I am writing an assembly program that capitalizes the first letter of each word in the sentence I input.
My problem is that it doesn't capitalize the first letter of the words. What's wrong with my code?
Below is my code
.model small
.stack 100h
.data
prompt1 db "Input String: $"
prompt2 db "Output String: $"
InputString db 21,?,21 dup("$")
newline db 10,13,"$"
.code
start:
mov ax, #data
mov ds, ax
; Getting input string
mov ah,09h
lea dx, prompt1
int 21h
lea si, InputString
mov ah, 0Ah
mov dx, si
int 21h
nextline:
mov ah, 09h
lea dx, newline
int 21h
loop1:
mov cl, [si+1]
mov ch,0
add si, cx
inc si
dec cx
cmp cx, 0
je exit
jmp checkspace
checkspace:
cmp si, " "
inc si
je checkletter
checkletter:
cmp si, "a"
jae checkletter2
checkletter2:
cmp si, "z"
jbe capital
capital:
mov ah, [si]
xor ah, 00100000b
mov [si], ah
jmp loop1
exit:
mov ah,09h
lea dx, prompt2
int 21h
mov ah, 09h
mov dx, offset InputString+2
int 21h
mov ah, 4ch
int 21h
end start
Below is the updated part of my code. It can now only capitalize the first letter of the word. I don't know if the problem is with the loop. Please help me figure out which part of my code is wrong. I'm so sorry I don't know how to use a debugger for checking. Thank you.
mov cl, [si+1]
mov ch,0
add si,2
jmp checkletter
loop1:
inc si
dec cx
cmp cx, 0
je exit
cmp si, " "
je checkletter3
jmp loop1
checkletter:
cmp si, 41h
jae checkletter2
checkletter2:
cmp si, 5Ah
jbe capital
checkletter3:
inc si
dec cx
jmp checkletter
capital:
mov ah, [si]
xor ah, 00100000b
mov [si], ah
jmp loop1
cmp si, " "
cmp si, "a"
cmp si, "z"
cmp si, 41h
cmp si, 5Ah
All of the above are comparing the address in SI with an immediate value. They are not comparing the byte that can be found at the address that SI is pointing to with the immediate! You need to dereference it.
cmp byte [si], " "
cmp byte [si], "a"
cmp byte [si], "z"
cmp byte [si], 41h
cmp byte [si], 5Ah
To solve the task of capitalizing just the first letter of each word
You first need to find the start of each word. This is done by skipping the whitespace in front of each word.
Once located, you compare to see if the starting character is in small caps, and only if it is, you convert it to uppercase by subtracting 32.
The rest of the word is again skipped until either the end of the line is encountered or another whitespace is found for which the process starts all over again from the top.
It look like this:
mov cl, [si+1]
test cl, cl ;Exit if the input was empty!!!
jz Exit
add si, 2 ;Point at first byte
SkipSpace:
mov al, [si]
cmp al, " "
jne FirstCharacter
inc si ;(*)
dec cl
jz Exit
jmp SkipSpace
FirstCharacter:
cmp al, "a"
jb SkipRemainingCharacters
cmp al, "z"
ja SkipRemainingCharacters
sub al, 32 ;Capitalize
mov [si], al ; and write back in string
SkipRemainingCharacters:
inc si
dec cl
jz Exit
mov al, [si]
cmp al, " " ;If not space then it's part of same word
jne SkipRemainingCharacters
jmp SkipSpace ;This could just as easily jump to (*)
Exit:
Related
I am a beginner in assembly and I am trying to make a program where I should input 2 strings from the keyboard. The first string should be the main string and the second input is the substring which I need to look for in the main string. If I find it, I should display that it was found, and if not, I should display that it wasn't found.
I tried to compare the lengths of the strings so that if the first one has less characters than the second, the message "Invalid" would be displayed. Then I tried to compare the substring with the string until the substring is found in the string and the message "string found" gets displayed, if not, the message : "string not found" gets displayed. No matter what words I input, it will always say "Invalid". How can I change that?
Here is my code:
.model small
.stack 200h
.data
prompt1 db "Input String: $"
prompt2 db 10,10, 13, "Input Word: $"
prompt3 db 10,10, 13, "Output: $"
found db "Word Found. $"
notfound db "Word Not Found. $"
invalid db 10,10, 13, "Invalid. $"
InputString db 21,?,21 dup("$")
InputWord db 21,?,21 dup("$")
actlen db ?
