Reversing a string in assembly language 80x86 - string

In a program I am currently doing, I have to reverse a string a user has entered. I have to leave the word that the user entered in where I prompted them to enter it and right below it I want to print out the word in reverse. When I try to run it in DOSBox with the Tasm compiler it gives me an error that says "Illegal memory reference" on line 189 which is the line that contains the variable I plan to put the reversed word in. Can someone help me find out what I am doing wrong? I would greatly appreciate it! Also only in my program there are 4 boxes. The first box I try to print the reverse word below the prompt. The rest of the boxes prints the user entered word instead of the reversed version of it.
title fill in title ;program name
;------------------------------------------------------------------
stacksg segment para stack 'Stack' ;define the stack
db 32 dup(0) ;32 bytes, might want larger
stacksg ends
;------------------------------------------------------------------
datasg segment para 'Data' ;data segment
paralst Label Byte
maxlen db 21
actlen db ?
dbdata db 21 dup('$')
outit db 'Enter String: $' ;14 chars minus $
switch db 21 dup('$')
datasg ends
;------------------------------------------------------------------
codesg segment para 'Code' ;code segment
main proc far ;main procedure
assume ss:stacksg, ds:datasg, cs:codesg ;define segment registers
mov ax, datasg ;initialize data segment register
mov ds, ax
;--------------- --------------------------top left corner
mov ah, 06h
mov al, 00
mov bh, 01000001b ; 4eh
mov ch, 0
mov cl, 0
mov dl, 39
mov dh, 12
int 10h
;-------------------------------------------top right corner
mov ah, 06h
mov al, 0
mov bh, 11110010b
;mov cx, 0c00h
;mov dx, 184fh
mov ch, 0
mov cl, 39
mov dh, 12
mov dl, 79
int 10h
;--------------------------------------------bottom left corner
mov ah, 06h
mov al, 0
mov bh, 11100100b ;yellow
mov ch, 12
mov cl, 0
mov dh, 24
mov dl, 39
int 10h
;------------------------------------------bottom right corner
mov ah, 06h
mov al, 0
mov bh, 01011111b ; magenta 80
mov ch, 12
mov cl, 39
mov dh, 24
mov dl, 79
int 10h
;--------------------------------------------------- 1st quad
mov ah, 02h
mov bh, 0
mov dh, 5
mov dl, 5
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 7
mov dl, 5
int 10h
call REVERSE
;--------------------------------------print output
mov ah, 09h
lea dx, switch
int 21h
;----------------------------------------------------2nd quad
mov ah, 02h
mov bh, 0
mov dh, 5
mov dl, 44
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 7
mov dl, 44
int 10h
;--------------------------------------print output
mov ah, 09h
lea dx, dbdata
int 21h
;------------------------------------------------------3rd quad
mov ah, 02h
mov bh, 0
mov dh, 17
mov dl, 5
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 19
mov dl, 5
int 10h
;--------------------------------------print output
mov ah, 09h
lea dx, dbdata
int 21h
;------------------------------------------------------4th quad
mov ah, 02h
mov bh, 0
mov dh, 17
mov dl, 44
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 19
mov dl, 44
int 10h
;--------------------------------------print output
mov ah, 09h
lea dx, dbdata
int 21h
mov ax, 4c00h ;end processing
int 21h
main endp ;end of procedure
;----------------------------------------reverse procedure
REVERSE PROC NEAR
mov cx, 0
;-----figure out actlen here
mov actlen, 0
lea bx, dbdata ;may need to use paralst instead
hi: cmp [bx], '$'
jne sup
inc actlen
inc bx
jmp hi
sup:
;------------
mov cx, 0
mov cl, actlen
lea bx, dbdata
add bx, cx
yo: cmp actlen, 0
je hola
mov switch, byte ptr[bx]
dec bx
inc switch
dec actlen
jmp yo
hola:
RET
REVERSE ENDP
codesg ends ;end of code segment
end main ;end of program

You can't do mov memory, memory. Move the source into a register first, then place it at the destination.Also, inc switch doesn't do what you seem to think it's doing. You can't change the address of switch at runtime, so what's actually happening there is that you're incrementing the first element at switch (as if you had written inc [switch]). If you need to modify an address; again, put it in a register.
So for example:
lea si, dbdata
add si,cx
lea di, switch
mov al,[bx]
jcxz hola ; jump if cx == 0
cld ; clear the direction flag; for stosb
yo: mov al,[si]
stosb ; [di++] = al
dec si
loop yo ; if (--cx) jmp yo
hola:
I didn't look at the entire piece of code, so it's unclear to me whether your reverse is supposed to copy the terminator ('$') or not. If it should, you should always execute the loop at least once (and then you don't need the jcxz). If it shouldn't, you should set the source address to dbdata + cx - 1 instead of dbdata + cx.

