Extracting digit at end of string assembly 8086 masm

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1
down vote

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I have to extract digit stored in si register. here is my code

lea si, userInput

inc si



mov bx, [si]



mov dx, [si+bx]

add dx, 30h

mov ah, 2h 

int 21h

So i think code works fine if i put hard value in dx register like mov dx, [si+2] but if try to use mov dx, [si+bx] it does not work and does not give ouput as expected

edited Nov 11 at 13:56

asked Nov 11 at 3:10

doey doe

I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doing inc si right after your lea. So you are skipping the first byte? Then you do mov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure what mov dx, [si+bx] is supposed to do. If the mov from the line above loads 1234 into bx, then this would read the byte at [si + 1234]. That can't be what you want? Then you do mov dx, 30h right before calling int 21. Do you really always intend to print 0? Or did you mean to add 30h?
– David Wohlferd
Nov 11 at 5:36

sorry I added mov dx, 30h by mistake it should have been add dx, 30h so bx contains the length of the address stored in si and I am using it to print specific index
– doey doe
Nov 11 at 13:54

add a comment |

up vote
1
down vote

favorite

I have to extract digit stored in si register. here is my code

lea si, userInput

inc si



mov bx, [si]



mov dx, [si+bx]

add dx, 30h

mov ah, 2h 

int 21h

So i think code works fine if i put hard value in dx register like mov dx, [si+2] but if try to use mov dx, [si+bx] it does not work and does not give ouput as expected

edited Nov 11 at 13:56

asked Nov 11 at 3:10

doey doe

I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doing inc si right after your lea. So you are skipping the first byte? Then you do mov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure what mov dx, [si+bx] is supposed to do. If the mov from the line above loads 1234 into bx, then this would read the byte at [si + 1234]. That can't be what you want? Then you do mov dx, 30h right before calling int 21. Do you really always intend to print 0? Or did you mean to add 30h?
– David Wohlferd
Nov 11 at 5:36

sorry I added mov dx, 30h by mistake it should have been add dx, 30h so bx contains the length of the address stored in si and I am using it to print specific index
– doey doe
Nov 11 at 13:54

add a comment |

up vote
1
down vote

favorite

I have to extract digit stored in si register. here is my code

lea si, userInput

inc si



mov bx, [si]



mov dx, [si+bx]

add dx, 30h

mov ah, 2h 

int 21h

So i think code works fine if i put hard value in dx register like mov dx, [si+2] but if try to use mov dx, [si+bx] it does not work and does not give ouput as expected

edited Nov 11 at 13:56

asked Nov 11 at 3:10

doey doe

I have to extract digit stored in si register. here is my code

lea si, userInput

inc si



mov bx, [si]



mov dx, [si+bx]

add dx, 30h

mov ah, 2h 

int 21h

So i think code works fine if i put hard value in dx register like mov dx, [si+2] but if try to use mov dx, [si+bx] it does not work and does not give ouput as expected

assembly dos x86-16

edited Nov 11 at 13:56

asked Nov 11 at 3:10

doey doe

edited Nov 11 at 13:56

asked Nov 11 at 3:10

doey doe

edited Nov 11 at 13:56

asked Nov 11 at 3:10

doey doe

asked Nov 11 at 3:10

doey doe

asked Nov 11 at 3:10

doey doe

I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doing inc si right after your lea. So you are skipping the first byte? Then you do mov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure what mov dx, [si+bx] is supposed to do. If the mov from the line above loads 1234 into bx, then this would read the byte at [si + 1234]. That can't be what you want? Then you do mov dx, 30h right before calling int 21. Do you really always intend to print 0? Or did you mean to add 30h?
– David Wohlferd
Nov 11 at 5:36

sorry I added mov dx, 30h by mistake it should have been add dx, 30h so bx contains the length of the address stored in si and I am using it to print specific index
– doey doe
Nov 11 at 13:54

add a comment |

I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doing inc si right after your lea. So you are skipping the first byte? Then you do mov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure what mov dx, [si+bx] is supposed to do. If the mov from the line above loads 1234 into bx, then this would read the byte at [si + 1234]. That can't be what you want? Then you do mov dx, 30h right before calling int 21. Do you really always intend to print 0? Or did you mean to add 30h?
– David Wohlferd
Nov 11 at 5:36

sorry I added mov dx, 30h by mistake it should have been add dx, 30h so bx contains the length of the address stored in si and I am using it to print specific index
– doey doe
Nov 11 at 13:54

