CODEWITHSUNDEEP
stringassemblyx86
KSuliguin
KSuliguin

Reputation: 21

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

Upvotes: 2

Views: 304

Answers (1)

Sep Roland
Sep Roland

Reputation: 39205

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

Upvotes: 1

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