TASM: Generating a simple calculator - arrays

i am trying to make a simple calculator in TASM, its not done yet, i am trying to figure out addition part but i am stuck because of Operands type do not match, here is what i have;
;FILENAME: SimpleClc.asm
;FILE FORMAT: EXE
PAGE 55,132
.386
STACK_SEG SEGMENT STACK USE16
DB 100 DUP(?)
STACK_SEG ENDS
DATA_SEG SEGMENT 'DATA' USE16
ADDITION MACRO Result,Char1,Char2
MOV AX,Char1
CWD
ADD AX,Char2,
MOV Result,AX
ENDM
SUBTRACTION MACRO Char1, Char2
SUB Char1,Char2
ENDM
DIVISION MACRO Char1, Char2
DIV Char1,Char2
ENDM
MULTIPLICATION MACRO Char1, Char2
MUL Char1,Char2
ENDM
Mainmsg DB 'Please enter a algebraic command line: $'
ErMessage DB 'Error!!', 0DH, 0AH
DB 'INPUT FORMAT:Operand1 Operator Operand',0DH, 0AH
DB 'Operand: Decimal Numbers',0DH, 0AH
DB 'Operator: + -'
INCHAR DB 21
Res DB 10 dup('$')
INCDAT DB 21 dup('$')
VarX DB 10 dup('$')
VarY DB 10 dup('$')
DATA_SEG ENDS
CODE_SEG SEGMENT PARA 'CODE' PUBLIC USE16
ASSUME CS:CODE_SEG, DS:DATA_SEG, SS:STACK_SEG
MAIN PROC FAR
PUSH DS ;INITIATE THE PROGRAM
XOR AX,AX
PUSH AX
MOV AX,DATA_SEG
MOV DS,AX
AGAIN:
LEA DX,Mainmsg ;PRINT MESSAGE
MOV AH,9
INT 21H
MOV DX, OFFSET INCHAR
MOV AH,0AH
INT 21H
MOV DX, OFFSET INCHAR
MOV AH,0AH
INT 21H
LEA DX,INCDAT ;Writing the incoming input
MOV AH,9
INT 21H
MOV DI,OFFSET INCDAT
MOV AL, [DI]
CMP AL, 9
JLE OPERAND ; jump if less or equal
JG ERRORMESSAGE ; jump if not less or equal
INC DI
OPERAND:
MOV AL, [DI]
CMP AL, '+'
JE LASTOPERAND
CMP AL, '-'
JE LASTOPERAND
CMP AL, '*'
JE LASTOPERAND
CMP AL, '/'
JE LASTOPERAND
JMP ERRORMESSAGE
INC DI
LASTOPERAND:
MOV AL, [DI]
CMP AL, 9
JLE OPERATION ; jump if less or equal
JG ERRORMESSAGE ; jump if not less or equal
OPERATION:
CMP AL, '+'
JE ADDITION1
JMP AGAIN
ADDITION1:
MOV DI,OFFSET INCDAT
MOV AL,[DI]
MOV VarX,AL
MOV AL,[DI+2]
MOV VarY,AL
ADDITION Res,VarX,VarY
JMP AGAIN
CMP AL, '-'
JE SUBTRACTION1
JMP AGAIN
SUBTRACTION1:
CMP AL, '*'
JE MULTIPLICATION1
JMP AGAIN
MULTIPLICATION1:
JMP AGAIN
CMP AL, '/'
JE DIVISION1
JMP AGAIN
DIVISION1:
ERRORMESSAGE:
LEA DX,ErMessage ;PRINT MESSAGE
MOV AH,9
INT 21H
MAIN ENDP
CODE_SEG ENDS
END MAIN
I know it looks quite messy right now, it is just because i am trying bunch of things at the same time, btw my calculator will not calculate results that are greater than 10. Thanks for help.Any comment will be appreciated.