I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doing inc si right after your lea. So you are skipping the first byte? Then you do mov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure what mov dx, [si+bx] is supposed to do. If the mov from the line above loads 1234 into bx, then this would read the byte at [si + 1234]. That can't be what you want? Then you do mov dx, 30h right before calling int 21. Do you really always intend to print 0? Or did you mean to add 30h?
– David Wohlferd
Nov 11 at 5:36

sorry I added mov dx, 30h by mistake it should have been add dx, 30h so bx contains the length of the address stored in si and I am using it to print specific index
– doey doe
Nov 11 at 13:54

add a comment |

1 Answer
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up vote
0
down vote

accepted

Assuming your userInput points at the input structure required for DOS's buffered input function 0Ah, these are my recommandations to correct your code:

The second byte holds the number of inputted characters. You mistakenly retrieve this as if it were a word.

The input consists of characters that are 1 byte wide. You currently retrieve it as if they were 2 bytes.

Since the input is already made up of characters, you don't need to do any conversion on it. (The attempt mov dx, 30h) The fact that this character could represent a (numerical) digit, a letter, a punctuation mark, or anything else, doesn't change this.

Your code then becomes:

lea si, userInput

inc si



mov bl, [si]     ;Number of inputted characters

mov bh, 0        ;Need to zero to be able to use the whole address-register BX next



mov dl, [si+bx]  ;Retrieve the last inputted character (right before the terminating CR)

mov ah, 02h 

int 21h          ;Display the character

answered Nov 11 at 13:57

Sep Roland

11.4k21843

thanks, I think I was putting everything into bx register directly that's why it was giving weird results
– doey doe
Nov 11 at 14:05

@doeydoe You can find some additional info about this DOS input mechanism in another post of mine : stackoverflow.com/questions/47379024/how-buffered-input-works. First chapter in the answer there.
– Sep Roland
Nov 11 at 14:09

thanks i will take a brief look
– doey doe
Nov 11 at 14:11

i have one more question i am using this loop to print everything where dx is 1 loop1: mov dx, [si+bx] cmp dx, 1 je match cmp bx, 1 je endLoop sub bx, 1 jmp loop1 match: mov ah, 2h int 21h jmp loop1 endLoop: but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

@doeydoe Please notice that you need to change all of those DX 's into DL. It's always a single byte! Also I think you want to compare with the character 1, so write cmp dl, '1'
– Sep Roland
Nov 11 at 14:36

|
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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

Assuming your userInput points at the input structure required for DOS's buffered input function 0Ah, these are my recommandations to correct your code:

The second byte holds the number of inputted characters. You mistakenly retrieve this as if it were a word.

The input consists of characters that are 1 byte wide. You currently retrieve it as if they were 2 bytes.

Since the input is already made up of characters, you don't need to do any conversion on it. (The attempt mov dx, 30h) The fact that this character could represent a (numerical) digit, a letter, a punctuation mark, or anything else, doesn't change this.

