Mixing
Arithmetic operation in NASM showing arbitrary character
Below is my code on NASM. I am taking two integers as user input (2 and 3) and want to add them then print to stdout. Input is taking correctly but as output it is showing some arbitrary character. What I am doing wrong? I believe something is wrong in adding input block but not sure what it is.
section .bss
fnum rest 255
lnum rest 255
section .text
global _start
_start:
; first input
mov ecx, fnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; second input
mov ecx, lnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; adding inputs 2 and 3
mov eax, fnum
add eax, ecx
add eax, 48
push eax
mov eax, esp
;print output
mov ecx, eax
mov edx, 2
mov ebx, 1
mov eax, 4
int 80h
assembly nasm
asked Nov 19 '18 at 19:25
al mamunal mamun
33
migrated from unix.stackexchange.com Nov 19 '18 at 19:52
This question came from our site for users of Linux, FreeBSD and other Un*x-like operating systems.
|
show 3 more comments
Below is my code on NASM. I am taking two integers as user input (2 and 3) and want to add them then print to stdout. Input is taking correctly but as output it is showing some arbitrary character. What I am doing wrong? I believe something is wrong in adding input block but not sure what it is.
section .bss
fnum rest 255
lnum rest 255
section .text
global _start
_start:
; first input
mov ecx, fnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; second input
mov ecx, lnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; adding inputs 2 and 3
mov eax, fnum
add eax, ecx
add eax, 48
push eax
mov eax, esp
;print output
mov ecx, eax
mov edx, 2
mov ebx, 1
mov eax, 4
int 80h
assembly nasm
asked Nov 19 '18 at 19:25
al mamunal mamun
33
migrated from unix.stackexchange.com Nov 19 '18 at 19:52
This question came from our site for users of Linux, FreeBSD and other Un*x-like operating systems.
The same arbitrary character every time, or different every time?
– Kusalananda
Nov 19 '18 at 19:35
same. something like this ▒, For simplicity I am taking inputs that will produce output less than 9.
– al mamun
Nov 19 '18 at 19:41
1
you'll need to reinventprintf("%dn", ...)to turn the contents of a register into a human-readable number, either by compiling in libc and callingprintfor by doing it manually via something like github.com/thrig/scripts/blob/master/asm/Darwin/x86_64/…
– thrig
Nov 19 '18 at 19:51
1
You're printing the low 2 bytes of the sum of the addresses, not the sum of the ASCII characters at the pointed-to locations.
– Peter Cordes
Nov 19 '18 at 19:57
@thrig I think the adding 48 accomplishes that, but I don't see the equivalent subtracting 48 from the inputs (or, more simply, subtracting 48 once and never adding it).
– Daniel H
Nov 19 '18 at 20:00
|
show 3 more comments
Below is my code on NASM. I am taking two integers as user input (2 and 3) and want to add them then print to stdout. Input is taking correctly but as output it is showing some arbitrary character. What I am doing wrong? I believe something is wrong in adding input block but not sure what it is.
section .bss
fnum rest 255
lnum rest 255
section .text
global _start
_start:
; first input
mov ecx, fnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; second input
mov ecx, lnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; adding inputs 2 and 3
mov eax, fnum
add eax, ecx
add eax, 48
push eax
mov eax, esp
;print output
mov ecx, eax
mov edx, 2
mov ebx, 1
mov eax, 4
int 80h
assembly nasm
asked Nov 19 '18 at 19:25
al mamunal mamun
33
Below is my code on NASM. I am taking two integers as user input (2 and 3) and want to add them then print to stdout. Input is taking correctly but as output it is showing some arbitrary character. What I am doing wrong? I believe something is wrong in adding input block but not sure what it is.
