Using the Keyboard Buffered Input of DOS
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> [....] My question, [how to] use the DOS function for buffered
> keyboard input (0Ah), then write a program that allows the user
> to type a string, then echo's the string back to the screen
> using function (09h).
>
> Thank you
>
> Bill
>
>
DOS's int 21h service 0Ah is one of the few that are truly services. It allows the user to input a string a chars while being able to use backspace to correct errors and other niceties as such.
First, service 0Ah needs a little bit of information before it can start working. It needs a small buffer to store the string once the user has pressed the enter key. This service also needs another byte of information. The working length of this buffer. i.e. The allowed length of the string.
Let us start.
; This assembler listing is assembled using NBASM ver 00.22.8x
; To get NBASM, a free assembler, go to:
; https://www.fysnet.net/newbasic.htm
.model tiny
.code
; first let us print a string to the screen letting the
; user no that we want him/her to enter a string
; I will discuss the print service later
mov dx,offset FirstS
mov ah,09h
int 21h
; Even though this service wants the location (offset) of
; our buffer in DX, we need a BASE register to access
; individual bytes in this buffer. So lets put the offset in
; BX. (read below on how large of a buffer we need)
mov bx,offset OurBuff
; However, for smaller code, let us put the offset
; in to DX also.
; (mov dx,bx is one byte smaller than mov dx,offset OurBuff)
mov dx,bx
; we need to tell service 0Ah that we want to allow up to
; 32 chars in our string. This is done by putting 33 in the
; first byte of the buffer. Why 33? 32 chars plus the
; <Enter> key.
mov byte [bx],33
; now we can call the service (interrupt) so DOS will do the
; user input routine.
mov ah,0Ah
int 21h
; Once the user enters the string and presses <ENTER>, DOS
; returns to this point. Now we can do what we want with this
; string.
; Since you mentioned that you wanted to print the string to the
; screen using service 09h, we will have to do some manipulation
; on are newly acquired string.
; First let us print another string denoting to the user, the
; string to be printed.
mov dx,offset SecndS
mov ah,09h
int 21h
; Service 09h also wants the offset to the buffer (string) in DX.
; But as stated before, we need a base register to access individual
; bytes of the string.
mov bx,offset OurBuff
; Also, service 09h needs a '$' (dollar sign) at the end of the string
; denoting 'end of string' for the service to stop printing here.
; Where do we put this '$'? Well we know that service 0Ah above
; returns the actual length of the string in byte position two of the
; working buffer.
mov al,[bx+1]
; AL now contains the actual length of the string.
; Remember that offsets are zero based and [bx+1] is the second
; position in the buffer.
; Now we want to put a '$' at the end of the string so that service
; 09h will stop printing at the end of our string.
; Remember that the string doesn't actually start at the beginning
; of our buffer. It starts at offset 02h or at the third position. So
; we need to add 2 to the length to get to the end of the string.
add al,02
; Now that we know the offset from the beginning of the buffer
; of where to put the '$', we need to put it there.
; Remember that with 8086 assembler we can not put an 8-bit register
; into a 16 bit register, so we must clear out AH first
; Then we need another Base (or index) register to hold this value.
xor ah,ah
mov si,ax
; Just a note: In assembler, there are many ways to do simple items
; like clearing AH.
; mov ah,00h (takes 3 bytes)
; xor ah,ah (takes 2 bytes and is faster)
; cbw (takes one byte, However, This clears AH only if bit 7 of )
; ( AL is clear (i.e. if AL < 128 ) )
; Now we can put the '$' at this position.
mov byte [bx+si],'$'
; Now we can point DX to the start of the string and call the service
; (interrupt). Remember that DX needs to point to offset 02h.
mov dx,offset OurBuff
add dx,02
mov ah,09h
int 21h
; Taadaa. Did it work? Now we must tell DOS to quit our program or
; the CPU will try to interpret the data in OurBuff as code and could
; crash the machine.
mov ah,4Ch
int 21h
; We need to have the first byte of our buffer hold the working
; size of the buffer. The second byte is returned by the service
; as the amount of chars entered. The we need 32 bytes of space
; because we wanted up to 32 chars. And finally the last byte is
; room for the CR (0Dh) that DOS appends to the string.
OurBuff db 00h,00h ; the working and actual lengths
dup 32 ; string (adjust length to same as code above)
db 00h ; the CR (ascii 13d)
; Now the two strings to print for a nice user interface.
; Note the 13,10 at the beginning. Service 09h takes the 13 as
; a carriage return (CR) and the 10 as a Line feed (LF).
FirstS db 13,10,'Please enter a string: $'
SecndS db 13,10,' String entered: $'
.end
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