.code
start:
mov ax, #data
mov ds, ax
mov es, ax
;Getting input string
mov ah,09h
lea dx, prompt1
int 21h
lea si, InputString
mov ah, 0Ah
mov dx, si
int 21h
;Getting input word
mov ah,09h
lea dx, prompt2
int 21h
lea di, InputWord
mov ah, 0Ah
mov dx, di
int 21h
;To check if the length of substring is shorter than the main string
mov cl, [si+1]
mov ch, 0
add si, 2
add di, 2
mov bl, [di+1]
mov bh, 0
cmp bx, cx
ja invalid_length
je valid
jb matching
valid:
cld
repe cmpsb
je found_display
jne notfound_display
mov bp, cx ;CX is length string (long)
sub bp, bx ;BX is length word (short)
inc bp
cld
lea si, [InputString + 2]
lea di, [InputWord + 2]
matching:
mov al, [si] ;Next character from the string
cmp al, [di] ;Always the first character from the word
je check
continue:
inc si ;DI remains at start of the word
dec bp
jnz matching ;More tries to do
jmp notfound_display
check:
push si
push di
mov cx, bx ;BX is length of word
repe cmpsb
pop di
pop si
jne continue
jmp found_display
again:
mov si, ax
dec dx
lea di, InputWord
jmp matching
invalid_length:
mov ah, 09h
lea dx, invalid
int 21h
jmp done
found_display:
mov dx, offset found
mov ah, 09h
int 21h
jmp done
notfound_display:
mov dx, offset notfound
mov ah, 09h
int 21h
;fallthrough is intentional
done:
mov ax,4C00h
int 21h ;exit program and return to DOS
end start
I see that you have tried to apply some of the advice I gave in the answer at Finding the substring in an input string.
But it's gone wrong mostly because you decided to special-case where the inputted string has the same length as the inputted word. That's not a special case at all! If it so happens, my calculation of the number of possible finds will remain valid and yield a 1 in the BP register. In short, your problems originate from having inserted that valid part and not having edited the program accordingly.
add si, 2
add di, 2
je valid
jb matching
valid:
cld
repe cmpsb
je found_display
jne notfound_display
You don't need all of the above once you drop the redundant valid part.
again:
mov si, ax
dec dx
lea di, InputWord
jmp matching
And don't forget to remove any code that you don't actually need in your program, especially when you use code that you found on the internet.
Solution
...
; To check if the length of substring is shorter than the main string
mov cl, [si+1]
mov ch, 0
mov bl, [di+1]
mov bh, 0
mov bp, cx ; CX is length string (long)
sub bp, bx ; BX is length word (short)
jb notfound_display
inc bp ; -> BP is number of possible finds 1+
cld
lea si, [InputString + 2]
lea di, [InputWord + 2]
matching:
mov al, [si] ; Next character from the string
cmp al, [di] ; Always the first character from the word
je check
continue:
inc si ; DI remains at start of the word
dec bp
jnz matching ; More tries to do
jmp notfound_display
check:
push si
push di
mov cx, bx ; BX is length of word
repe cmpsb
pop di
pop si
jne continue
jmp found_display
...
Some optimization
I have absorbed the instructions cmp bx, cx ja invalid_length in the calculation of the number of possible finds (shaving off 2 bytes). If the subtraction produces a borrow, you know the word is longer than the string and so you can branch away. Whether you jump to invalid_length or notfound_display is up to you...
You can shorten this program by 2 bytes if you replace lea si, [InputString + 2] lea di, [InputWord + 2] by add si, 2 add di, 2.
This should work:
.model small
.stack 100h
print macro p
lea dx,p
mov ah,09h
int 21h
endm
.data
cn db 0
pn db 0
space db 10,13, " $"
msg db 10,13, "hjut$"
msg1 db "Introduceti primul sir:$"
msg2 db "Introduceti al doilea sir:$"
msg3 db "Al doilea sir nu se gaseste in primul.$"
msg4 db "Al doilea sir se gaseste in primul. $"
ar db 20 dup("$")
br db 20 dup("$")
.code
start:
mov ax,#data
mov ds,ax
mov si,01h
mov di,00h
mov cn,00h
print msg1
read1:mov ah,01h
int 21h
mov ar[si],al
inc si
cmp al,0dh
jnz read1
mov si,00h
print msg2
read2:mov ah,01h
int 21h