Your code then becomes:

lea si, userInput

inc si



mov bl, [si]     ;Number of inputted characters

mov bh, 0        ;Need to zero to be able to use the whole address-register BX next



mov dl, [si+bx]  ;Retrieve the last inputted character (right before the terminating CR)

mov ah, 02h 

int 21h          ;Display the character

answered Nov 11 at 13:57

Sep Roland

11.4k21843

thanks, I think I was putting everything into bx register directly that's why it was giving weird results
– doey doe
Nov 11 at 14:05

@doeydoe You can find some additional info about this DOS input mechanism in another post of mine : stackoverflow.com/questions/47379024/how-buffered-input-works. First chapter in the answer there.
– Sep Roland
Nov 11 at 14:09

thanks i will take a brief look
– doey doe
Nov 11 at 14:11

i have one more question i am using this loop to print everything where dx is 1 loop1: mov dx, [si+bx] cmp dx, 1 je match cmp bx, 1 je endLoop sub bx, 1 jmp loop1 match: mov ah, 2h int 21h jmp loop1 endLoop: but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

@doeydoe Please notice that you need to change all of those DX 's into DL. It's always a single byte! Also I think you want to compare with the character 1, so write cmp dl, '1'
– Sep Roland
Nov 11 at 14:36

|
show 7 more comments

up vote
0
down vote

accepted

Assuming your userInput points at the input structure required for DOS's buffered input function 0Ah, these are my recommandations to correct your code:

The second byte holds the number of inputted characters. You mistakenly retrieve this as if it were a word.

The input consists of characters that are 1 byte wide. You currently retrieve it as if they were 2 bytes.

Since the input is already made up of characters, you don't need to do any conversion on it. (The attempt mov dx, 30h) The fact that this character could represent a (numerical) digit, a letter, a punctuation mark, or anything else, doesn't change this.

Your code then becomes:

lea si, userInput

inc si



mov bl, [si]     ;Number of inputted characters

mov bh, 0        ;Need to zero to be able to use the whole address-register BX next



mov dl, [si+bx]  ;Retrieve the last inputted character (right before the terminating CR)

mov ah, 02h 

int 21h          ;Display the character

answered Nov 11 at 13:57

Sep Roland

11.4k21843

thanks, I think I was putting everything into bx register directly that's why it was giving weird results
– doey doe
Nov 11 at 14:05

@doeydoe You can find some additional info about this DOS input mechanism in another post of mine : stackoverflow.com/questions/47379024/how-buffered-input-works. First chapter in the answer there.
– Sep Roland
Nov 11 at 14:09

thanks i will take a brief look
– doey doe
Nov 11 at 14:11

i have one more question i am using this loop to print everything where dx is 1 loop1: mov dx, [si+bx] cmp dx, 1 je match cmp bx, 1 je endLoop sub bx, 1 jmp loop1 match: mov ah, 2h int 21h jmp loop1 endLoop: but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

@doeydoe Please notice that you need to change all of those DX 's into DL. It's always a single byte! Also I think you want to compare with the character 1, so write cmp dl, '1'
– Sep Roland
Nov 11 at 14:36

|
show 7 more comments

up vote
0
down vote

accepted

Assuming your userInput points at the input structure required for DOS's buffered input function 0Ah, these are my recommandations to correct your code:

The second byte holds the number of inputted characters. You mistakenly retrieve this as if it were a word.

The input consists of characters that are 1 byte wide. You currently retrieve it as if they were 2 bytes.

Since the input is already made up of characters, you don't need to do any conversion on it. (The attempt mov dx, 30h) The fact that this character could represent a (numerical) digit, a letter, a punctuation mark, or anything else, doesn't change this.

Your code then becomes:

lea si, userInput

inc si



mov bl, [si]     ;Number of inputted characters

mov bh, 0        ;Need to zero to be able to use the whole address-register BX next



mov dl, [si+bx]  ;Retrieve the last inputted character (right before the terminating CR)

mov ah, 02h 

int 21h          ;Display the character

answered Nov 11 at 13:57

Sep Roland

11.4k21843

Assuming your userInput points at the input structure required for DOS's buffered input function 0Ah, these are my recommandations to correct your code:

The second byte holds the number of inputted characters. You mistakenly retrieve this as if it were a word.

The input consists of characters that are 1 byte wide. You currently retrieve it as if they were 2 bytes.