Ups, it is hard to digit all the linenumbers.
To place code inside the data segment is not the best idea.
For what this following instruction are good?:
PUSH DS ; pushing the old address of DS to the stack
XOR AX,AX
PUSH AX ; pushing a zero word to the stack
And at last how to terminate the programm?
I am not sure, but maybe the errors occur for using labelnames that starts similar like the names of assembler mnemonics.
And why using the input function one after another for to override the same buffer?
MOV DX, OFFSET INCHAR
MOV AH,0AH
INT 21H
MOV DX, OFFSET INCHAR
MOV AH,0AH
INT 21H

Related

Define a string array using NASM - Prototype give the number of days in a month

Well as the title said , I am working on a NASM project: the idea is simple , I need to take a month from input (string) and give the number of day (using cases regarding numbers). Until now I spent the day dealing with reading/printing and finally comparing 2 strings , which I could do after some struggles and thanks to some old questions I've found here and on other forums.My current problem is that I need to put the months (names) in an array so I could do the comparing with a loop, I saw on another answer that I could label the 'array' like this:
label: db str1,str2
I tried that and when I try printing with just the label I get only the last month (I tried label+i but I still get the same thing)
Well, here is a part of my code:
segment .data
org 100h
msg db "a"
mon1 db "janvier",0
mon2 db "fevrier",0
mon3 db "mars",0
mon4 db "avril",0
mon5 db "mai",0
mon6 db "juin",0
mon7 db "juillet",0
mon8 db "aout",0
mon9 db "septembre",0
mon10 db "octobre",0
mon11 db "novembre",0
mon12 db "decembre",24h
mo dw 1,2,3,4,5,6,7,8,9,10,11,12
mon:
dw mon1,mon2,mon3,mon4,mon5,mon6,mon7,mon8,mon9,mon10,mon11,mon12
segment .code
mov dx,mon
mov ah,09h
int 21h
edit2: I tried the solution given by #ecm , I had to make a few change because It gave me some errors then after i could finally run it It went to error, Here is the whole code:
segment .code
display_month:
; takes month 1 to 12 in ax
dec ax ; make number 0-based
cmp ax,amount
jae error ; if out of range -->
add ax, ax ; make it an index into a word array
mov bx, ax ; use bx for addressing memory
mov dx, word [mon+ bx] ; access array entry
mov di, dx
mov cx, -1
mov al, 0
repne scasb
not cx
dec cx ; string length
mov bx, 1
mov ah, 40h
int 21h ; write to stdout
clc ; indicate not error
mov dx,msg1
mov ah,09h
int 21h
mov ah,0Ah
mov dx, len ;start of buffer
int 21h
mov ah,02h
mov dl,10
int 21h
mov dl,13
int 21h
mov bx,act
mov dx,buffer
add dl,byte [bx]
mov bx,dx ; move pointer into BX
mov byte [bx],24h ; put the $ there.
; compare input with msg variable(a placeholder for the moment ) I want to compare with the mon array values , and then use the index as an argument to call the cond procedure.
mov ax,msg
mov si,ax
mov ax,buffer
mov di,ax
cmpsb
jz Yep ; if strings equal goto Yep
jmp Nope ; goto Nope
Yep:
mov dx,good
mov ah,9
int 21h ; tell the user it's good
Nope:
mov dx,bad
mov ah,9
int 21h ; tell the user it's bad
end:
mov ah,4Ch
int 21h
ret
error:
stc
retn
cond:
cmp ax,2
je fev
cmp ax,7
jg odd
jle even
else: mov bx,31
jmp endif
fev: mov bx,28
odd: test ax,1
jnz trente
jmp else
even: test ax,1
jp trente
jmp else
trente: mov bx,30
endif: ret
segment .data
org 100h
;; variables declaration
msg1 db "Veullez entrer un mois:",24h
msg db "a",24h
mon1: db "janvier",0
mon2: db "fevrier",0
mon3: db "mars",0
mon4: db "avril",0
mon5: db "mai",0