mov br[si],al
inc si
cmp al,0dh
jnz read2
mov si,00h
mov di,00h
jmp lop1
lop1: mov di,00h
inc si
mov bh,ar[si]
cmp bh,0dh
jz disp
mov bh,br[di]
cmp ar[si],bh
jnz lop1
jz lop2
lop2:inc si
inc di
mov bh,br[di]
cmp bh,0dh
jz l1
mov bh,br[di]
cmp ar[si],bh
jz lop2
jmp lop1
l1:
add cn,01h
dec si
jmp lop1
disp:
cmp cn,00h
jz disp1
print msg4
add cn,30h
mov dl,cn
mov ah,02h
int 21h
jmp exit
disp1:print msg3
exit:mov ah,4ch
int 21h
end start
I am trying to make a program where the user have to enter an input, for example: Hello World and get an output: 'DLROw OLLEh'. Here is my program
org 100h
include emu8086.inc
.DATA
STR1 DB 0DH, 0AH, 'Input: $'
STR2 DB 0DH, 0AH, 'Output: $'
Nl DB 0Dh, 0Ah,'$'
.CODE
START:
MOV AX, #DATA
MOV DS, AX
DISP:
LEA DX,STR1
MOV AH,09H
INT 21H
MOV CL,00
MOV AH,01H
READ:
INT 21H
MOV BL, AL
PUSH BX
INC CX
CMP AL, 0DH
JZ DISPLAY
CMP AL, 'A' ; < then A
JB NotALetter
CMP AL, 'Z' ; > then Z
JA AGAIN ; repeat again
JMP CONTINUE1
AGAIN:
CMP AL, 'a' ; < then a
JB NotALetter
CMP AL, 'z' ; > then z
JA NotALetter
CONTINUE1:
JMP READ
DISPLAY:
LEA DX, STR2
MOV AH, 09h
INT 21H
LEA DX, NL
MOV AH, 09h
INT 21h
POP BX ; pop enter key
ANS:
MOV AH, 02h
POP BX ; pop the character
CMP BL, 'a' ; check if its in upper case
JB toLower ; if yes then jmp to toLower
SUB BL, 32 ; if not in upper case then convert to upper case
JMP CONTINUE2
toLower:
ADD BL, 32 ; convert to lower case
CMP BL, 20h
;SUB BL, 32
CONTINUE2:
MOV DL, BL
INT 21H
LOOP ANS
JMP EXIT ; if everything is fine jmp to exit
NotALetter:
printn
print "The input character is not a letter."
EXIT:
hlt
.EXIT
END START
I can enter any input but as soon as I enter any symbol, I am getting a message that this is a symbol, then program ends whereas I want to get the same output but still allow to enter a space character. I am really new in Assembly and moreover while I was trying to figure everything out I got even more lost.
If I comment out JB NotALetter and JA NotALetter, my space character becomes # probably because I am adding 20 to the ASCII hex number. Can someone please help to figure out this problem?
I can enter any input but as soon as I enter any symbol, I am getting
a message that this is a symbol, then program ends whereas I want to
get the same output but still allow to enter a space character.
As OP wants to capture the space without messing with symbol message. This can be achieved with the following:
In the READ label after you compare for enter key add this:
CMP AL, ' ' ; compare for space
JZ CONTINUE1
And in the ANS label after you pop bx add this:
CMP BL, ' ' ; if equal to space
JZ CONTINUE2 ; then print it by going to CONTINUE2 label
Just add an extra comparison to your toLower method as follows:
toLower:
CMP BL, 'A'
JL CONTINUE2
ADD BL, 32 ; convert to lower case
Complete code:
org 100h
include emu8086.inc
.DATA
STR1 DB 0DH, 0AH, 'Input: $'
STR2 DB 0DH, 0AH, 'Output: $'
Nl DB 0Dh, 0Ah,'$'
.CODE
START:
MOV AX, #DATA
MOV DS, AX
DISP:
LEA DX,STR1
MOV AH,09H
INT 21H
MOV CL,00
MOV AH,01H
READ:
INT 21H
MOV BL, AL
PUSH BX
INC CX
CMP AL, 0DH
JZ DISPLAY
CMP AL, 'A' ; < then A
JB CONTINUE1
CMP AL, 'Z' ; > then Z
JA AGAIN ; repeat again
JMP CONTINUE1
AGAIN:
CMP AL, 'a' ; < then a
JB CONTINUE1
CMP AL, 'z' ; > then z
JA CONTINUE1
CONTINUE1:
JMP READ
DISPLAY:
LEA DX, STR2
MOV AH, 09h
INT 21H
LEA DX, NL
MOV AH, 09h
INT 21h
POP BX ; pop enter key
ANS:
MOV AH, 02h
POP BX ; pop the character
CMP BL, 'a' ; check if its in upper case
JB toLower ; if yes then jmp to toLower
SUB BL, 32 ; if not in upper case then convert to upper case
JMP CONTINUE2
toLower:
CMP BL, 'A'
JL CONTINUE2
ADD BL, 32 ; convert to lower case
CONTINUE2:
MOV DL, BL
INT 21H
LOOP ANS
JMP EXIT ; if everything is fine jmp to exit
;NotALetter:
; printn
; print "The input character is not a letter."
EXIT:
hlt
.EXIT
END START
Input Hello World,
Output DLROw OLLEh
Also, you don't really need NotALetter method as you can notice, I just commented out.
