🚀 go-pugleaf

In article <90351.150210TOMIII@MTUS5.BITNET> TOMIII@MTUS5.BITNET (Thomas Dwyer III) writes:
>Hi there.  Would some kind soul please tell me why my machine hangs when
>I run this TSR?  What am I doing wrong?
>
>------ Code Begins ------
>
>INT_NUM equ     9h
>
>code    segment
>        org     0100h
>        assume  cs:code,ds:code,es:code
	[Init Code Deleted]
>new_int proc    far
>        jmp     short go
>
>old_int dd      ?
>
>go:
>        jmp     [old_int]                       ; Jump to old int routine

I'm not sure, but I think that this is DS relative. In other words, when
your ISR is called, DS may be pointing elsewhere.

>new_int endp
>
>
>code    ends
>        end     start
>
>----- Code Ends ------

A much better (IMHO) way to do this, is to borrow the interrupt from the
system, call the original ISR, do your own dirty work, and return.
The way I do this in a couple programs is to save the old ISR address
in the same way you have, but call it differently.

new_int proc    far
        jmp     short go

old_int dd      ?

go:	pushf		; Pushing the flags before a FAR call simulates
			; an INT instruction. The ISR being called will
			; then IRET to here.

	call	DWORD PTR cs:[old_isr] 
	.
	.		; Do something usefull here.
	.
	IRET
new_int	endp

This gives you the flexibility of handling the original ISR, as well as
your own code.

Hope this helps....
Steve



	
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