Related

Assembly 8086 TASM - 0Ah 21h remembers last entry

So I'm writing a uni assignment - a program that subtracts two entered decimals (max 10 characters ea). The first iteration works as intended. However, when I restart the program, for some reason the second operand is remembered.
The prompt to enter it does come up, but is then skipped as if I've entered something already - the same thing I entered the first iteration, in fact.
The question is: why does it happen and how do I fix it? The first prompt works correctly.
The prompt is under INPUT_2:
.model small
.386
stack 100h
dataseg
inputMsg1 db 0Ah, 0Dh, 'Enter first operand', 0Ah, 0Dh, '$'
inputMsg2 db 0Ah, 0Dh, 'Enter second operand', 0Ah, 0Dh, '$'
inputMax1 db 11
inputLen1 db ?
input1 db 12 dup(?)
input1Packd db 5 dup(0)
inputMax2 db 11
inputLen2 db ?
input2 db 12 dup(?)
input2Packd db 5 dup(0)
packMode db 0 ;Режим упаковки: 1 - первая цифра, 2 - вторая
resMsg db 0Ah, 0Dh, 'Result: $'
res db 9 dup(' '),'$'
retryMsg db 0Ah, 0Dh
db 'Press Any Key to continue, ESC to quit'
db '$'
errorMsg db 0Ah, 0Dh, 'Something went wrong. Try again$'
codeseg
START:
startupcode
jmp INPUT_1
INPUT_1_ERROR:
lea DX, errorMsg
mov AH, 09h
int 21h
INPUT_1:
lea DX, inputMsg1
mov AH, 09h
int 21h
lea DX, inputMax1
mov AH, 0Ah
int 21h
cmp inputLen1, 0
jz INPUT_1_ERROR
INPUT_1_PROCESS:
lea BX, input1
lea DX, input1Packd
xor CX, CX
mov CL, inputLen1
mov SI, CX
dec SI
mov DI, 4
INPUT_1_LOOP:
mov AL, [BX][SI]
cmp AL, '0'
jb INPUT_1_ERROR
cmp AL, '9'
ja INPUT_1_ERROR
and AL, 0Fh
mov AH, packMode
cmp AH, 0
jnz INPUT_1_PACK_SECOND
INPUT_1_PACK_FIRST:
inc AH
push BX
mov BX, DX
mov [BX][DI], AL
pop BX
jmp INPUT_1_PACK_FINISHED
INPUT_1_PACK_SECOND:
dec AH
shl AL, 4
push BX
mov BX, DX
or [BX][DI], AL
pop BX
dec DI
INPUT_1_PACK_FINISHED:
mov packMode, AH
dec SI
loop INPUT_1_LOOP
mov packMode, 0
jmp INPUT_2
INPUT_2_ERROR:
lea DX, errorMsg
mov AH, 09h
int 21h
INPUT_2:
lea DX, inputMsg2
mov AH, 09h
int 21h
lea DX, inputMax2
mov AH, 0Ah
int 21h
cmp inputLen2, 0
jz INPUT_2_ERROR
INPUT_2_PROCESS:
lea BX, input2
lea DX, input2Packd
xor CX, CX
mov CL, inputLen2
mov SI, CX
dec SI
mov DI, 4
INPUT_2_LOOP:
mov AL, [BX][SI]
cmp AL, '0'
jb INPUT_2_ERROR
cmp AL, '9'
ja INPUT_2_ERROR
and AL, 0Fh
mov AH, packMode
cmp AH, 0
jnz INPUT_2_PACK_SECOND
INPUT_2_PACK_FIRST:
inc AH
push BX
mov BX, DX
mov [BX][DI], AL
pop BX
jmp INPUT_2_PACK_FINISHED
INPUT_2_PACK_SECOND:
dec AH
shl AL, 4
push BX
mov BX, DX
or [BX][DI], AL
pop BX
dec DI
INPUT_2_PACK_FINISHED:
mov packMode, AH
dec SI
loop INPUT_2_LOOP
MATH_SETUP:
mov SI, 4
mov CX, 4
mov DI, 7
MATH:
