Objectives. Saving Interrupt Vectors. Writing a Custom Interrupt Handler. Examples of use of System functions for Input-Output and Interrupts

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1 ICT106 Fundamentals of Computer Systems Week 11 Practical Examples of use of System functions for Input-Output and Interrupts Objectives To illustrate how to write interrupt service routine (ISR) for Intel microprocessors through examples: Ctrl-Break Handler Example (Ch 16.4) Use of User Timer Interrupt (INT 1Ch) Writing a Custom Interrupt Handler Interrupt vectors stored in an interrupt vector table are replaceable; E.g. one can manipulate an input keystroke before displaying on the monitor, by replacing the address of INT 9 It is IMPORTANT to save the original interrupt vector prior to any vector replacement Saving Interrupt Vectors.data int9save LABEL WORD DWORD? ; store old INT 9 address here.code mov ah, 35h ; get interrupt vector mov al, 9h ; for INT 9 ; call MS-DOS mov int9save,bx ; store the offset mov int9save+2, ES ; store the segment 1

2 Restoring Interrupt Vectors mov ax, int23save+2 ; get old segment and ; store in ds mov dx, int23save ; get old offset mov ah, 25h ; set interrupt vector mov al, 23h ; restore old interrupt address Example: Ctrl-Break Handler The program consists of two parts: The main function is to echo every input key till a ESC key is pressed; The handler displays a string BREAK when Ctrl-Break (Ctrl- C) is pressed. INTERRUPT DISPLAY BREAK END NO START DISPLAY GREETING INSTALL HANDLER WAIT FOR A KEY ESC KEY? YES END Ctrl-Break Handler Code TITLE Control-Break Handler ; This program installs its own Ctrl-Break hander and ; prevents the user from using Ctrl-Break (or Ctrl-C) ; to halt the program. The program inputs and echoes ; keystrokes until the Esc key is pressed..model SMALL ; each of data and code segment <= 64K.STACK 100h ; reserve 256 bytes of stack space.data breakmsg BYTE "BREAK", 0dh, 0ah, '$' msg BYTE "Ctrl-Break demonstration." BYTE 0dh, 0ah BYTE "This program disables Ctrl-Break (Ctrl-C). Press any" BYTE 0dh, 0ah BYTE "keys to continue, or press ESC to end the program." BYTE 0dh, 0ah, '$' int23save LABEL WORD DWORD?.code main PROC mov ; store old INT 23 address here mov dx,offset msg ; display greeting message call WriteString 2

3 mov ax,3523h ; get INT 23 vector ; AH=35, AL=23 mov word ptr int23save,bx mov word ptr int23save+2,es ; save INT 23 vector install_handler: push ds ; save DS mov ax,@code ; initialise DS to code segment mov ds,ax mov ah,25h ; set interrupt vector mov al,23h ; for interrupt 23h L1: mov ah, 1 ; wait for a key, echo it cmp al,1bh ; ESC pressed? jnz L1 mov ax, word ptr int23save+2 ; get old segment and ; store in ds mov dx, word ptr int23save ; get old offset mov ax, 2523h ; restore old interrupt address mov dx,offset break_handler pop ds ; restore DS mov ax, 4c00h main ENDP ; terminate program WriteString Procedure ; The following procedure displays message string ; to the console using INT 21 Function 9. ; Receives: DS:DX = address of string WriteString PROC push ax ; save ax push dx ; save dx mov ah, 9 ; call INT 21 Function 9 ; call MS-DOS pop dx ; restore dx pop ax ; restore ax ret ; return from a procedure WriteString ENDP Interrupt Service Routine ; The following procedure executes when Ctrl-Break is ; pressed. All registers must be preserved. break_handler PROC push ax ; save ax push dx ; save dx mov dx,offset breakmsg ; get offset of breakmsg call WriteString ; variable then call WriteString ; procedure pop dx ; restore dx pop ax ; restore ax iret ; return from interrupt break_handler ENDP END main 3

