Reversing a string in assembly language 80x86 - string

Question

In a program I am currently doing, I have to reverse a string a user has entered. I have to leave the word that the user entered in where I prompted them to enter it and right below it I want to print out the word in reverse. When I try to run it in DOSBox with the Tasm compiler it gives me an error that says "Illegal memory reference" on line 189 which is the line that contains the variable I plan to put the reversed word in. Can someone help me find out what I am doing wrong? I would greatly appreciate it! Also only in my program there are 4 boxes. The first box I try to print the reverse word below the prompt. The rest of the boxes prints the user entered word instead of the reversed version of it.
title fill in title ;program name
;------------------------------------------------------------------
stacksg segment para stack 'Stack' ;define the stack
db 32 dup(0) ;32 bytes, might want larger
stacksg ends
;------------------------------------------------------------------
datasg segment para 'Data' ;data segment
paralst Label Byte
maxlen db 21
actlen db ?
dbdata db 21 dup('$')
outit db 'Enter String: $' ;14 chars minus $
switch db 21 dup('$')
datasg ends
;------------------------------------------------------------------
codesg segment para 'Code' ;code segment
main proc far ;main procedure
assume ss:stacksg, ds:datasg, cs:codesg ;define segment registers
mov ax, datasg ;initialize data segment register
mov ds, ax
;--------------- --------------------------top left corner
mov ah, 06h
mov al, 00
mov bh, 01000001b ; 4eh
mov ch, 0
mov cl, 0
mov dl, 39
mov dh, 12
int 10h
;-------------------------------------------top right corner
mov ah, 06h
mov al, 0
mov bh, 11110010b
;mov cx, 0c00h
;mov dx, 184fh
mov ch, 0
mov cl, 39
mov dh, 12
mov dl, 79
int 10h
;--------------------------------------------bottom left corner
mov ah, 06h
mov al, 0
mov bh, 11100100b ;yellow
mov ch, 12
mov cl, 0
mov dh, 24
mov dl, 39
int 10h
;------------------------------------------bottom right corner
mov ah, 06h
mov al, 0
mov bh, 01011111b ; magenta 80
mov ch, 12
mov cl, 39
mov dh, 24
mov dl, 79
int 10h
;--------------------------------------------------- 1st quad
mov ah, 02h
mov bh, 0
mov dh, 5
mov dl, 5
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 7
mov dl, 5
int 10h
call REVERSE
;--------------------------------------print output
mov ah, 09h
lea dx, switch
int 21h
;----------------------------------------------------2nd quad
mov ah, 02h
mov bh, 0
mov dh, 5
mov dl, 44
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 7
mov dl, 44
int 10h
;--------------------------------------print output
mov ah, 09h
lea dx, dbdata
int 21h
;------------------------------------------------------3rd quad
mov ah, 02h
mov bh, 0
mov dh, 17
mov dl, 5
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 19
mov dl, 5
int 10h
;--------------------------------------print output
mov ah, 09h
lea dx, dbdata
int 21h
;------------------------------------------------------4th quad
mov ah, 02h
mov bh, 0
mov dh, 17
mov dl, 44
int 10h
mov ah, 09h
lea dx, outit
int 21h
;------------------------------------input
mov ah, 0ah
lea dx, paralst
int 21h
; -----------------------------------move cursor
mov ah, 02h
mov bh, 0
mov dh, 19
mov dl, 44
int 10h
;--------------------------------------print output
mov ah, 09h
lea dx, dbdata
int 21h
mov ax, 4c00h ;end processing
int 21h
main endp ;end of procedure
;----------------------------------------reverse procedure
REVERSE PROC NEAR
mov cx, 0
;-----figure out actlen here
mov actlen, 0
lea bx, dbdata ;may need to use paralst instead
hi: cmp [bx], '$'
jne sup
inc actlen
inc bx
jmp hi
sup:
;------------
mov cx, 0
mov cl, actlen
lea bx, dbdata
add bx, cx
yo: cmp actlen, 0
je hola
mov switch, byte ptr[bx]
dec bx
inc switch
dec actlen
jmp yo
hola:
RET
REVERSE ENDP
codesg ends ;end of code segment
end main ;end of program

Answer 1

You can't do mov memory, memory. Move the source into a register first, then place it at the destination.Also, inc switch doesn't do what you seem to think it's doing. You can't change the address of switch at runtime, so what's actually happening there is that you're incrementing the first element at switch (as if you had written inc [switch]). If you need to modify an address; again, put it in a register.
So for example:
lea si, dbdata
add si,cx
lea di, switch
mov al,[bx]
jcxz hola ; jump if cx == 0
cld ; clear the direction flag; for stosb
yo: mov al,[si]
stosb ; [di++] = al
dec si
loop yo ; if (--cx) jmp yo
hola:
I didn't look at the entire piece of code, so it's unclear to me whether your reverse is supposed to copy the terminator ('$') or not. If it should, you should always execute the loop at least once (and then you don't need the jcxz). If it shouldn't, you should set the source address to dbdata + cx - 1 instead of dbdata + cx.