more on interrupts

More on InterruptsMore on Interrupts

Interrupt service routinesInterrupt service routines

o DOS facilities to install ISRs

o Restrictions on ISRso Currently running program should have no

idea that it was interrupted. o ISRs should be as short as possible

because lower priority interrupts are blocked from executing until the higher priority ISR completes

Function Action

INT 21h Function 25h Set Interrupt vectorINT 21h Function 35h Get Interrupt vectorINT 21h Function 31h Terminate and stay resident

Installing ISRsInstalling ISRs

Let N be the interrupt to serviceo Read current function pointer in vector table

o Use DOS function 35h o Set AL = N o Call DOS Function AH = 35h, INT 21h o Returns: ES:BX = Address stored at vector N

o Set new function pointer in vector tableo Use DOS function 25h o Set DS:DX = New Routine o Set AL = N o DOS Function AH = 25h, INT 21h

Installing ISRInstalling ISRo Interrupts can be

installed, chained, or called

o Install New interruptreplace old interrupt

o Chain into interruptService myCode first

MyIntVector Save Registers Service Hardware Reset PIC Restore Registers IRET

MyIntVector Save Registers MyCode Restore Registers JMP CS:Old_Vector

MyIntVector PUSHF CALL CS:Old_Vector Save Registers MyCode Restore Registers IRET

oCall Original InterruptService MyCode last

Timer interrupt exampleTimer interrupt example

o In this example we will patch the ISR for the Timer Interrupt

o Our ISR will count the number of timer interrupts received

o Our main program will use this count to display elapsed time in minutes and seconds

About the PC timerAbout the PC timero We’ll examine it in detail when we look at the

8253o You can view it as a clock that ticks and sends

a signal to your processor approximately every 1/18.2 seconds

o It is also called the “eighteenth of a second” o We’ll find out where this 18.2 comes from latero You can approximately count minutes and

seconds as followso For every 18 ticks of the timer we add 1

secondo For every 60 seconds we add one minute

and reset the counters for seconds

ISR overviewISR overview

o Install the ISR for vector 08ho Call original timer interrupto Increment a counter on every interrupt from

the timer (count)o Increment a second counter (scount) when

count=18 and reset counto Increment a minute counter (mcount) when

scount=60 and reset the second countero The program will print the scount, mcount

and count values in red, green and blue colors

o It will look as a (somewhat low-resolution) timer printed in your screen

Timer interrupt – Timer interrupt – main proc skeletonmain proc skeleton

;====== Variables ===================

; Old Vector (far pointer to old interrupt function)

oldv RESW 2count DW 0 ;Interrupt

counter (1/18 sec)scount DW 0 ;Second counter

mcount DW 0 ;Minute counterpbuf DB 8 ;Minute counter;====== Main procedure =====..start…;----Install Interrupt Routine-----call Install ;Main program (print count values) .showcMov ax, [mcount] ;Minute Count…

call pxymov ax, [scount] ;Second Count…call pxymov ax,[count] ;Interrupt Count

(1/18 sec)… call pxymov ah,1int 16h ;Check for key

press jz .showc ;Quit on any key

;---- Uninstall Interrupt Routine-----call UnInst ;Restore original INT8…call mpxit

Timer interrupt – complete main Timer interrupt – complete main procproc

..startmov ax, cs ;Initialize DS=CS mov ds, axmov ax, 0B800h;ES=VideoTextSegmentmov es, axcall install ;Insert my ISR

showc:

mov di,12 ;Column 6 (DI=12/2)mov ah,00001010b;Intense Green call pxy

mov ax,[count] ;Int Count (1/18th sec)mov bx,pbufcall binasc mov bx, pbuf

Contd…

Timer interrupt – complete main Timer interrupt – complete main procproc

.. mov ax, [mcount];Minute Countmov bx, pbufcall binascmov bx, pbuf mov di,0 ;Column 0mov ah,00001100b;Intense Redcall pxymov ax,[scount];Second Countmov bx,pbufcall binascmov bx, pbuf

mov ah,00000011b;Cyanmov di,24 ;Column 12 (DI=24/2)call pxymov ah,1int 16h ;Key Pressed ?jz showc

Call UnInst ;Restore original INT8mov ax,4c00h;Normal DOS Exitint 21h

Timer interrupt – PXY Timer interrupt – PXY and Install interruptand Install interrupt

;pxy (bx = *str, ah = color, di = column)

pxymov al, [bx]cmp al, ‘$'je .pxydonemov es:[di+2000], axinc bxadd di,2jmp pxy

.pxydoneret

mov al, 8 ;INT = 8

mov ah, 35h ;Read Vector Subfunction

int 21h ;DOS Service

mov word [oldv+0], bx

mov word [oldv+2], es

mov al, 8 ;INT = 8

Contd…

Timer interrupt – PXY Timer interrupt – PXY and Install interruptand Install interrupt

;====== Install Interrupt =====

install ;Install new INT 8 vector

push es

push dx

push ax

push bx

mov ah, 25h ;Set Vector Subfunction

mov dx, myint

;DS:DX point to functionint 21h ;DOS Service

pop bx pop axpop dxpop esret

Timer interrupt – uninstall interruptTimer interrupt – uninstall interrupt

;====== Uninstall Interrupt ===========UnInst ; Uninstall Routine (Reinstall old vector)

push dspush dxpush axmov dx, word [oldv+0]mov ds, word [oldv+2]

mov al, 8 ; INT = 8mov ah, 25h ; Subfunction = Set Vectorint 21h ; DOS Service

pop axpop dxpop dsret

Timer interrupt – ISR codeTimer interrupt – ISR code

;====== ISR Code =========

myintpush ds ;Save all registerspush axmov ax, cs ;Load default segmentmov ds, axpushf ;Call Orig Function w/flagscall far [oldv] ;Far Call to existing routine

inc word [count];Increment

Interrupt countcmp word [count],18jne .myintdone

inc word [scount] ;Next second mov word [count], 0cmp word [scount], 60jne .myintdoneinc word [mcount] ; Next minutemov word [scount], 0