Since the input is already made up of characters, you don't need to do any conversion on it. (The attempt mov dx, 30h) The fact that this character could represent a (numerical) digit, a letter, a punctuation mark, or anything else, doesn't change this.

Your code then becomes:

lea si, userInput

inc si



mov bl, [si]     ;Number of inputted characters

mov bh, 0        ;Need to zero to be able to use the whole address-register BX next



mov dl, [si+bx]  ;Retrieve the last inputted character (right before the terminating CR)

mov ah, 02h 

int 21h          ;Display the character

answered Nov 11 at 13:57

Sep Roland

11.4k21843

answered Nov 11 at 13:57

Sep Roland

11.4k21843

answered Nov 11 at 13:57

Sep Roland

11.4k21843

answered Nov 11 at 13:57

Sep Roland

11.4k21843

thanks, I think I was putting everything into bx register directly that's why it was giving weird results
– doey doe
Nov 11 at 14:05

@doeydoe You can find some additional info about this DOS input mechanism in another post of mine : stackoverflow.com/questions/47379024/how-buffered-input-works. First chapter in the answer there.
– Sep Roland
Nov 11 at 14:09

thanks i will take a brief look
– doey doe
Nov 11 at 14:11

i have one more question i am using this loop to print everything where dx is 1 loop1: mov dx, [si+bx] cmp dx, 1 je match cmp bx, 1 je endLoop sub bx, 1 jmp loop1 match: mov ah, 2h int 21h jmp loop1 endLoop: but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

@doeydoe Please notice that you need to change all of those DX 's into DL. It's always a single byte! Also I think you want to compare with the character 1, so write cmp dl, '1'
– Sep Roland
Nov 11 at 14:36

|
show 7 more comments

thanks, I think I was putting everything into bx register directly that's why it was giving weird results
– doey doe
Nov 11 at 14:05

@doeydoe You can find some additional info about this DOS input mechanism in another post of mine : stackoverflow.com/questions/47379024/how-buffered-input-works. First chapter in the answer there.
– Sep Roland
Nov 11 at 14:09

thanks i will take a brief look
– doey doe
Nov 11 at 14:11

i have one more question i am using this loop to print everything where dx is 1 loop1: mov dx, [si+bx] cmp dx, 1 je match cmp bx, 1 je endLoop sub bx, 1 jmp loop1 match: mov ah, 2h int 21h jmp loop1 endLoop: but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

@doeydoe Please notice that you need to change all of those DX 's into DL. It's always a single byte! Also I think you want to compare with the character 1, so write cmp dl, '1'
– Sep Roland
Nov 11 at 14:36

thanks, I think I was putting everything into bx register directly that's why it was giving weird results
– doey doe
Nov 11 at 14:05

@doeydoe You can find some additional info about this DOS input mechanism in another post of mine : stackoverflow.com/questions/47379024/how-buffered-input-works. First chapter in the answer there.
– Sep Roland
Nov 11 at 14:09

thanks i will take a brief look
– doey doe
Nov 11 at 14:11

i have one more question i am using this loop to print everything where dx is 1

loop1:     mov dx, [si+bx]      cmp dx, 1     je match      cmp bx, 1     je endLoop      sub bx, 1     jmp loop1   match:     mov ah, 2h      int 21h     jmp loop1  endLoop:

but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

i have one more question i am using this loop to print everything where dx is 1

loop1:     mov dx, [si+bx]      cmp dx, 1     je match      cmp bx, 1     je endLoop      sub bx, 1     jmp loop1   match:     mov ah, 2h      int 21h     jmp loop1  endLoop:

but it is not entering match loop when dx is 1
– doey doe
Nov 11 at 14:29

@doeydoe Please notice that you need to change all of those DX 's into DL. It's always a single byte! Also I think you want to compare with the character 1, so write cmp dl, '1'
– Sep Roland
Nov 11 at 14:36

|
show 7 more comments

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