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Input & Output Instructions. CPU communicates with the peripherals through I/O registers called I/O ports . There are 2 instructions, IN & OUT , that access the ports directly. These instructions are used when fast I/O is essential ……. in a game program. IN & OUT. - PowerPoint PPT Presentation
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21/11/2005CAP2411
Input & Output Instructions
• CPU communicates with the peripherals through I/O registers called I/O ports.
• There are 2 instructions, IN & OUT, that access the ports directly.
• These instructions are used when fast I/O is essential ……. in a game program.
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IN & OUT
• Most application programs do not use IN and OUT instructions.
• Why?
1) port addresses vary among computer models
2) easier to program I/O with service routines
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2 categories of I/O service routine
1. The BIOS routines.
2. The DOS routines.
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BIOS routines
• Are stored in ROM and interact directly with I/O ports.
• Used to carry basic screen operations such as moving the cursor & scrolling the screen.
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DOS routines
• Can carry out more complex tasks.
• Printing a character string…. They use the BIOS routines to perform direct I/O operations.
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The INT Instruction.
• To invoke a DOS or BIOS routine , the INT (interrupt) instruction is used.
• FORMAT
INT interrupt_number
is a number that specifies a routine.
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Example
• INT 16h
invokes a BIOS routine that performs
keyboard input.
• We will use a particular DOS routine
INT 21h
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INT 21h
• Used to invoke a large number of DOS functions.
• Put the function number in AH register and then invoke INT 21h
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FUNCTIONS
Function number Routines
1 single-key input
2 single-character output
9 character string output
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INT 21h functions
• Input values are to be in certain registers & return output values in other registers.
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Function 1
• Single-Key Input
Input : AH = 1Output : AL = ASCII code if character key is
pressed.
= o if non-character is pressed.
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Example
MOV AH,1 ; input key function
INT 21h ; ASCII code in AL
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Example
• If character k is pressed, AL gets its ASCII code; the character is also displayed on the screen
• If Arrow key or F1-F10, AL will contain 0
• The instructions following the INT 21h can examine AL and take appropriate action.
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Function 2
• INT 21h, function 1 …. doesn’t prompt the user for input, he might not know whether the computer is waiting for input or it is occupied by some computation.
• Function 2 can be used to prompt the user
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Function 2• Display a character or execute a control function
Input : AH = 2
DL = ASCII code for the display character
or control character
Output : AL = ASCII code of the display character
or control character
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Example
MOV AH,2 ; display character function
MOV DL, ‘?’ ; character is ‘?’
INT 21h ; display character
• After the character is displayed, the cursor advances to the next position on the line.
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Control functions
• Function 2 may also be used to perform control functions.
• If DL contains the ASCII code of a control character, INT 21h causes the control function to be performed.
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Control functions
ASCII code (hex) Symbol Function
7 BEL beep
8 BS backspace
9 HT tab
A LF line feed (new
line)
D CR carriage return
(start of current
line)
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A First Program
Read a character and display it at the beginning of the next line
1-We start by displaying a question mark:MOV AH,2 ; display character function
MOV DL,'?‘ ; character is ‘?’
INT 21H ; display character
Move 3Fh, the ASCII code for “?” , into DL
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Read a character
MOV AH,1 ; read character function
INT 21H ; character in AL
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Display the character on next line
• First , the character must be saved in another register.
MOV BL , AL ; save it in BL
• This because the INT 21h , function 2 , changes AL
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Display the character on next line
-Move cursor to the beginning of the next line:
• Execute carriage return & line feed.
• Put their ASCII codes in DL & execute INT 21h.
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Move cursor to the beginning of the next line
MOV AH , 2 ; display character function
MOV DL , 0Dh ; carriage return
INT 21h ; execute carriage return
MOV DL , 0Ah ; line feed
INT 21h ; execute line feed
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Display the character
MOV DL , BL ; get character
INT 21h ; and display it
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Program ListingTITLE PGM4_1 : Echo PROGRAM
.MODEL SMALL.STACK 100H
.CODEMAIN PROC
;display promptMOV AH,2
MOV DL'?',INT 21H
;input a characterMOV AH,1INT 21HMOV BL,AL
;go to a new lineMOV AH,2MOV DL,0DHINT 21HMOV DL,0AHINT 21H
;display charactersMOV DL,BLINT 21H
;return to DOSMOV AH,4CHINT 21HMAIN ENDPEND MAIN
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When a program terminates, it should return control to DOS
MOV AH,4CH ; DOS exit function
INT 21H ; exit to DOS
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Displaying a string
INT 21h , Function 9:
Display a string
Input : DX = offset address of string.