section .bss
fnum rest 255
lnum rest 255
section .text
global _start
_start:
; first input
mov ecx, fnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; second input
mov ecx, lnum
mov edx, 255
mov ebx, 0
mov eax, 3
int 80h
; adding inputs 2 and 3
mov eax, fnum
add eax, ecx
add eax, 48
push eax
mov eax, esp
;print output
mov ecx, eax
mov edx, 2
mov ebx, 1
mov eax, 4
int 80h
assembly nasm
assembly nasm
asked Nov 19 '18 at 19:25
al mamunal mamun
33
asked Nov 19 '18 at 19:25
al mamunal mamun
33
asked Nov 19 '18 at 19:25
al mamunal mamun
33
asked Nov 19 '18 at 19:25
al mamunal mamun
33
asked Nov 19 '18 at 19:25
al mamunal mamun
33
33
migrated from unix.stackexchange.com Nov 19 '18 at 19:52
This question came from our site for users of Linux, FreeBSD and other Un*x-like operating systems.
migrated from unix.stackexchange.com Nov 19 '18 at 19:52
This question came from our site for users of Linux, FreeBSD and other Un*x-like operating systems.
The same arbitrary character every time, or different every time?
– Kusalananda
Nov 19 '18 at 19:35
same. something like this ▒, For simplicity I am taking inputs that will produce output less than 9.
– al mamun
Nov 19 '18 at 19:41
1
you'll need to reinventprintf("%dn", ...)to turn the contents of a register into a human-readable number, either by compiling in libc and callingprintfor by doing it manually via something like github.com/thrig/scripts/blob/master/asm/Darwin/x86_64/…
– thrig
Nov 19 '18 at 19:51
1
You're printing the low 2 bytes of the sum of the addresses, not the sum of the ASCII characters at the pointed-to locations.
– Peter Cordes
Nov 19 '18 at 19:57
@thrig I think the adding 48 accomplishes that, but I don't see the equivalent subtracting 48 from the inputs (or, more simply, subtracting 48 once and never adding it).
– Daniel H
Nov 19 '18 at 20:00
|
show 3 more comments
The same arbitrary character every time, or different every time?
– Kusalananda
Nov 19 '18 at 19:35
same. something like this ▒, For simplicity I am taking inputs that will produce output less than 9.
– al mamun
Nov 19 '18 at 19:41
1
you'll need to reinventprintf("%dn", ...)to turn the contents of a register into a human-readable number, either by compiling in libc and callingprintfor by doing it manually via something like github.com/thrig/scripts/blob/master/asm/Darwin/x86_64/…
– thrig
Nov 19 '18 at 19:51
1
You're printing the low 2 bytes of the sum of the addresses, not the sum of the ASCII characters at the pointed-to locations.
– Peter Cordes
Nov 19 '18 at 19:57
@thrig I think the adding 48 accomplishes that, but I don't see the equivalent subtracting 48 from the inputs (or, more simply, subtracting 48 once and never adding it).
– Daniel H
Nov 19 '18 at 20:00
The same arbitrary character every time, or different every time?
– Kusalananda
Nov 19 '18 at 19:35
The same arbitrary character every time, or different every time?
– Kusalananda
Nov 19 '18 at 19:35
same. something like this ▒, For simplicity I am taking inputs that will produce output less than 9.
– al mamun
Nov 19 '18 at 19:41
same. something like this ▒, For simplicity I am taking inputs that will produce output less than 9.
– al mamun
Nov 19 '18 at 19:41
1
1
you'll need to reinvent
printf("%dn", ...) to turn the contents of a register into a human-readable number, either by compiling in libc and calling printf or by doing it manually via something like github.com/thrig/scripts/blob/master/asm/Darwin/x86_64/…– thrig
Nov 19 '18 at 19:51
you'll need to reinvent
printf("%dn", ...) to turn the contents of a register into a human-readable number, either by compiling in libc and calling printf or by doing it manually via something like github.com/thrig/scripts/blob/master/asm/Darwin/x86_64/…– thrig
Nov 19 '18 at 19:51
1
1
You're printing the low 2 bytes of the sum of the addresses, not the sum of the ASCII characters at the pointed-to locations.
– Peter Cordes
Nov 19 '18 at 19:57
You're printing the low 2 bytes of the sum of the addresses, not the sum of the ASCII characters at the pointed-to locations.
– Peter Cordes
Nov 19 '18 at 19:57
@thrig I think the adding 48 accomplishes that, but I don't see the equivalent subtracting 48 from the inputs (or, more simply, subtracting 48 once and never adding it).
