Why won't the program print the sum of the array?
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So I'm building a simple 8086 Assembly program that allows the user to input 4 digits, store them in an array and print out the sum of those digits (The sum must be a one digit number):
data segment
i db ?
array db 20 dup(?)
sum db ?
ends
stack segment
dw 128 dup(0)
ends
code segment
mov ax, data
mov ds, ax
mov es, ax
mov i, 0
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
mov array[bx], al
inc i
cmp i, 4
jne Enter
mov sum, 0
mov i, 0
Calc:
mov bl, i
mov bh, 0
mov al, array[bx]
add sum, al
inc i
cmp i, 4
jne Calc
mov dl, sum
mov ah, 2
int 21h
mov ax, 4c00h
int 21h
ends
However when I input the numbers 1 1 2 5 instead of giving me 9 it gives me some random character.
Any ideas?
assembly x86-16
edited Nov 17 at 14:57
Fifoernik
7,52111223
asked Nov 11 at 19:44
David Daniels
595
add a comment |
up vote
2
down vote
favorite
So I'm building a simple 8086 Assembly program that allows the user to input 4 digits, store them in an array and print out the sum of those digits (The sum must be a one digit number):
data segment
i db ?
array db 20 dup(?)
sum db ?
ends
stack segment
dw 128 dup(0)
ends
code segment
mov ax, data
mov ds, ax
mov es, ax
mov i, 0
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
mov array[bx], al
inc i
cmp i, 4
jne Enter
mov sum, 0
mov i, 0
Calc:
mov bl, i
mov bh, 0
mov al, array[bx]
add sum, al
inc i
cmp i, 4
jne Calc
mov dl, sum
mov ah, 2
int 21h
mov ax, 4c00h
int 21h
ends
However when I input the numbers 1 1 2 5 instead of giving me 9 it gives me some random character.
Any ideas?
assembly x86-16
edited Nov 17 at 14:57
Fifoernik
7,52111223
asked Nov 11 at 19:44
David Daniels
595
1
You need to convert input from ascii and result to ascii.
– Jester
Nov 11 at 19:49
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So I'm building a simple 8086 Assembly program that allows the user to input 4 digits, store them in an array and print out the sum of those digits (The sum must be a one digit number):
data segment
i db ?
array db 20 dup(?)
sum db ?
ends
stack segment
dw 128 dup(0)
ends
code segment
mov ax, data
mov ds, ax
mov es, ax
mov i, 0
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
mov array[bx], al
inc i
cmp i, 4
jne Enter
mov sum, 0
mov i, 0
Calc:
mov bl, i
mov bh, 0
mov al, array[bx]
add sum, al
inc i
cmp i, 4
jne Calc
mov dl, sum
mov ah, 2
int 21h
mov ax, 4c00h
int 21h
ends
However when I input the numbers 1 1 2 5 instead of giving me 9 it gives me some random character.
Any ideas?
assembly x86-16
edited Nov 17 at 14:57
Fifoernik
7,52111223
asked Nov 11 at 19:44
David Daniels
595
So I'm building a simple 8086 Assembly program that allows the user to input 4 digits, store them in an array and print out the sum of those digits (The sum must be a one digit number):
data segment
i db ?
array db 20 dup(?)
sum db ?
ends
stack segment
dw 128 dup(0)
ends
code segment
mov ax, data
mov ds, ax
mov es, ax
mov i, 0
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
mov array[bx], al
inc i
cmp i, 4
jne Enter
mov sum, 0
mov i, 0
Calc:
mov bl, i
mov bh, 0
mov al, array[bx]
add sum, al
inc i
cmp i, 4
jne Calc
mov dl, sum
mov ah, 2
int 21h
mov ax, 4c00h
int 21h
ends
However when I input the numbers 1 1 2 5 instead of giving me 9 it gives me some random character.
Any ideas?
assembly x86-16
assembly x86-16
edited Nov 17 at 14:57
Fifoernik
7,52111223
asked Nov 11 at 19:44
David Daniels
595
edited Nov 17 at 14:57
Fifoernik
7,52111223
asked Nov 11 at 19:44
David Daniels
595
edited Nov 17 at 14:57
Fifoernik
7,52111223
edited Nov 17 at 14:57
Fifoernik
7,52111223
edited Nov 17 at 14:57
Fifoernik
7,52111223
7,52111223
asked Nov 11 at 19:44
David Daniels
595
asked Nov 11 at 19:44
David Daniels
595
asked Nov 11 at 19:44
David Daniels
595
595
1
You need to convert input from ascii and result to ascii.
– Jester
Nov 11 at 19:49
add a comment |
1
You need to convert input from ascii and result to ascii.
– Jester
Nov 11 at 19:49
1
1
You need to convert input from ascii and result to ascii.
– Jester
Nov 11 at 19:49
You need to convert input from ascii and result to ascii.
– Jester
Nov 11 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
The DOS character input function gives you characters.
