DAVID%UCONNVM.BITNET@cunyvm.cuny.edu ( carl david) (09/20/90)
I am trying to write an interupt handler for interupt 9, the keyboard interupt, which accepts two keystrokes, swallows the first, and processes the second. This means that the code is re-entrant (I think), in that I stay in the interupt code after the first keystroke, and then at the second one, re-enter the interupt code again. The problem is that when I try to chain from my second) interupt back to the original interupt 9 handler, after doing "my thing", some stack somewhere appears to be goofed. What I would like to know is 1) Where is the stack which is pushed, pulled, popped or otherwise manipulated, when the interupt happens, 2) what happens to that stack at each interupt, and 3) how can I reset it properly so that I can get back to the standard interupt handler after processing the second keystroke. Any help would be greatly appreciated. I am using TurboPower 4.0 to handle the interupts, etc., although it is clear that this is not the problem, it is the re-entrancy that is killing me. Thank you in advance for any help. Carl David DAVID at UCONNVM(bitnet)
ts@uwasa.fi (Timo Salmi LASK) (09/20/90)
In article <24551@adm.BRL.MIL> DAVID%UCONNVM.BITNET@cunyvm.cuny.edu ( carl david) writes: > >I am trying to write an interupt handler for interupt 9, the keyboard >interupt, which accepts two keystrokes, swallows the first, and >processes the second. ... much deleted ... But why interrupt 9, and not 16Hex. Have you considered the latter. You might have better success. I've used Int 16Hex in my own Turbo Pascal keyboard routines with good success. (Or maybe I do not understand your problem correctly.) ................................................................... Prof. Timo Salmi (Moderating at anon. ftp site 128.214.12.3) School of Business Studies, University of Vaasa, SF-65101, Finland Internet: ts@chyde.uwasa.fi Funet: gado::salmi Bitnet: salmi@finfun
jc58+@andrew.cmu.edu (Johnny J. Chin) (09/20/90)
First of all why are you using INT 09h? INT 09h is generated everytime a key
is pressed.
I suggest that you use INT 16h and look into the keyboard buffer for the next
keystroke. This way the interrupt routine doesn't have to be reenterant.
If you choose to use INT 09h, then I suggest you use a flag of some sort so
that the interrupt routine knows of a previous keystroke situation has occured.
Simply reserve a byte somewhere in your TSR and refer to it in your routine.
NOTE: The original INT 09h allows for multiple calls to itself (ie. type-ahead
will cause an INT 09h to occur even if the current procedure running is the
interrupt handler for INT 09h).
I strongly suggest that you look into using INT 16h. Hope this piece of info.
helps you in your program.
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