mon6: db "juin",0
mon7: db "juillet",0
mon8: db "aout",0
mon9: db "septembre",0
mon10: db "octobre",0
mon11: db "novembre",0
mon12: db "decembre",0
good db "Bon choix!",24h
bad db "Mauvais choix!",24h
mon:
start:
dw mon1,mon2,mon3,mon4,mon5,mon6,mon7,mon8,mon9,mon10,mon11,mon12
tend:
len db 254 ; a fair amount of space
act db 0 ; will be filled with actual amount of chars received
buffer times 254 db 0
size equ tend - start
amount equ size / 2
I forgot to add how I am compiling , well I am using dosbox to run .com files , because they don't work on windows 10 , and I use the command : nasm name.asm -o name.com to create the .com file and then I just open it in dos.
EDIT: Well I struggled a lot to make that happen , I couldn't do it even thought I tried different ways , the last thing I went with was to use the first month and keeping adding the size of months to pass to the other ones (i.e mon1+8 gives me mon2 ...) but then I encountered with the difference in sizes , so I changed all the months to just 3 letters (4 for june) so I could move with multiples of 4 but then I couldn't go all the way to the end .. So after that I just decided to use the mon(i) by calling the names and repeating the manoeuvre , which seemed to work thought I encountered a problem with comparing (I am still trying to figure out) , well here is the last version until I figure out how to fix that.The problem at this moment is that even if I type something different than the mon1 to mon12 I get this:Le mois de 'input' est de 30 jours while normally it should take me back to the start of the nope label.
segment .code
org 100h
mov dx,msg1
mov ah,09h
int 21h
here: mov ah,0Ah
mov dx, len
int 21h
mov ah,02h
mov dl,10
int 21h
mov dl,13
int 21h
mov bx,act
mov dx,buffer
add dl,byte [bx]
mov bx,dx
mov byte [bx],24h
jmp comp
yep:
mov dx,g1
mov ah,09h
int 21h
mov dx,buffer
int 21h
mov dx,g2
int 21h
mov ax, [i]
call cond
mov ah,02h
mov dl,bh
int 21h
mov dl,bl
int 21h
mov dx,g3
mov ah,09h
int 21h
jmp end
comp: mov bx,0
mov [i],bx
jan: call inct
mov bx,buffer
mov si,bx
mov bx,mon1
mov di,bx
cmpsb
jz yep
fevr: call inct
mov bx,buffer
mov si,bx
mov bx,mon2
mov di,bx
cmpsb
jz yep
mar: call inct
mov bx,buffer
mov si,bx
mov bx,mon3
mov di,bx
cmpsb
jz yep
avr: call inct
mov bx,buffer
mov si,bx
mov bx,mon4
mov di,bx
cmpsb
jz yep
mai: call inct
mov bx,buffer
mov si,bx
mov cx,mon5
mov di,cx
cmpsb
jz yep
juin: call inct
mov bx,buffer
mov si,bx
mov bx,mon6
mov di,bx
cmpsb
jz yep
jmp jui
jui: call inct
mov cx,buffer
mov si,cx
mov cx,mon7
mov di,cx
cmpsb
jz yep
aout: call inct
mov cx,buffer
mov si,cx
mov cx,mon8
mov di,cx
cmpsb
jz yep
jmp sep
sep: call inct
mov bx,buffer
mov si,bx
mov cx,mon9
mov di,cx
cmpsb
jz yep
jmp oct
oct: call inct
mov bx,buffer
mov si,bx
mov cx,mon10
mov di,cx
cmpsb
jz yep
jmp nov
nov: call inct
mov bx,buffer
mov si,bx
mov cx,mon11
mov di,cx
cmpsb
jz yep
jmp dect
dect: call inct
mov bx,buffer
mov si,bx
mov cx,mon12
mov di,cx
cmpsb
jz yep
jmp nope
nope:
mov dx,bad
mov ah,9
int 21h
jmp here
end:
mov ah,4Ch
int 21h
ret
inct:
push bx
mov bx,[i]
inc bx
mov [i],bx
pop bx
ret
cond:
cmp ax,2
je fev
cmp ax,6
je th
cmp ax,8
je th
cmp ax,7
jg odd
jle even
else: mov bl,'1'
mov bh,'3'
jmp endif
fev: mov bl, '8'
mov bh,'2'
jmp endif
th: jmp else
odd: test ax,1
jnz trente
jmp else
even: test ax,1
jp trente
jmp else
trente: mov bh,'3'
mov bl,'0'
endif: ret
segment .data
i db 0
msg1 db "Veullez entrer un mois:",24h
mon1: db "janvier",24h
mon2: db "fevrier",24h
mon3: db "mars",24h