lea BX, input1Packd
mov AL, [BX][SI]
lea BX, input2Packd
mov AH, [BX][SI]
sbb AL, AH
pushf
das
dec SI
mov AH, AL
lea BX, res
and AL, 0Fh
or AL, 30h
mov [BX][DI], AL
dec DI
shr AH, 4
or AH, 30h
mov [BX][DI], AH
dec DI
popf
loop MATH
lea BX, res
mov CX, 7
mov SI, 0
SHORTEN:
mov AL, [BX][SI]
cmp AL, '0'
jnz WRAPUP
inc SI
loop SHORTEN
WRAPUP:
push CX
lea DX, resMsg
mov AH, 09h
int 21h
lea DX, res
pop CX
cmp CX, 0
jz SKIP_SHORTEN
PRINT_SHORTEN:
add DX, 7
sub DX, CX
jmp FINISH_SHORTEN
SKIP_SHORTEN:
add DX, 6
FINISH_SHORTEN:
mov AH, 09h
int 21h
lea DX, retryMsg
mov AH, 09h
int 21h
mov AH, 01h
int 21h
cmp AL, 1Bh
jz QUIT
lea BX, input1Packd
mov DX, 1
BCD_CLEANUP:
mov DI, 0
mov CX, 5
BCD_CLEANUP_LOOP:
mov [BX][DI], 0
inc DI
loop BCD_CLEANUP_LOOP
lea BX, input2Packd
cmp DX, 1
mov DX, 0
jz BCD_CLEANUP
jmp START
QUIT:
exitcode 0
end START
Any better code suggestions welcome as well, but not needed if you don't have an answer for the question.
The DOS buffered input function 0Ah allows you to have a preset text in the storage space of the input buffer that you provide. For a complete explanation of this DOS function see how buffered input works.
The first time that your program runs the inputLen1 and inputLen2 fields are empty because of how you defined them in the source using db ? which translates to zero.
But when you re-run the code this is no longer the case! The lengths still show what you got in the previous run. You need to zero both these fields before invoking function 0Ah again on the same input buffers.
mov DX, 0
jz BCD_CLEANUP
mov inputLen1, DL ;DL=0
mov inputLen2, DL ;DL=0
jmp START
The MATH loop has a couple of problems regarding the CF.
The sbb al, ah instruction depends on the value in the carry flag, but you neglect to make sure it is off on the first iteration of this loop. Just add clc:
clc
MATH:
lea BX, input1Packd
mov AL, [BX][SI]
lea BX, input2Packd
mov AH, [BX][SI]
sbb AL, AH
The das instruction consumes the carry flag that you get from the sbb instruction, but it's the carry flag that you get from the das instruction that you need to preserve/restore to have it propagate through the loop.
sbb AL, AH
das
pushf
...a program that subtracts two entered decimals (max 10 characters ea)...
If ever you enter 9 or 10 characters, those most significant digits will not be taken into account because MATH_SETUP really limits you to 8 characters (which in turn is a good thing since the res buffer only has room to show 8 characters)!
MATH_SETUP:
mov SI, 4 <-- Could permit 10 packed BCD digits
mov CX, 4 <-- Max 8 characters
mov DI, 7 <-- Max 8 characters

