It seems that I'm not getting what I want. Perhaps I shall fast forward.

Now I am able to share data at 56K with the PC if I make a "monitor" program. (one that interfaces with the terminal).

Here is the code I have so far when trying to make my dial-up server (on the 8051). Remember, this "server" will make the PC think it is connected to the internet.


;--------------------------------------------
; Timer 0 interrupt (Serial mini-buffer to XRAM)
;--------------------------------------------
org 0Bh
clr TR0 ;suspend timer when in INT
push DPL
push DPH
mov DPTR,#8000h
mov R1,#30h
moremini:
mov A,@R1
movx @DPTR,A
inc DPTR
inc R1
mov A,R1
cjne A,R0,moremini
mov R0,#30h
pop DPH
pop DPL
setb TR0
reti


;--------------------------------------------
; serial interrupt
;--------------------------------------------
org 23h
jbc RI,smrecv
jbc TI,smxmit
reti

smxmit:
setb C
reti

smrecv:
mov @R0,SBUF
inc R0
cjne R0,#80h,recvadd
dec R0
recvadd:
reti

Obviously I haven't finished the code. R0 and R1 are set to #30h so that I make use of the internal ram locations #30h-#7Fh

My intent is to use the fewest number of clock cycles possible to store a byte in IRAM.
"PUSH" won't work because when the serial routine ends, two POPS are internally executed (for the RETI instruction).

A timer is active which will activate after a certain delay. this "certain" delay will be long enough so that a byte is in memory before the timer interrupt starts.

Once the interrupt starts, data is transferred from IRAM to XRAM.

I was just wondering, for the interrupts, since I want the serial interrupt to be of high priority, If within the serial interrupt, the "timer 0 interrupt" is called because of timer overflow, does it wait in line to be processed, or does it get ignored?