mon4: db "avril",24h
mon5: db "mai",24h
mon6: db "juin",24h
mon7: db "juillet",24h
mon8: db "aout",24h
mon9: db "septembre",24h
mon10: db "octobre",24h
mon11: db "novembre",24h
mon12: db "decembre",24h
g1 db "Le mois de",20h,24h
g2 db 20h,"est de",20h,24h
g3 db 20h,"jours",24h
bad db "Le mois saisi n'est pas correct!",10,13,"Veuillez entrer un autre mois:",24h
len db 254 ; a fair amount of space
act db 0 ; will be filled with actual amount of chars received
buffer times 254 db 0
Try this:
segment .code
display_month:
; takes month 1 to 12 in ax
dec ax ; make number 0-based
cmp ax, montab.amount
jae .error ; if out of range -->
add ax, ax ; make it an index into a word array
mov bx, ax ; use bx for addressing memory
mov dx, word [montab + bx] ; access array entry
mov di, dx
mov cx, -1
mov al, 0
repne scasb
not cx
dec cx ; string length
mov bx, 1
mov ah, 40h
int 21h ; write to stdout
clc ; indicate not error
retn
.error:
stc
retn
segment .data
mon1: db "janvier",0
mon2: db "fevrier",0
mon3: db "mars",0
mon4: db "avril",0
mon5: db "mai",0
mon6: db "juin",0
mon7: db "juillet",0
mon8: db "aout",0
mon9: db "septembre",0
mon10: db "octobre",0
mon11: db "novembre",0
mon12: db "decembre",0
align 2
montab:
.:
dw mon1
dw mon2
dw mon3
dw mon4
dw mon5
dw mon6
dw mon7
dw mon8
dw mon9
dw mon10
dw mon11
dw mon12
.end:
.size equ .end - .
.amount equ .size / 2
You could use 12 as the hardcoded array length in this case. But using equates for the length and amount of entries is useful for static data arrays more generally.
ETA: I added alignment for the table, which is good for performance. Not needed but it doesn't cost much.
Note that I dropped the org directive from my example. This is because I am presenting just one function that should be called from other program logic. If you're assembling into a simple style flat .COM executable for 86-DOS, you still need to include the org 256 at some point.
Here's a breakdown of your question's attempt:
msg db "a"
This seems like an unused leftover.
mon1 db "janvier",0
mon2 db "fevrier",0
mon3 db "mars",0
mon4 db "avril",0
mon5 db "mai",0
mon6 db "juin",0
mon7 db "juillet",0
mon8 db "aout",0
mon9 db "septembre",0
mon10 db "octobre",0
mon11 db "novembre",0
These are all fine, they define ASCIZ strings. (The Z stands for zero-terminated.)
mon12 db "decembre",24h
This one uses a different terminator, in this case the dollar sign (24h = 36). It is not appropriate to mix terminators between array entries. Either use all ASCIZ or all CP/M-style dollar-terminated strings.
mo dw 1,2,3,4,5,6,7,8,9,10,11,12
This is useless. If you wanted to map a number from 1 to 12 as index into this array, the original value would be found at that array entry. If the entries had different numbers this type of array could be useful, but not with identity mapping.
mon:
dw mon1,mon2,mon3,mon4,mon5,mon6,mon7,mon8,mon9,mon10,mon11,mon12
This is essentially correct. (I put each entry on its own line and prepended the align directive, but neither is absolutely necessary.)
mov dx,mon
mov ah,09h
int 21h
This loads the address of your message table into dx then passes dx to interrupt 21h function 09h. This function expects a dollar-terminated string, so only your December string would work. More importantly, it will try to display the literal bytes that make up your array. You need to, instead, dereference a pointer to one of your array's entries to load the address stored therein, which is the address of the associated message string.

