Beginner: need help understand this code sample

[dantytektips]

Hi,
I am new to programming. I recently started reading few assembly tutorials on the net and I must say I am enjoying it. I found this nice tutorial below but i don't quite understand this program. Could some one please add more comments to the code lines for me to understand what is going on starting from start:. Here is the code sample from

http://www.skynet.ie/~darkstar/assembler/tut5.html

------------------------------
jmp start
;****************************
;* Program to read in two *
;* numbers and add them *
;* and print out the result *
;****************************
number db 7 dup 0 ; string which will store input and output
n1 dw 0 ; two input variables
n2 dw 0
res dw 0 ; one output variable
cr dw 13,10,"$" ; carriage return, line feed
start:
mov dx,offset number
mov bx,dx
mov b[bx],5 ; maximum 5 characters to read
mov ah,0ah
int 21h ; read in a string from keyboard
mov bx,offset number +1
mov cx,00
mov cl,[bx] ; cl now contains number of digits
mov ax,00 ; ax will contain the number input
usedigit:
inc bx ; get next digit
shl ax,1 ; multiply by 10 using 2 shift ops and an add...
mov dx,ax ; ... x*8 + x*2 = x*10 is the principle.
shl ax,2
add ax,dx ; ax is now multiplied by 10
mov dx,00
mov dl,[bx] ; dl has new character
sub dx,48 ; subtract 48 = ascii('0') to get number value
add ax,dx ; add to ax
loop usedigit ; loop statement= jmp if cx > 0
cmp n1,00 ; see if this is first or second number read
jnz second
mov n1,ax ; assign it to the first variable
jmp start ; read in another number
second:
mov n2,ax ; or assign to second variable and continue
print_cr:
mov ah,09
mov dx,offset cr ; print out a carriage return character
int 21h
addnos:
mov ax,n1 ; move numbers to registers ...
mov bx,n2
add ax,bx ; ...and add
mov res,ax ; store the result
mov cx,00
setup_string:
mov bx,offset number+7 ; put a $ at end of buffer.
mov b[bx],'$' ; we will fill buffer from back forwards
dec bx
mov ax,res
convert_decimal:
mov dx,10
div dl ; divide by 10
add ah,48 ; convert remainder to character
mov [bx],ah ; and move to buffer for output
dec bx
mov ah,00 ; quotient becomes new value
cmp ax,00 ; if we haven't got all digits divide again
jnz convert_decimal
printout:
mov dx,bx
inc dx ; we decremented once too many, go forward one.
mov ah,09
int 21h ; output the string
close:
mov ah,4ch
mov al,00
int 21h ; end program
------------------------------------

Thank you all in advance.