECTOR
DISP:
CALL CRLF
PUSH CX
MOV CX,3
NUM:
MOV AH,2
MOV DL,[SI]
INT 21h ; afiseaza sirul de numere IN ASCII
INC SI
LOOP NUM
POP CX
LOOP DISP
RET
CRLF:
MOV AH,2
MOV DL,0Ah
INT 21h
MOV AH,2
MOV DL,0Dh
INT 21h
RET
ASCBIN: MOV CX, 5
ASCBIN2:
PUSH CX
MOV CX, 2 ; but if i have a number on 3 digits??
MOV BX, 10
MOV SI, OFFSET VECTOR
AGAIN2: MOV AX, [NUMAR]
MUL BX ; inmulteste suma partiala cu 10
MOV DL,[SI]
MOV DH,0
AND DL,0FH ; conversie ASCII binar pentru cifra curenta
ADD AX,DX ; aduna cifra curenta
MOV [NUMAR],AX
INC SI
LOOP AGAIN2
POP CX
RET
END START
I need to iterate each element and convert it in binary. i dont know how to count the digits of every number
can someone modify my function ASCBIN2??
Your concern is about ASCBIN2 only, and more precisely about MOV CX, 2 ; but if i have a number on 3 digits??.
Because of the definition VECTOR DB 5 DUP(20H,20H,20H) and the fact that you insert the numbers using right-alignment and space-padding on the left, you have to deal with next cases:
0-digit number eg. " "
1-digit number eg. " 7"
2-digit number eg. " 94"
3-digit number eg. "208"
The solution must always process all 3 bytes, and simply skip the byte if it contains a space character:
ASCBIN2:
PUSH CX
mov cx, 3 ; ALWAYS ALL THREE BYTES
MOV BX, 10
MOV SI, OFFSET VECTOR
AGAIN2:
cmp byte ptr [si], " " ; SKIP LEADING SPACE(S)
je SkipSpace
MOV AX, [NUMAR]
MUL BX
MOV DL,[SI] ; THIS IS "0" ... "9"
MOV DH,0
AND DL,0FH ; CONVERT INTO 0 ... 9
ADD AX,DX
MOV [NUMAR],AX
SkipSpace:
INC SI
LOOP AGAIN2
POP CX
Just like I wrote in the answer to another question of yours: Allocate and initialize a vector of 20 elements in the range [20-200]. Calculate and print the average of elements divisible by 3, you still need to zero the NUMAR variable before doing this calculation. Adding the zeroing and coming up with some improvements like combining 'Test For Space' with 'Conversion From ASCII':
ASCBIN2:
push cx
xor bx, bx ; So BH is zero
mov cx, 3 ; ALWAYS ALL THREE BYTES
mov si, OFFSET VECTOR
xor di, di ; Let DI play the part of NUMAR
AGAIN2:
mov bl, [si] ; BH=0 BL={" ", "0", "1", ... , "9"}
sub bl, "0" ; From ["0","9"] to [0,9], BUT ...
jb SkipSpace ; produces a borrow if BL was a space character
mov ax, 10 ; No need to waste a register on the CONST 10
mul di
mov di, ax
add di, bx
SkipSpace:
inc si
dec cx
jnz AGAIN2
mov NUMAR, di
pop cx
If allowable for the task at hand, replace the widening multiplication mul di by the non-widening multiplication IMUL DI, 10 so that DX comes available as the counter:
ASCBIN2:
mov dx, 3 ; ALWAYS ALL THREE BYTES
mov si, OFFSET VECTOR
xor di, di ; Let DI play the part of NUMAR
AGAIN2:
movzx bx, byte ptr [si] ; BX={" ", "0", "1", ... , "9"}
sub bx, "0" ; From ["0","9"] to [0,9], BUT ...
jb SkipSpace ; produces a borrow if BX was a space character
imul di, 10
add di, bx
SkipSpace:
inc si
dec dx
jnz AGAIN2
mov NUMAR, di