4 Creating a Timer A timer can be created by using polling or interrupt. Polling approach is based on a simple assumption that it takes roughly the same time to execute the same instruction, e.g. Provide that it takes 5 nsec to execute the instruction inc ax, then one can generate 5msec by looping the same instruction 1000 times Disadvantages of Polling Approach However, such approach has a serious accuracy problem because its accuracy is easily influent by other factors, e.g. interrupts Other issues include Waste of CPU resource Lack of multitasking support User Timer Interrupt The BIOS provides a regular 18.2Hz heartbeat for OS and applications by executing the User Timer Interrupt (INT 1Ch) every 55 milliseconds (roughly) Thus, 91 heartbeats are required in order to generate 5 seconds, i.e Hz * 5 sec = 91 cycles We increment a timer by one on every interrupt. If the timer reaches 91, it indicates 5 seconds have already passed. TITLE Timer Handler ; This program displays one "Tick" strings every 5 ; seconds, up to 10 strings..model SMALL ; each of data and code segment <= 64K.STACK 100h ; reserve 256 bytes of stack space.data timemsg BYTE "Tick", 0dh, 0ah, '$' waitmsg BYTE "Waiting", 0dh, 0ah, '$' msg BYTE "This program displays 'Tick'" BYTE 0dh, 0ah BYTE "every 5 seconds for 10 times" BYTE 0dh, 0ah, '$' timer dw 0 ; time counter tickcount dw 0 ; run for 50 seconds intsave LABEL WORD DWORD? ; store old INT 1C address here 4

5 .code main PROC mov mov dx,offset msg ; display greeting message call WriteString mov ax,351ch ; get INT 1C vector mov word ptr intsave,bx mov word ptr intsave+2,es ; save INT 1C vector mov cx,0 ; use cx as 5-second flag L1: L2: cmp cx,1 ; Has lasted for 5 seconds? jnz L2 mov cx,0 ; reset flag mov dx,offset timemsg ; displays "tick" string call WriteString inc tickcount ; increment tick counter cmp tickcount,10 ; Has ticked for 10 times? jnz L1 install_handler: push ds ; save DS mov ax,@code ; initialise DS to code segment mov ds,ax mov ah,25h ; set interrupt vector mov al,1ch ; for interrupt 1C mov dx,offset timer_handler pop ds ; restore DS mov ax, word ptr intsave+2 mov dx, word ptr intsave mov ax, 251Ch mov ax, 4c00h main ENDP ; get old segment and ; store in ds ; get old offset ; restore old interrupt address ; The following procedure is executed 18.2 times per sec. ; It sets the flag when 5 seconds is passed. timer_handler PROC pushf ; save flags cmp timer,91 ; check if 5 seconds has passed jnz skip ; no: skip mov timer,0 ; yes: reset timer mov cx,1 ; set flag skip: inc timer ; increment timer popf ; restore flags iret ; interrupt return timer_handler ENDP ; The following procedure displays message string ; to the console using INT 21 Function 9. ; Receives: DS:DX = address of string WriteString PROC push ax push dx mov ah, 9 ; call INT 21 Function 9 pop dx pop ax ret WriteString ENDP END main 5

6 Can we create a 4-second timer? Solution!!! To measure 4 second, one needs to cycle: 18.2 Hz * 4 sec = 72.8 (Not a whole number!) If we choose 72, we will lost 0.04 sec (= 72 / 18.2) in every 4 second, or 36 second slower each day; OR We will be 0.01 second faster in every 4 second if we choose 73, or 9 second faster every day We re-adjust the timer on every 16 th second Thus, 1 st 4 sec: 73 2 nd 4 sec: 73 3 rd 4 sec: 73 4 th 4 sec: 73 5 th 4 sec: 72 Is it correct? We can check the number of counted cycles after 20 seconds 72.8 * 5 = 364 AND = 364!! Problem solved! 6