How to convert String to Number in 8086 assembly?

I have to build a Base Converter in 8086 assembly .
The user has to choose his based and then put a number,
after then , the program will show him his number in 3 more bases[he bring a decimal number, and after this he will see his number in hex, oct, and bin.
This first question is, how can I convert the number he gave me, from string, to a number?
the sec question is, how can i convert? by RCR, and then adc some variable?
Here is my code:
data segment
N=8
ERROR_STRING_BASE DB ,10,13, " THIS IS NOT A BASE!",10,13, " TRY AGINE" ,10,13," $"
OPENSTRING DB " Welcome, to the Base Convertor",10,13," Please enter your base to convert from:",10,13," <'H'= Hex, 'D'=Dec, 'O'=oct, 'B'=bin>: $"
Hex_string DB "(H)" ,10,13, "$"
Octalic_string DB "(O) ",10,13, "$"
Binar_string DB "(B)",10,13, "$"
Dece_string DB "(D)",10,13, "$"
ENTER_STRING DB ,10,13, " Now, Enter Your Number (Up to 4 digits) ",10,13, "$"
Illegal_Number DB ,10,13, " !!! This number is illegal, lets Start again" ,10,13,"$"
BASED_BUFFER DB N,?,N+1 DUP(0)
Number_buffer db N, ? ,N+1 DUP(0)
TheBase DB N DUP(0)
The_numer DB N DUP(0)
The_binNumber DB 16 DUP(0)
data ends
sseg segment stack
dw 128 dup(0)
sseg ends
code segment
assume ss:sseg,cs:code,ds:data
start: mov ax,data
mov ds,ax
MOV DX,OFFSET OPENSTRING ;PUTS THE OPENING SRTING
MOV AH,9
INT 21H
call EnterBase
CALL CheckBase
HEXBASE: CALL PRINTtheNUMBER
MOV DX,OFFSET Hex_string
MOV AH,9
INT 21h
JMP I_have_the_numberH
oCTALICbASE: CALL PRINTtheNUMBER
MOV DX,OFFSET Octalic_string
MOV AH,9
INT 21h
JMP I_have_the_numberO
BINBASE:CALL PRINTtheNUMBER
MOV DX,OFFSET Binar_string
MOV AH,9
INT 21h
JMP I_have_the_numberB
DECBASE: CALL PRINTtheNUMBER
MOV DX,OFFSET Dece_string
MOV AH,9
INT 21h
JMP I_have_the_numberD
I_have_the_numberH: CALL BINcalculation
CALL OCTcalculation
CALL DECcalculation
I_have_the_numberO: CALL BINcalculation
CALL DECcalculation
CALL HEXcalculation
I_have_the_numberB: CALL OCTcalculation
CALL DECcalculation
CALL HEXcalculation
I_have_the_numberD: CALL BINcalculation
CALL OCTcalculation
CALL HEXcalculation
exit: mov ax, 4c00h
int 21h
EnterBase PROC
MOV DX,OFFSET BASED_BUFFER ; GETS THE BASE
MOV AH,10
INT 21H
LEA DX,BASED_BUFFER[2]
MOV BL,BASED_BUFFER[1]
MOV BH,0
MOV BASED_BUFFER[BX+2],0
LEA SI, BASED_BUFFER[2]
XOR CX, CX
MOV CL, BASED_BUFFER[1]
LEA DI, TheBase
LOL_OF_BASE: MOV DL, [SI]
MOV [DI], DL
INC SI
INC DI
INC AL
RET
EnterBase ENDP
CheckBase proc
CMP TheBase,'H'
JE HEXBASE
CMP TheBase,'h'
JE HEXBASE
CMP TheBase,'O'
JE oCTALICbASE
CMP TheBase,'o'
JE oCTALICbASE
CMP TheBase,'B'
JE BINBASE
CMP TheBase,'b'
JE BINBASE
CMP TheBase,'D'
JE DECBASE
CMP TheBase,'d'
JE DECBASE
CMP TheBase, ' '
je ERRORoFBASE
ERRORoFBASE: MOV DX,OFFSET ERROR_STRING_BASE ;PUTS WORNG BASE Illegal_Number
MOV AH,9
INT 21H
JMP START
CheckBase ENDP
PRINTtheNUMBER PROC
MOV DX,OFFSET ENTER_STRING
MOV AH,9
INT 21h
MOV DX,OFFSET Number_buffer ; GETS THE number
MOV AH,10
INT 21H
LEA DX,Number_buffer[2]
MOV BL,Number_buffer[1]
MOV BH,0
MOV Number_buffer[BX+2],0
LEA SI, Number_buffer[2]
XOR CX, CX
MOV CL, Number_buffer[1]
LEA DI, The_numer
xor AL,AL
LOL_OF_NUMBER_CHECK: MOV DL, [SI]
MOV [DI], DL
INC SI
INC DI
INC AL
CMP AL,5
JE ERRORofNUMBER
LOOP LOL_OF_NUMBER_CHECK
RET
ERRORofNUMBER: MOV DX,OFFSET Illegal_Number ;PUTS WORNG BASE Illegal_Number
MOV AH,9