When you key in 1 DOS presents you with AL='1' meaning you get 49 where you might expect 1.
When you key in 2 DOS presents you with AL='2' meaning you get 50 where you might expect 2.
When you key in 5 DOS presents you with AL='5' meaning you get 53 where you might expect 5.
That's why we subtract 48 in these cases.
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
SUB AL, '0' ;Same as SUB AL, 48
mov array[bx], al
This way your array will contain the values 1, 1, 2, and 5 (No longer the characters '1', '1', '2', and '5')
Now you can safely do the additions, yielding 9.
Because sum now holds the value 9, but you need the character '9', you simple add 48 to do the conversion:
mov dl, sum
ADD DL, '0' ;Same as ADD DL, 48
mov ah, 02h
int 21h
answered Nov 11 at 20:06
Sep Roland
11.6k21845
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
up vote
2
down vote
The DOS character input function gives you characters.
When you key in 1 DOS presents you with AL='1' meaning you get 49 where you might expect 1.
When you key in 2 DOS presents you with AL='2' meaning you get 50 where you might expect 2.
When you key in 5 DOS presents you with AL='5' meaning you get 53 where you might expect 5.
That's why we subtract 48 in these cases.
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
SUB AL, '0' ;Same as SUB AL, 48
mov array[bx], al
This way your array will contain the values 1, 1, 2, and 5 (No longer the characters '1', '1', '2', and '5')
Now you can safely do the additions, yielding 9.
Because sum now holds the value 9, but you need the character '9', you simple add 48 to do the conversion:
mov dl, sum
ADD DL, '0' ;Same as ADD DL, 48
mov ah, 02h
int 21h
answered Nov 11 at 20:06
Sep Roland
11.6k21845
add a comment |
up vote
2
down vote
The DOS character input function gives you characters.
When you key in 1 DOS presents you with AL='1' meaning you get 49 where you might expect 1.
When you key in 2 DOS presents you with AL='2' meaning you get 50 where you might expect 2.
When you key in 5 DOS presents you with AL='5' meaning you get 53 where you might expect 5.
That's why we subtract 48 in these cases.
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
SUB AL, '0' ;Same as SUB AL, 48
mov array[bx], al
This way your array will contain the values 1, 1, 2, and 5 (No longer the characters '1', '1', '2', and '5')
Now you can safely do the additions, yielding 9.
Because sum now holds the value 9, but you need the character '9', you simple add 48 to do the conversion:
mov dl, sum
ADD DL, '0' ;Same as ADD DL, 48
mov ah, 02h
int 21h
answered Nov 11 at 20:06
Sep Roland
11.6k21845
add a comment |
up vote
2
down vote
up vote
2
down vote
The DOS character input function gives you characters.
When you key in 1 DOS presents you with AL='1' meaning you get 49 where you might expect 1.
When you key in 2 DOS presents you with AL='2' meaning you get 50 where you might expect 2.
When you key in 5 DOS presents you with AL='5' meaning you get 53 where you might expect 5.
That's why we subtract 48 in these cases.
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
SUB AL, '0' ;Same as SUB AL, 48
mov array[bx], al
This way your array will contain the values 1, 1, 2, and 5 (No longer the characters '1', '1', '2', and '5')
Now you can safely do the additions, yielding 9.
Because sum now holds the value 9, but you need the character '9', you simple add 48 to do the conversion:
mov dl, sum
ADD DL, '0' ;Same as ADD DL, 48
mov ah, 02h
int 21h
answered Nov 11 at 20:06
Sep Roland
11.6k21845
The DOS character input function gives you characters.
When you key in 1 DOS presents you with AL='1' meaning you get 49 where you might expect 1.
When you key in 2 DOS presents you with AL='2' meaning you get 50 where you might expect 2.
When you key in 5 DOS presents you with AL='5' meaning you get 53 where you might expect 5.
That's why we subtract 48 in these cases.
Enter:
mov ah, 1
int 21h
mov bl, i
mov bh, 0
SUB AL, '0' ;Same as SUB AL, 48
mov array[bx], al
This way your array will contain the values 1, 1, 2, and 5 (No longer the characters '1', '1', '2', and '5')
Now you can safely do the additions, yielding 9.
Because sum now holds the value 9, but you need the character '9', you simple add 48 to do the conversion:
mov dl, sum
ADD DL, '0' ;Same as ADD DL, 48
mov ah, 02h
int 21h
answered Nov 11 at 20:06
Sep Roland
11.6k21845
edited Nov 11 at 20:12
answered Nov 11 at 20:06
Sep Roland
11.6k21845
answered Nov 11 at 20:06
Sep Roland
11.6k21845
answered Nov 11 at 20:06
Sep Roland
11.6k21845
11.6k21845
add a comment |
add a comment |
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1
You need to convert input from ascii and result to ascii.
– Jester
Nov 11 at 19:49