Reversing an input string in Assembly - string

Question

I have this assembly code that reverses a string that I input. It only accepts maximum 20 characters. My problem is that when I hit enter to see the output there is an extra character at the end of the reversed string.
Please help me understand why that does occur and how I can remove that in the output.
We're required to only use function 09H int 21h to display the string and function 0Ah int 21h to input the string. We're using TASM.
Your help would be very much appreciated. Thank you.
Here is my code:
.model small
.stack 100h
.data
MSG DB "Input String(max 20 chars): ", 10, 13, "$"
Sentence1 DB 21,?,21 dup("$")
str2 dw 21 dup("$")
.code
start:
mov ax, #data
mov ds, ax
;Getting the string input
mov ah,09h
lea dx, MSG
int 21h
lea si,Sentence1
mov ah,0ah
mov dx,si
int 21h
;Reverse String
mov cl,Sentence1
add cl,1
add si,2
loop1:
inc si
cmp byte ptr[si],"$"
jne loop1
dec si
lea di,str2
loop2:
mov al,byte ptr[si]
mov byte ptr[di],al
dec si
inc di
loop loop2
;Printing the reverse string
mov ah,09h
lea dx,str2
int 21h
mov ah, 4ch
int 21h
end start

Answer 1

str2 dw 21 dup("$")
Normally this would be using the db directive.
mov cl,Sentence1
add cl,1
The reversal loop uses CX as its loop counter, but you don't set it correctly!
The 2nd byte of the "Sentence1" input structure, contains the value that you want in the CX register. You don't need to search for any terminating character. Moreover if you did, you'd rather have to look for ASCII code 13 (carriage return) instead of '$'.
mov cl, [si + 1] ;Number of characters in the string
mov ch, 0 ;Make it a word because LOOP depends on CX (not just CL)
Setting up SI then becomes:
add si, 2 ;To the start of the string
add si, cx ;To the position after the string
dec si ;To the last character of the string
but shorter:
add si, cx
inc si
If ever the user didn't input any text, you will want to by-pass the reversal entirely! That's what the jcxz is for in next code:
lea si, Sentence1
mov ah, 0Ah
mov dx, si
int 21h
;Reverse String
mov cl, [si + 1]
mov ch, 0
add si, cx
inc si
lea di, str2
jcxz EmptyString ;By-pass the reversal entirely!
loop2:
mov al, byte ptr[si]
mov byte ptr[di], al
dec si
inc di
loop loop2
EmptyString:
;Printing the reverse string (could be empty)
mov ah, 09h
lea dx, str2
int 21h