INT 21H
JMP START
PRINTtheNUMBER ENDP
PROC BINcalculation
XOR CX,CX
XOR AX,AX
MOV CX,4
MOV AX,16
LEA SI, The_binNumber[0]
TheBinarLoop: RCL The_numer,1
ADC [SI],0
INC SI
LOOP TheBinarLoop
ENDP
PROC OCTcalculation
ENDP
PROC DECcalculation
ENDP
PROC HEXcalculation
ENDP
code ends
end start
It should be look like this:
thanks!
שלו לוי
the algorighm to decode ascii strings from ANY base to integer is the same:
result = 0
for each digit in ascii-string
result *= base
result += value(digit)
for { bin, oct, dec } value(digit) is ascii(digit)-ascii('0')
hex is a bit more complicated, you have to check if the value is 'a'-'f', and convert this to 10-15
converting integer to ascii(base x) is similar, you have to divide the value by base until it's 0, and add ascii representation of the remainder at the left
e.g. 87/8= 10, remainder 7 --> "7"
10/8= 1, remainder 2 --> "27"
1/8= 0, remainder 1 --> "127"
Copy-paste next little program in EMU8086 and run it : it will capture a number as string from keyboard, then convert it to numeric in BX. To store the number in "The_numer", you have to do mov The_numer, bl :
.stack 100h
;------------------------------------------
.data
;------------------------------------------
msj1 db 'Enter a number: $'
msj2 db 13,10,'Number has been converted',13,10,13,10,'$'
string db 5 ;MAX NUMBER OF CHARACTERS ALLOWED (4).
db ? ;NUMBER OF CHARACTERS ENTERED BY USER.
db 5 dup (?) ;CHARACTERS ENTERED BY USER.
;------------------------------------------
.code
;INITIALIZE DATA SEGMENT.
mov ax, #data
mov ds, ax
;------------------------------------------
;DISPLAY MESSAGE.
mov ah, 9
mov dx, offset msj1
int 21h
;------------------------------------------
;CAPTURE CHARACTERS (THE NUMBER).
mov ah, 0Ah
mov dx, offset string
int 21h
;------------------------------------------
call string2number
;------------------------------------------
;DISPLAY MESSAGE.
mov ah, 9
mov dx, offset msj2
int 21h
;------------------------------------------
;STOP UNTIL USER PRESS ANY KEY.
mov ah,7
int 21h
;------------------------------------------
;FINISH THE PROGRAM PROPERLY.
mov ax, 4c00h
int 21h
;------------------------------------------
;CONVERT STRING TO NUMBER IN BX.
proc string2number
;MAKE SI TO POINT TO THE LEAST SIGNIFICANT DIGIT.
mov si, offset string + 1 ;<================================ YOU CHANGE THIS VARIABLE.
mov cl, [ si ] ;NUMBER OF CHARACTERS ENTERED.
mov ch, 0 ;CLEAR CH, NOW CX==CL.
add si, cx ;NOW SI POINTS TO LEAST SIGNIFICANT DIGIT.
;CONVERT STRING.
mov bx, 0
mov bp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:
;CONVERT CHARACTER.
mov al, [ si ] ;CHARACTER TO PROCESS.
sub al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
mov ah, 0 ;CLEAR AH, NOW AX==AL.
mul bp ;AX*BP = DX:AX.
add bx,ax ;ADD RESULT TO BX.
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
mov ax, bp
mov bp, 10
mul bp ;AX*10 = DX:AX.
mov bp, ax ;NEW MULTIPLE OF 10.
;CHECK IF WE HAVE FINISHED.
dec si ;NEXT DIGIT TO PROCESS.
loop repeat ;COUNTER CX-1, IF NOT ZERO, REPEAT.
ret
endp
The proc you need is string2number. Pay attention inside the proc : it uses a variable named "string", you have to change it by the name of your own variable. After the call the result is in BX: if the number is less than 256, you can use the number in BL.
By the way, the string is ALWAYS converted to a DECIMAL number.

