8086 programming using TASM: pc to pc communication

.model small
.stack 100
.data
.code

mov ah,00h
mov al,0e3h
mov dx,00h
int 14h

back: nop

l1: mov ah,03h
    mov dx,00h
    int 14h

    and ah,01h
    cmp ah,01h
    jne l1

    mov ah,02h
    mov dx,00h
    int 21h

mov dl,al
mov ah,02h
int 21h

jmb back
mov ah,4ch
int 21h

end

This a pc to pc commnicat开发者_如何转开发ion receiver program. I would like to know why it uses the mov dx,00h command and what mov al,0e3h means?

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Take a look here. AX will contain the transmission parameters (baud rate etc) and DX chooses the port number. E3 = 9600 rate, no parity, two stop bits, 8 bits char size.

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According to the docs I could find on int 14h,

dx determines the port numbber. So if you're using port one, you put 00h into dx. al is used for parameters of the serial communication. Check the docs for more details on the parameters.

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dx is used to select the com ports. 00=com1, 01=com2. al is used to select character size(0 and 1 bit), stop bit(2nd bit), parity bits (3rd and 4th bit) and baud rate(5,6,7 bit no.)

al=11100011=e3=8bits: no parity, one stop bit, 9600 baud rate