I have this assembly program where I need to find the substring in the main string I input. My problem is that it always outputs the "word found" even if I typed two completely different words. I don't know which part of my loop or condition is wrong. Please help me figure it out. Also, please suggest some string instructions that could be used in checking for a substring so that I can shorten my code. I am really confused with how the cmpsb works, I only tried to use it. Btw, I don't know how to use a debugger that's why I can't debug my code and I am just a newbie in assembly language.
Below is the logic part of my code.
.data
prompt1 db "Input String: $"
prompt2 db 10,10, 13, "Input Word: $"
prompt3 db 10,10, 13, "Output: $"
found db "Word Found. $"
notfound db "Word Not Found. $"
invalid db 10,10, 13, "Invalid. $"
InputString db 21,?,21 dup("$")
InputWord db 21,?,21 dup("$")
actlen db ?
strlen dw ($-InputWord)
.code
start:
mov ax, #data
mov ds, ax
mov es, ax
;Getting input string
mov ah,09h
lea dx, prompt1
int 21h
lea si, InputString
mov ah, 0Ah
mov dx, si
int 21h
;Getting input word
mov ah,09h
lea dx, prompt2
int 21h
lea di, InputWord
mov ah, 0Ah
mov dx, di
int 21h
;To check if the length of substring is shorter than the main string
mov cl, [si+1]
mov ch, 0
add si, cx
mov bl, [di+1]
mov bh, 0
cmp bx, cx
ja invalid_length
je valid
jb matching
valid:
cld
repe cmpsb
je found_display
jne notfound_display
matching:
mov al, [si]
mov ah, [di]
cmp al, ah
je check
jne iterate
iterate:
inc si
mov dx, strlen
dec dx
cmp dx, 0
je notfound_display
jmp matching
check:
mov cl, [di+1]
mov ch, 0
mov ax, si
add ax, 1
cld
repe cmpsb
jne again
jmp found_display
again:
mov si, ax
dec dx
lea di, InputWord
jmp matching
invalid_length:
mov ah, 09h
lea dx, invalid
int 21h
strlen dw ($-InputWord)
This does nothing useful. The length that it calculate can not help you in any way!
;To check if the length of substring is shorter than the main string
mov cl, [si+1]
mov ch, 0
add si, cx
mov bl, [di+1]
mov bh, 0
cmp bx, cx
Here (as Jester told you) the add si, cx instruction is wrong. You need add si, 2 to set SI to the start of the string. You will also need to add add di, 2 to set DI to the start of the word. Do this and the valid part of your program will work correctly.
For the matching part:
Consider the case where the string has 7 characters and the word that you're looking for has 6 characters. You can find the word in at most 2 ways.
Consider the case where the string has 8 characters and the word that you're looking for has 6 characters. You can find the word in at most 3 ways.
Consider the case where the string has 9 characters and the word that you're looking for has 6 characters. You can find the word in at most 4 ways.
Notice the regularity? The number of possible finds is equal to the difference in length plus 1.
mov bp, cx ;CX is length string (long)
sub bp, bx ;BX is length word (short)
inc bp
This sets BP to the number of tries in your matching routine.
cld
lea si, [InputString + 2]
lea di, [InputWord + 2]
matching:
mov al, [si] ;Next character from the string
cmp al, [di] ;Always the first character from the word
je check
continue:
inc si ;DI remains at start of the word
dec bp
jnz matching ;More tries to do
jmp notfound_display
The check part will use repe cmpsb to test for a match, but in the event that the match is not found, you must be able to return to the matching code at the continue label. You have to preserve the registers.
check:
push si
push di
mov cx, bx ;BX is length of word
repe cmpsb
pop di
pop si
jne continue
jmp found_display
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
; multi-segment executable file template.
data segment
string db "THis is LuxUR in Summer."
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax, data
mov ds, ax
mov es, ax
mov bx, offset string
mov al, 0 ; lower letters in word
mov dl,0 ; maximum letters
check:
mov cl, 41h ; from A-Z
mov ch, 5Ah
mov ah, [bx]
cmp ah, "."
je dot
cmp ah, " "
je empty
jne letters
letters:
cmp ah, cl
je uppercase
inc cl
cmp cl, ch
jne letters
mov cl, 61h ; a-z
mov ch, 7Ah
lowercase:
inc al
cmp dl,al
jl maksimum
inc bx
jmp check
maksimum:
mov dl, al
inc bx
jmp check
uppercase:
inc bx
jmp check
empty:
mov al, 0
inc bx
jmp check
dot:
My program count lowercases in a word in al. and then puts in dl. (maximum lowercases)
I have label which name is dot. there I have to put some instruction by which I can print my result:
Summer is the word with the most lower cases 5
I try few instructions to do that but it doesnt work.
If you're using Windows, the easiest way is to use DOS Interrupts. Specifically, try interrupt 09. This takes a string and outputs it to the standard output.