Assembly language string reversal per letter and comparison with original string

how can I reverse a string letter by letter and compare it with the first string letter by letter using SI? I originally thought about adding SI to Cl for the loop but found that it won't let me add SI to Cl. Any suggestions would be greatly appreciated. This is the code that I used and just used the same string to test the comparison.
.model small
.stack 100h
.data
input db 'Input string: $'
display db 10,10,13,'String is $'
length db 10,10,13,'String length is $'
character db 10,10,13,'Characters are:$'
equaldata db 'Equal$'
notequaldata db 'Not Equal$'
string db 20 dup('$')
.code
mov ax, #data
mov ds, ax
lea si, string
mov ah, 09h
mov dx, offset input
int 21h
mov ah, 0Ah ;request to input string
lea dx, string
int 21h
mov ah, 09h
mov dx, offset display
int 21h
lea dx, string + 2
int 21h
mov ah, 09h
mov dx, offset length
int 21h
mov bl, string + 1 ;length of string
mov ax, 0
mov al, bl ;length in hexadecimal
aam ;length in decimal
mov ch, ah ;tens digit of length
mov cl, al ;ones digit of length
mov ah, 02h
add ch, 30h
mov dl, ch ;display tens digit of length
int 21h
add cl, 30h
mov dl, cl ;display ones digit of length
int 21h
mov ah, 09h
mov dx, offset character
int 21h
mov cx, 0
mov cl, bl ;counter for loop
mov dh, cl
print_character:
mov bh, si + 2
mov ah, 02h
mov dl, 0Ah ;newline
int 21h
mov dl, 0Dh ;carriage return
int 21h
mov dl, bh ;character of string
int 21h
mov dl, 20h ;spaces
int 21h
int 21h
int 21h
int 21h
;----------------------second letter--------------
mov bl, si + 2
mov dl, bl
int 21h
mov dl, 20h
int 21h
int 21h
int 21h
int 21h
;---------------------equal or not equal-----------
cmp bh, bl
je equal
jne notequal
equal:
mov ah, 09h
mov dx, offset equaldata ;display equal data
int 21h
jmp lineend
notequal:
mov ah, 09h
mov dx, offset notequaldata ;display not equal
int 21h
jmp lineend
lineend:
inc si
dec dh
loop print_character
int 20h
end
reverse a string letter by letter and compare it with the first string letter by letter
This is pretty useless because only in the case of a palindrome the compare will result in equality. Testing for a palindrome doesn't need reversing at all! Use 2 pointers, one at the start of the string, one at te end of the string. Stop at the first inequality. If you arrive in the middle, you know the string is a palindrome.
Since you've got the string length in BL, this code will setup for the loop:
mov bh, 0
dec bx
lea di, [si+2]
Again:
mov al, [di] ;Character from the front of the string
mov ah, [di+bx] ;Character from the rear of the string
cmp al, ah
...
inc di
sub bx, 2
jnbe Again
IsPalindrome:
In your current code you inevitably get equality since you compare identical memory content.
print_character:
mov bh, si + 2
...
;----------------------second letter--------------
mov bl, si + 2
...
;---------------------equal or not equal-----------
cmp bh, bl
je equal
jne notequal

Garbage in output and wrong strings printed

I'm using nasm , and I have to do this assembly program for class that is supposed to receive a password, compare with the one stored in pass, and write whatever the password is valid or not, however for some odd reason, when printing cade it yields garbage along with the string, and when to print the other two it prints cade instead along with contr and garbage, so I'm not really sure what's happening, and if someone could help me I would really appreciate it.
Here is the macro I use to print the string
%macro inout 2
mov ah,%2
mov dx,%1
int 21h
%endmacro
And here is the program
[bits 16]
%include "macros2.inc"
[segment .data]
cade: db 'Enter the password',0ah, 0dh,'$'
invalida: db 'wrong password',0ah, 0dh,'$'
valida: db 'valid password',0ah, 0dh,'$'
pass: db '12345$'
band: db 0
contr: resb 10
[segment .code]
..start
mov ax,data
mov dx,ax
inout cade,09h
mov bx,contr
mov ah,09h
mov dx,cade
int 21h
leer:
mov ah,08h
int 21h
cmp al,0dh
je checa
mov [bx],al
inc bx
mov ah,02h
mov dl, "*"
int 21h
jmp leer
checa:
lea di,[contr]
lea si, [pass]
dec si
revisa:
loadsb
inc di
cmp [di],al
jne error
inc bx
inc di
cmp si,'$'
jl revisa
mov ah,09h
mov dx,valida
int 21h
jmp fin
error:
mov ah,09h
mov dx,invalida
int 21h
fin:
mov ah,4ch
int 21h
[bits 16]
%include "macros2.inc"
[segment .data]
cade: db 'Introduzca la contrsena',0ah, 0dh,'$'
invalida: db 'Contraseña no Valida',0ah, 0dh,'$'
valida: db 'Contraseña valida',0ah, 0dh,'$'
pass: db '12345$'
band: db 0
contr: resb 10
[segment .code]
..start
mov ax,data
mov dx,ax
inout cade,09h
mov bx,contr
mov ah,09h
mov dx,cade
int 21h
leer:
mov ah,08h
int 21h
cmp al,0dh
je checa
mov [bx],al
inc bx
mov ah,02h
mov dl, "*"
int 21h
jmp leer
checa:
lea di,[contr]
lea si, [pass]
dec si
revisa:
loadsb
inc di
cmp [di],al
jne error
inc bx
inc di
cmp si,'$'
jl revisa
mov ah,09h
mov dx,valida
int 21h
jmp fin
error:
mov ah,09h
mov dx,invalida
int 21h
fin:
mov ah,4ch
int 21h}
Doesn't take long to find your first problem...
[segment .code]
..start
mov ax,data
mov dx,ax
I think you want that to be ds, not dx.
I see you reinvent gets(). How about checking that you don't overflow your buffer?
It isn't too clear what you're trying to do after that. How about some comments?