– Daniel H
Nov 19 '18 at 20:00
@thrig I think the adding 48 accomplishes that, but I don't see the equivalent subtracting 48 from the inputs (or, more simply, subtracting 48 once and never adding it).
– Daniel H
Nov 19 '18 at 20:00
|
show 3 more comments
1 Answer
1
active
oldest
votes
When you read in the inputs, you ask for at least 255 bytes of text to be put into the locations fnum and lnum. You never read that text, and instead add the two memory addresses together (just the addresses, not the contents), and then add 48. You then print out the first two bytes of the result of that addition, which isn't actually what you want and isn't even meaningful.
You need to read the bytes read from those memory addresses ([fnum] instead of fnum), turn them into integers you can do arithmetic on, add those together instead of the memory addresses, turn the resulting number back into text, and then print the text.
For the simplest case, where both inputs and the output are single-digit nonnegative numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), you can convert from the character to the number by subtracting 0x30 (decimal 48), and you can convert from the number to the character by adding 0x30. This is because the ASCII digits are consecutive, starting at character 0x30 being the digit 0. I believe your third line after the comment "adding inputs 2 and 3" is supposed to be this sort of conversion, but the thing you're converting isn't the output you want.
In NASM, you can write 0x30 as '0', which makes the purpose more obvious to human readers.
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
As a simplification, you can just subtract 48 once and not add it back, since[fnum]+[lnum]- 48 is mathematically equal to[fnum]- 48 +[lnum]- 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.
– Daniel H
Nov 20 '18 at 17:49
add a comment |
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When you read in the inputs, you ask for at least 255 bytes of text to be put into the locations fnum and lnum. You never read that text, and instead add the two memory addresses together (just the addresses, not the contents), and then add 48. You then print out the first two bytes of the result of that addition, which isn't actually what you want and isn't even meaningful.
You need to read the bytes read from those memory addresses ([fnum] instead of fnum), turn them into integers you can do arithmetic on, add those together instead of the memory addresses, turn the resulting number back into text, and then print the text.
For the simplest case, where both inputs and the output are single-digit nonnegative numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), you can convert from the character to the number by subtracting 0x30 (decimal 48), and you can convert from the number to the character by adding 0x30. This is because the ASCII digits are consecutive, starting at character 0x30 being the digit 0. I believe your third line after the comment "adding inputs 2 and 3" is supposed to be this sort of conversion, but the thing you're converting isn't the output you want.
In NASM, you can write 0x30 as '0', which makes the purpose more obvious to human readers.
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
As a simplification, you can just subtract 48 once and not add it back, since[fnum]+[lnum]- 48 is mathematically equal to[fnum]- 48 +[lnum]- 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.
– Daniel H
Nov 20 '18 at 17:49
add a comment |
When you read in the inputs, you ask for at least 255 bytes of text to be put into the locations fnum and lnum. You never read that text, and instead add the two memory addresses together (just the addresses, not the contents), and then add 48. You then print out the first two bytes of the result of that addition, which isn't actually what you want and isn't even meaningful.
You need to read the bytes read from those memory addresses ([fnum] instead of fnum), turn them into integers you can do arithmetic on, add those together instead of the memory addresses, turn the resulting number back into text, and then print the text.
For the simplest case, where both inputs and the output are single-digit nonnegative numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), you can convert from the character to the number by subtracting 0x30 (decimal 48), and you can convert from the number to the character by adding 0x30. This is because the ASCII digits are consecutive, starting at character 0x30 being the digit 0. I believe your third line after the comment "adding inputs 2 and 3" is supposed to be this sort of conversion, but the thing you're converting isn't the output you want.
In NASM, you can write 0x30 as '0', which makes the purpose more obvious to human readers.
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
As a simplification, you can just subtract 48 once and not add it back, since[fnum]+[lnum]- 48 is mathematically equal to[fnum]- 48 +[lnum]- 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.
– Daniel H
Nov 20 '18 at 17:49
add a comment |
When you read in the inputs, you ask for at least 255 bytes of text to be put into the locations fnum and lnum. You never read that text, and instead add the two memory addresses together (just the addresses, not the contents), and then add 48. You then print out the first two bytes of the result of that addition, which isn't actually what you want and isn't even meaningful.