The string must end with a ‘$’ character.
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• If the string contains the ASCII code of a control character , the control function is performed.
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Example
• Print HELLO! on the screen.
MSG DB ‘HELLO!$’
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The LEA instruction
• INT 21h, function 9, expects the offset address of the character string to be in DX.
• To get it there, we use
LEA destination , source
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LEA destination , source
• LEA ….. Load Effective Address
• Destination … is a general register.
• Source ………… is a memory location.
• It puts a copy of the source offset address into destination.
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Example
MSG DB ‘HELLO!$’LEA DX , MSG ; puts the offset
;address of variable ; MSG in DX
• This example contains data segments initialize DS.
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Program Segment Prefix
• When a program is loaded in memory, DOS prefaces it with PSP. The PSP contains information about program.
• DOS places in DS & ES segment # of PSP.
• DS must be loaded with the segment # of data segment
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DS initialization
• A program containing a data segment begins with:
MOV AX,@DATA
MOV DS,AX
• @Data is the name of the data segment defined by .DATA. It is translated into a segment #.
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Print the message
• With DS initialized, we may print the “HELLO!” message:
LEA DX,MSG ;get message
MOV AH,9;display string function
INT 21h ;display string
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TITLE PGM4-2: PRINT STRING PROGRAM
; This program displays “Hello!”
.MODEL SMALL
.STACK 100H
program title (comment)
comment line
memory model: small programs use at most 64K code and 64K data
set the stack size
Sample Program
directive giving title for printed listings
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Sample Program
.DATAMSG DB “HELLO!”,0DH,0AH,’$’
.CODEMAIN PROC MOV AX,@DATA MOV DS,AX ;initialize DS
LEA DX,MSG ;get message MOV AH,9 ;display string function INT 21H ;display message
MOV AH,4CH INT 21H ;DOS exitMAIN ENDPEND MAIN
starts the data segment where variables are stored
starts the code segment
reserve room for some bytes
variable name
carriage return and line feed
Declares the beginning of the procedure which is called main
marks the end of the current proceduremarks the end of the program. “main” specifies the program execution is to begin with the procedure “main”
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• Sample execution:
A> PGM4_2
HELLO!
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Case Conversion Program
ENTER A LOWER CASE LETTER : a
IN UPPER CASE IT IS : A
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Case Conversion Program
• Use EQU to define CR & LF as names for the constants 0DH & 0AH.
CR EQU 0DHLF EQU 0AH
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The messages and input character can be stored in the Data Segment like this:
MSG1 DB 'ENTER A LOWER CASE LETTER : $‘
MSG2 DB CR,LF , 'IN UPPER CASE IT IS : ‘
CHARDB ?,'$'
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Our program begins by displaying the first message and reading the character:
LEA DX,MSG1 ; get first message
MOV AH,9 ; display string function
INT 21H ;display first message
MOV AH,1 ; read character function
INT 21H ; read a small letter into AL
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Convert to upper case
SUB AL,20H ; convert into uppercase
MOV CHAR,AL ; and store it
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Display second message & uppercase
LEA DX,MSG2 ; get second message
MOV AH,9 ; display string function
INT 21H ;display message & uppercase letter
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Program Listing
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.MODEL SMALL
.STACK 100H
.DATACR EQU 0DHLF EQU 0AH
MSG1 DB 'ENTER A LOWER CASE LETTER : $'MSG2 DB CR,LF,'IN UPPER CASE IT IS : 'CHAR DB ?,'$'.CODEMAIN PROC; initialize DS
MOV AX,@DATAMOV DS,AX
;print user promptLEA DX,MSG1MOV AH,9INT 21H
; input a character and convert to upper case
MOV AH,1INT 21HSUB AL,20HMOV CHAR,AL
; display on the next lineLEA DX,MSG2MOV AH,9INT 21H
; return TO DOSMOV AH,4CHINT 21H
MAIN ENDPEND MAIN