how to check compare if palindrome or not?

I don't know much of assembly language but I tried making this palindrome and it's quite hard. First I have to enter a string then show its original and reversed string then show if its a palindrome or not.
I already figured out how to show the reverse string by pushing and popping it through a loop from what I have read in another forum and figured it out
now the only problem for me is to compare the reverse string and original string to check if its a palindrome or not.
call clearscreen
mov dh, 0
mov dl, 0
call cursor
mov ah, 09h
mov dx, offset str1
int 21h
palin db 40 dup(?)
mov ah, 0ah
mov palin, 40
mov dx, offset palin
int 21h
mov dh, 1
mov dl, 0
call cursor
mov ah, 09h
mov dx, offset str2
int 21h
mov si, 2
forward:
mov al, palin + si
cmp al, 13
je palindrome
mov ah, 02h
mov dl, al
push ax 'push the letter to reverse
int 21h
inc si
jmp forward
palindrome:
mov dh, 2
mov dl, 0
call cursor
mov ah, 09h
mov dx, offset str3
int 21h
mov cx, 40 'pop each letter through a loop to show its reverse
reverse:
mov ah, 02h
pop ax
mov dl, al
int 21h
loop reverse
int 20h
clearscreen:
mov ax, 0600h
mov bh, 0Eh
mov cx, 0
mov dx, 8025
int 10h
ret
cursor:
mov ah, 02h
mov bh, 0
int 10h
ret
str1: db "Enter A String : $"
str2: db "Forward : $"
str3: db "Backward : $"
str4: db "Its a Palindrome! $"
str5: db "Not a Palindrome!$"
You have a string the user typed in, what you need to do is compare the first byte with the last byte, do the same for the 2nd one and the 2nd to last one. Keep doing this for the whole string. You also need the length of the string. To make life easier, you should convert the string to all UPPER case letters or all lowercase letters to make comparison easier.
It is unfortunate they are still teaching 16bit DOS code. This is a sample with the word defined in the data section. You will have to modify it to receive input and work on that string
.data
pal db "racecar"
pal_len equ $ - pal - 1
szYes db "yes$"
szNo db "no$"
.code
start:
mov ax,#data
mov ds,ax
call IsPalindrome
mov ah,4ch
int 21h
IsPalindrome:
lea si, pal
lea di, pal
add di, pal_len
mov cx, 0
CheckIt:
mov al, byte ptr [si]
mov dl, byte ptr [di]
cmp al, dl
jne No
inc si
dec di
inc cx
cmp cx, pal_len
jne CheckIt
mov ah,9
lea dx,szYes
int 21h
ret
No:
mov ah,9
lea dx,szNo
int 21h
ret
end start
For completeness and to bring us into the 21st century, 32bit NASM code:
section .data
fmt db "%s", 0
szPal db "RACECAR"
Pal_len equ $ - szPal - 1
szYes db "Yes", 10, 0
szNo db "No", 10, 0
extern printf, exit
global _start
section .text
_start:
call IsPalindrome
call exit
IsPalindrome:
mov ecx, 0
mov ebx, Pal_len
mov esi, szPal
.CheckIt:
mov al, byte [esi + ecx]
mov dl, byte [esi + ebx]
cmp al, dl
jne .No
inc ecx
dec ebx
jns .CheckIt
push szYes
push fmt
call printf
add esp, 4 * 2
mov eax, 1
jmp Done
.No:
push szNo
push fmt
call printf
add esp, 4 * 2
xor eax, eax
Done:
ret

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