.myintdonepop ax ;Restore all Registerspop dsiret ;Return from Interrupt

Replacing An Replacing An Interrupt HandlerInterrupt Handler

;install new interrupt vector

%macro setInt 3 ;Num, OffsetInt, SegmentInt

push ax

push dx

push ds

;store old interrupt vector

%macro getInt 3 ;Num, OffsetInt, SegmentInt

push bx

push es

mov al, %1

Contd…


mov dx, %2mov ax, %3mov ds, axmov al, %1mov ah, 25h ;set interrupt vectorint 21h

pop dspop dxpop ax

%endmacro

mov ah, 35h ;get interrupt vector

int 21h

mov %2, bx

mov %3, es

pop es

pop bx

%endmacro

Replacing An Interrupt Replacing An Interrupt HandlerHandler

CREQU 0dhLF EQU 0ah

SEGMENT stkseg STACK resb 8*64stacktop:

SEGMENT code

Warning DB “Overflow - Result Set to ZERO!!!!”,CR,LF,0

msgOK DB “Normal termination”, CR, LF, 0

old04hOffset RESWold04hSegment RESW

New04h ;our new ISR for int 04

;occurs on overflowsti ;re-enable interrupts

mov ax, Warningpush axcall putStr ;display

messagexor ax, ax ;set result to

zerocwd ;AX to DX:AX

iret


mov ax, msgOKpush axcall putStr

Error:;restore original int handlersetInt 04h, [old04hOffset], [old04hSegment]

mov ax, 4c00hint 21h

..startmov ax, csmov ds, ax

;store old vectorgetInt 04h, [old04hOffset], [old04hSegment]

;replace with address of new int handlersetInt 04h, New04h, cs

mov al, 100add al, alinto ;calls int 04 if an overflow occurredtest ax, 0FFhjz Error

Contd…


NOTESINTO is a conditional instruction that acts only when the overflow flag is setWith INTO after a numerical calculation the control can be automatically routed to a handler routine if the calculation results in a numerical overflow.By default Interrupt 04h consists of an IRET, so it returns without doing anything.

ReentrancyReentrancyo What happens if

o An ISR for some devices is executing and it has enabled interrupts

o Another interrupt from the same device comes along…The program may not behave correctly

o Assume that the ISRs modify some register and store its value in a memory location

o The previous ISR might have already modified the register but didn’t have the time to store the value in the memory locations

o The new ISR will save this register but it will try to update the memory location, which is in an inconsistent state

ReentrancyReentrancy

AnISR; assume scount=3, MSEC=950push dspush axmov ax, csmov ds, axmov ax, [MSEC] add ax, 55 ; ax=1005cmp ax, 1000jb SetMsec

Contd…


; Assume that another interrupt occurs at this pointinc [scount] ; the second interrupt will set scount=4 sub ax, 1000 ; mov [MSEC], ax; the second interrupt will set MSEC=5…; but the interrupted ISR will reexecute the inc so scount will be; set to 5 although the timer has not ticked 55 times!!!!pop axpop ds


o The code between the mov ax, [MSEC] and the mov [MSEC], ax must be executed atomically, without any interruptions

o This is called a critical regiono You can protect a critical region from being

interrupted by using:pushf ;preserve the current I flag statecli ; turn off interrupts… ; critical regionpopf ; restore the I flag state

Reentrancy in practiceReentrancy in practice

o First thing to remember: don’t call DOS from your ISRs, DOS is not reentranto DOS subroutines assume to be

entered by a single point at any timeo If you write an ISR and attempt to

call DOS you will most likely hang the machine forever

o Second thing to remember: BIOS is not reentrant

Contd…

Reentrancy in practiceReentrancy in practice

o There are ways to check if you’re executing inside DOS by testing a flag (function code 34h)o If the flag is 0 it is safe to call DOSo If the flag is 1 it might not be safe to call

DOSo You can also check if DOS is “idling”

(function code 28h), in that case it is safe to call DOS

DOS Memory UsageDOS Memory Usage

00000h

Interrupt vector003FFh

Various DOS/BIOS vars

Free memory area for use

by programs

0BFFFh

High memory area

Video, ROM, adapter memory

0FFFFh

Free memory pointer

DOS Memory UsageDOS Memory Usage

00000h



Memory in use by your program

0BFFFh

High memory area


0FFFFh

Free memory pointerFree memory area

Terminate and Stay Terminate and Stay Resident (TSR)Resident (TSR)

00000h



Marked by the program as

resident and protected by DOS

0BFFFh

High memory area


0FFFFh

Free memory pointer

Free memory area


o Your program can have a resident portion and a transient portion

o The main program, normal data, support routines etc, are usually transient

o You can define ISRs and maintain them in memory after the program terminates using the resident portion

o Use DOS function 31h with the size of the resident portion passed in dx


o If you want to use resident ISR’s you should define them in the lower parts of your memory address space

o You have to set your DS properlyo The resident code has no idea about the

values of the segment registerso You have to set the data segment to the

value of your code segmentpush dspush cspop ds ; this moves cs to ds…pop ds

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