Extracting digit at end of string assembly 8086 masm
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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
assembly dos x86-16
asked Nov 11 at 3:10
doey doe
94
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
assembly dos x86-16
asked Nov 11 at 3:10
doey doe
94
I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doinginc siright after yourlea. So you are skipping the first byte? Then you domov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure whatmov 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 domov dx, 30hright before calling int 21. Do you really always intend to print 0? Or did you mean toadd30h?
– David Wohlferd
Nov 11 at 5:36
sorry I addedmov dx, 30hby mistake it should have beenadd dx, 30hso 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
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
assembly dos x86-16
asked Nov 11 at 3:10
doey doe
94
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
assembly dos x86-16
asked Nov 11 at 3:10
doey doe
94
asked Nov 11 at 3:10
doey doe
94
edited Nov 11 at 13:56
asked Nov 11 at 3:10
doey doe
94
asked Nov 11 at 3:10
doey doe
94
asked Nov 11 at 3:10
doey doe
94
94
I'm not quite clear on what you are trying to do. But there are a few things I notice. For example you are doinginc siright after yourlea. So you are skipping the first byte? Then you domov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure whatmov 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 domov dx, 30hright before calling int 21. Do you really always intend to print 0? Or did you mean toadd30h?
– David Wohlferd
Nov 11 at 5:36
sorry I addedmov dx, 30hby mistake it should have beenadd dx, 30hso 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 doinginc siright after yourlea. So you are skipping the first byte? Then you domov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure whatmov 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 domov dx, 30hright before calling int 21. Do you really always intend to print 0? Or did you mean toadd30h?
– David Wohlferd
Nov 11 at 5:36
sorry I addedmov dx, 30hby mistake it should have beenadd dx, 30hso 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
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
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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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 intobxregister 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 1loop1: 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 enteringmatchloop when dx is 1
– doey doe
Nov 11 at 14:29
@doeydoe Please notice that you need to change all of thoseDX's intoDL. It's always a single byte! Also I think you want to compare with the character 1, so writecmp dl, '1'
– Sep Roland
Nov 11 at 14:36
|
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 intobxregister 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 1loop1: 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 enteringmatchloop when dx is 1
– doey doe
Nov 11 at 14:29
@doeydoe Please notice that you need to change all of thoseDX's intoDL. It's always a single byte! Also I think you want to compare with the character 1, so writecmp 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 intobxregister 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 1loop1: 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 enteringmatchloop when dx is 1
– doey doe
Nov 11 at 14:29
@doeydoe Please notice that you need to change all of thoseDX's intoDL. It's always a single byte! Also I think you want to compare with the character 1, so writecmp dl, '1'
– Sep Roland
Nov 11 at 14:36
|
show 7 more comments
up vote
0
down vote
accepted
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
11.4k21843
thanks, I think I was putting everything intobxregister 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 1loop1: 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 enteringmatchloop when dx is 1
– doey doe
Nov 11 at 14:29
@doeydoe Please notice that you need to change all of thoseDX's intoDL. It's always a single byte! Also I think you want to compare with the character 1, so writecmp dl, '1'
– Sep Roland
Nov 11 at 14:36
|
show 7 more comments
thanks, I think I was putting everything intobxregister 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 1loop1: 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 enteringmatchloop when dx is 1
– doey doe
Nov 11 at 14:29
@doeydoe Please notice that you need to change all of thoseDX's intoDL. It's always a single byte! Also I think you want to compare with the character 1, so writecmp 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
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
@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
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
@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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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 siright after yourlea. So you are skipping the first byte? Then you domov bx, [si]. That reads 2 bytes from [si] (since bx is 16bits long). I'm not sure whatmov 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 domov dx, 30hright before calling int 21. Do you really always intend to print 0? Or did you mean toadd30h?– David Wohlferd
Nov 11 at 5:36
sorry I added
mov dx, 30hby mistake it should have beenadd dx, 30hso 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