I got stuck at "Enter a string:" and doesn't take any input-8086 programming

In the program for to find out no. of vowels in a string. I got stuck at "Enter a string:" ?? why? even tho the compiler says everything okay.
program to count the no. of vowels in a string.
;;;;;;;PROGRAM TO CHECK NO. OF VOWELS IN A STRING;;;;;
.model small
.stack 100h
.data
vowels db 'AEIOUaeiou$'
msg1 db 'Enter a string:$'
msg2 db 'The string is:$'
msg3 db 'No of vowels are:$'
string db 50 dup('$')
count db ?
.code
main proc
mov ax, #data
mov ds, ax
mov es, ax
lea dx,msg1
mov ah,09h ;for displaying enter a string
int 21h
lea di,string
cld
xor bl,bl
input:mov ah,01 ; stuck here, at taking input, why??
cmp al, 13
je endinput
stosb
inc bl
jmp input
endinput:cld
xor bh,bh
lea si,string
vowelornot: mov cx,11
lodsb
lea di,vowels
repne scasb
cmp cx, 00
je stepdown
inc bh
stepdown: dec bl
jnz vowelornot
mov ah,06 ;;;THIS FOR CLEARING SCREEN I GUESS
mov al,0 ;;;
int 10h
lea dx,msg2
mov ah,09
int 21h
mov dl, 13 ;;; NEW LINE
mov ah, 2
int 21h
mov dl, 10
mov ah, 2
int 21h
lea dx,string
mov ah, 09
int 21h
mov dl,13 ;;;NEW LINE
mov ah,2
int 21h
mov dl,10
mov ah,2
int 21h
lea dx, msg3
mov ah,09
int 21h
mov dl,13 ;;;NEW LINE
mov ah,2
int 21h
mov dl,10
mov ah,2
int 21h
mov count, bh
mov dh, count ;;; DH = VOWEL COUNT
mov ah,09
int 21h
mov ah, 4ch ;;; EXIT
int 21h
main endp
end
Multiple errors
input:mov ah,01 ; stuck here, at taking input, why??
cmp al, 13
Here your code is missing the int 21h instruction!
input:
mov ah,01
int 21h
cmp al, 13
xor bl,bl
input:
stosb
inc bl
jmp input
You're using BL to count the number of characters in the input string but the example you wrote needs much more than the 255 maximum that this byte-sized register can give you. This must fail!
Moreover the buffer that you've setup is limited to 50 bytes. No way you could store there such a long input.
lea si,string
vowelornot: mov cx,11
lodsb
lea di,vowels
repne scasb
cmp cx, 00
je stepdown
inc bh
stepdown: dec bl
jnz vowelornot
This is too complicated. Just interpret the ZeroFlag and don't look at CX at all. You don't need to terminate the vowels text with a "$" anymore (use CX=10).
lea si,string
vowelornot:
lodsb
mov cx,10
lea di,vowels
repne scasb
jne stepdown
inc bh ;Vowel found +1
stepdown:
dec bl
jnz vowelornot
mov ah,06 ;;;THIS FOR CLEARING SCREEN I GUESS
mov al,0 ;;;
int 10h
Sure, function 06h can clear the screen but you need to provide all of the required arguments. Upperleftcorner in CX, Lowerrightcorner in DX and Displaypage in BH.
mov dx, 184Fh ;(79,24) If screen is 80x25
xor cx, cx ;(0,0)
mov bh, 0
mov ax, 0600h
int 10h
lea dx,string
mov ah, 09
int 21h
This will fail because you did not put a "$" character at the end of the inputted string.
And if you're going to output a CRLF directly afterwards, why not add it to the buffer?
jmp input
endinput:
mov ax, 0A0Dh <-- ADD THIS
stosw <-- ADD THIS
mov al, "$" <-- ADD THIS
stosb <-- ADD THIS
xor bh,bh
You print msg2 and msg3 followed by a CRLF. Why don't you append it in the definition? No more need to output it separately.
msg2 db 'The string is:', 13, 10, '$'
msg3 db 'No of vowels are:', 13, 10, '$'
mov count, bh
mov dh, count ;;; DH = VOWEL COUNT
mov ah,09
int 21h
To output the count and provided it is a number in the range from 0 to 9, you need to convert the number into a character. Just add 48 or '0'.
Don't use function 09h. It requires an address and you clearly want to use a character.
mov dl, bh ;Count of vowels [0,9]
add dl, '0'
mov ah, 02h
int 21h