You need to read the bytes read from those memory addresses ([fnum] instead of fnum), turn them into integers you can do arithmetic on, add those together instead of the memory addresses, turn the resulting number back into text, and then print the text.
For the simplest case, where both inputs and the output are single-digit nonnegative numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), you can convert from the character to the number by subtracting 0x30 (decimal 48), and you can convert from the number to the character by adding 0x30. This is because the ASCII digits are consecutive, starting at character 0x30 being the digit 0. I believe your third line after the comment "adding inputs 2 and 3" is supposed to be this sort of conversion, but the thing you're converting isn't the output you want.
In NASM, you can write 0x30 as '0', which makes the purpose more obvious to human readers.
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
When you read in the inputs, you ask for at least 255 bytes of text to be put into the locations fnum and lnum. You never read that text, and instead add the two memory addresses together (just the addresses, not the contents), and then add 48. You then print out the first two bytes of the result of that addition, which isn't actually what you want and isn't even meaningful.
You need to read the bytes read from those memory addresses ([fnum] instead of fnum), turn them into integers you can do arithmetic on, add those together instead of the memory addresses, turn the resulting number back into text, and then print the text.
For the simplest case, where both inputs and the output are single-digit nonnegative numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), you can convert from the character to the number by subtracting 0x30 (decimal 48), and you can convert from the number to the character by adding 0x30. This is because the ASCII digits are consecutive, starting at character 0x30 being the digit 0. I believe your third line after the comment "adding inputs 2 and 3" is supposed to be this sort of conversion, but the thing you're converting isn't the output you want.
In NASM, you can write 0x30 as '0', which makes the purpose more obvious to human readers.
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
edited Nov 20 '18 at 1:58
Peter Cordes
120k16181311
120k16181311
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
answered Nov 20 '18 at 1:51
Daniel HDaniel H
5,07211830
5,07211830
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
As a simplification, you can just subtract 48 once and not add it back, since[fnum]+[lnum]- 48 is mathematically equal to[fnum]- 48 +[lnum]- 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.
– Daniel H
Nov 20 '18 at 17:49
add a comment |
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
As a simplification, you can just subtract 48 once and not add it back, since[fnum]+[lnum]- 48 is mathematically equal to[fnum]- 48 +[lnum]- 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.
– Daniel H
Nov 20 '18 at 17:49
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
Thank you so much for the clarification. It worked like a charm. I've assigned like mov eax, [fnum] then sub eax, 48. did same for lnum and finaly during print added 48.
– al mamun
Nov 20 '18 at 15:34
As a simplification, you can just subtract 48 once and not add it back, since
[fnum] + [lnum] - 48 is mathematically equal to [fnum] - 48 + [lnum] - 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.– Daniel H
Nov 20 '18 at 17:49
As a simplification, you can just subtract 48 once and not add it back, since
[fnum] + [lnum] - 48 is mathematically equal to [fnum] - 48 + [lnum] - 48 + 48. I wouldn't recommend it because that would make the code harder to read, though.– Daniel H
Nov 20 '18 at 17:49
add a comment |
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The same arbitrary character every time, or different every time?
– Kusalananda
Nov 19 '18 at 19:35
same. something like this ▒, For simplicity I am taking inputs that will produce output less than 9.
– al mamun
Nov 19 '18 at 19:41
1
you'll need to reinvent
printf("%dn", ...)to turn the contents of a register into a human-readable number, either by compiling in libc and callingprintfor by doing it manually via something like github.com/thrig/scripts/blob/master/asm/Darwin/x86_64/…– thrig
Nov 19 '18 at 19:51
1
You're printing the low 2 bytes of the sum of the addresses, not the sum of the ASCII characters at the pointed-to locations.
– Peter Cordes
Nov 19 '18 at 19:57
@thrig I think the adding 48 accomplishes that, but I don't see the equivalent subtracting 48 from the inputs (or, more simply, subtracting 48 once and never adding it).
– Daniel H
Nov 19 '18 at 20:00