masm palindrome check output dilemma

I was trying to do a palindrome check program using masm and used the below code but I always got not palindrome as output, not able to figure it out why. Please help to modify the code. The code is given below for further clarification.
Thanks in advance.
read macro
mov ah,01h
int 21h
endm
write macro
mov ah,02h
int 21h
endm
assume cs:code,ds:data
data segment
a db 20 dup(?)
b db 20 dup(?)
msg1 db "enter text: $"
msg2 db "reverse string is: $"
msg3 db " palindrome$"
msg4 db " not palindrome$"
data ends
code segment
start:mov ax,data
mov ds,ax
mov es,ax
mov cx,0000h
lea si,a
lea dx,msg1
mov ah,09h
int 21h
r:read
cmp al,0dh
je l
mov [si],al
inc si
inc cx
jmp r
l:dec si
lea di,b
mov ax,cx
l1:mov dl,[si]
mov [di],dl
inc di
dec si
loop l1
l2:
mov cx,ax
lea si,b
lea dx,msg2
mov ah,9h
int 21h
w:mov dl,[si]
write
inc si
loop w
mov cx,ax
cld
lea si,a
lea di,b
l4:cmpsb
jne l5
loop l4
lea dx,msg3
mov ah,09h
int 21h
mov ah,4ch
int 21h
l5:lea dx,msg4
mov ah,09h
int 21h
mov ah,4ch
int 21h
code ends
end start
whatever is input string it gives not a palindrome.
Instead of doing mov ax,cx use mov bx,cx as ah is used in interrupts it might cause some problems. And every where mov cx,ax is there use mov cx,bx ....now it will be correct.
It might seem funny that I answered my own question but I figured this now and didn't delete cause it might help others from not doing the same...

8086 Assembly - Better data storage/manipulation?

I am working on a project for class, and it works as required by the rubric, though I am wondering if there is a slightly better way to implement a few things. I was docked a few points for an unnecessary 'mov' in another project. Here is problem 1.
"If—else (34 points): Write a program that asks the user to enter a single digit. If that digit is less 5, you will state so and you will add 5 to it and store it in a variable; if the digit is greater than 5, you will state so and then subtract 5 from it and store it in a variable; if the digit is 5, you will add 3 to it and state so and store it in a variable."
org 100h
mov dx, offset start ;move start of string address 'start' into dx
mov ah, 09h
int 21h ;print the string stored at DS:DX
mov ah, 01h ;function for getting keyboard input
int 21h
sub al, 30h ;subtract 30h to store our number as hexadecimal
mov bl, al ;copying data to BL as the following commands manipulate the data
;at AL.
cmp bl, 5 ;BL = AL
jz ifZero ;jump to ifZero if BL = 5
jl ifLess ;jump to isLess if BL < 5
jg ifGreater ;jump to ifGreater if BL > 5
ifZero: ;direct console output function
mov ah, 06h
mov dl, 0Ah
int 21h
mov dl, 0Dh
int 21h ;print newline and character return
add bl, 03h ;add 3 to BL, BL = 8
mov temp, bl
mov dx, offset eq ;move start of string address 'eq' into dx
mov ah, 09h
int 21h ;print string
jmp exit ;unconditional jump to end program
ifLess: ;direct console output function
mov ah, 06h
mov dl, 0aH
int 21h
mov dl, 0Dh
int 21h ;print newline and character return
add bl, 05h ;add 5 to BL
mov temp, bl
mov dx, offset less ;move start of string address 'less' into dx
mov ah, 09h
int 21h ;print string
jmp exit ;unconditional jump to end program
ifGreater:
mov ah, 06h
mov dl, 0ah
int 21h
mov dl, 0dh
int 21h ;print newline and character return
sub bl, 05h ;subtract 5 from BL
mov temp, bl
mov dx, offset great ;move starting address of string 'great' into dx
mov ah, 09h
int 21h ;print string
jmp exit ;unconditional jump to end program
exit:
ret
temp db ?
start db "Please enter a number: $"
less db "Less than 5... adding 5 $"
great db "Greater than 5... subtracting 5 $"
eq db "Equal to 5... adding 3 $"
in this case, would 'mov bl, al' be not needed? Running through the disassembler shows that the data in AL changes after most of these commands. Is this supposed to happen? Is there a better way to do so?
Problem 3:
Counter-controlled loop. The program will ask the user to enter a character and then it will display
that character with a label five times
For example:
Enter a character: A
You entered: A
org 100h
mov cx, 05h ;counter controlled loop, start as 5
LabelLoop:
mov dx, offset prompt ;move string offset to dx
mov ah, 09h ;function for printing string from dx
int 21h
mov ah, 01h ;function to read character from keyboard
int 21h
mov bl, al ;preserving character read by copying to BL
;as register data for AL will be changing
;due to various functions
mov ah, 06h ;function for direct console output
mov dl, 0ah
int 21h
mov dl, 0dh ;these just make the text appear on a new
;line
int 21h
mov dx, offset output ;move the memory offset of output to dx
mov ah, 09h ;printing another string
int 21h
mov ah, 02h ;function to write a character to console
;gets the value from DL
mov dl, bl ;so we copy BL to DL and print it
int 21h
jmp newLine ;we unconditionally jump to the newLine
;label and print a new line for the program
;to run again
newLine:
mov ah, 06h
mov dl, 0ah
int 21h
mov dl, 0dh
int 21h
loop LabelLoop ;we jump to LabelLoop and CX = CX - 1
mov dx, offset goodbye
mov ah, 09h
int 21h
ret
prompt db 'Enter a character: $'
output db 'You entered: $'
goodbye db 'Good bye!$'
So, my question for these problems, is there a better way to do this? keyboard input is stored in AL, but the register value changes for each time i do a mov function into AH, whether string printing or character printing. in order to avoid a variable (since it wasn't a part of the requirements) or allocating it to memory (which we haven't learned), I moved the data to a different register. Is this an unnecessary 'mov' for either program?
edit: i realize that AL = DL after
mov ah, 06h
mov dl, 0ah
int 21h ;AL = DL after execution
cmp bl, 5 ;BL = AL
jz ifZero ;jump to ifZero if BL = 5
jl ifLess ;jump to isLess if BL < 5
jg ifGreater ;jump to ifGreater if BL > 5
ifZero:
The first improvement you should do is making use of the possibility to fall-through in the code. Don't use jz ifZero but rather fall through as equality is the only state that remains after jl and jg. Also ifEqual would be a more correct name for this state.
cmp bl, 5 ;BL = AL
jl ifLess ;jump to isLess if BL < 5
jg ifGreater ;jump to ifGreater if BL > 5
ifEqual:
A second optimization will be to get rid of all of those direct console outputs for CR and LF. You should incorporate these in the messages that you will print. Doing so will also remove the need to copy AL to BL using mov bl, al (you specifically asked this):
less db 13,10,"Less than 5... adding 5 $"
great db 13,10,"Greater than 5... subtracting 5 $"
eq db 13,10,"Equal to 5... adding 3 $"
Here's another opportunity to fall through:
jmp exit ;unconditional jump to end program
exit:
Your second program can benefit from these advices too.

How do i print a unsigned array of numbers in assembly?

this is my code:
.model small
.stack 100h
.data
A db 2,-5,3,4,-8
N equ 5
.code
mov ax, #data
mov ds, ax
mov si,offset A
mov cl,1
start:
cmp cl,N
je sof
mov al,[si]
cmp al,[si+1]
jg change
jmp next
change:
mov ah,al
mov al,[si+1]
mov [si],ah
mov [si+1],al
jmp next
next:
inc si
inc cl
jmp start
sof:
mov ah,9
mov cx, offset A
int 21h
.exit
end
i get a weird out put a lot of weird characters :/
You need to add 30h to a number in range 0 through 9 in order to display an ascii character. To display a character, use
;al = digit to display
mov dl,al
add dl,030h
mov ah,02 ;display character (in dl)
int 21h

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