6-Input & Output Instructions-ch4

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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.
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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
7
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
9
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 = 1
Output : AL = ASCII code if character key is
pressed.
= o if non-character is pressed.
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Example
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MOV
AH,1 ; input key function
INT
21h
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; 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
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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
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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
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MOV
DL , BL
; get character
INT
21h
; and display it
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Program Listing
TITLE PGM4_1 : Echo PROGRAM
.MODEL SMALL
.STACK 100H
.CODE
MAIN PROC
;display prompt
MOV AH,2
MOV DL,'?'
INT 21H
; input a character
MOV AH,1
INT 21H
MOV BL,AL
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; go to a new line
MOV AH,2
MOV DL,0DH
INT 21H
MOV DL,0AH
INT 21H
; display characters
MOV DL,BL
INT 21H
; return to DOS
MOV AH,4CH
INT 21H
MAIN ENDP
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When a program terminates, it
should return control to DOS
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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 ‘$’ character.
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• If the string contains the ASCII
code of a control character , the
control function is performed.
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Example
• Print
MSG
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HELLO!
on the screen.
DB ‘HELLO!$’
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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
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destination , source
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LEA
•
•
•
•
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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
LEA
DB ‘HELLO!$’
DX , MSG
; puts the offset
;address of variable
; MSG in DX
• This example contains data segments initialize
DS.
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Program Segment Prefix
• When a program is loaded in memory,
DOS prefaces it with PSP. The PSP
contains information about program.
• DOS places in DS & ES segment # of
PSP.
• DS must be loaded with the segment # of
data segment
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DS initialization
• A program containing a data segment
begins with:
MOV AX,@DATA
MOV DS,AX
• @Data is the name of the data segment
defined by .DATA. It is translated into a
segment #.
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Print the message
• With DS initialized, we may print the
“HELLO!” message:
LEA
MOV
INT
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DX,MSG ;get message
AH,9;display string function
21h
;display string
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Sample Program
directive giving title for printed listings
program title (comment)
TITLE PGM4-2: PRINT STRING PROGRAM
; This program displays “Hello!”
comment line
.MODEL SMALL
.STACK 100H
memory model: small
programs use at most 64K code
and 64K data
set the stack size
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Sample Program
starts the data segment where variables
are stored
reserve room for some bytes carriage return and line feed
.DATA
MSG DB “HELLO!”,0DH,0AH,’$’
variable name
starts the code segment
.CODE
Declares the beginning of the procedure which is
MAIN PROC
MOV AX,@DATA called main
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 exit
marks the end of the current procedure
MAIN ENDP
marks the end of the program. “main” specifies
END MAIN
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
LF EQU
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0DH
0AH
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The messages and input character can be
stored in the Data Segment like this:
MSG1
MSG2
CHAR
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DB
DB
DB
'ENTER A LOWER CASE LETTER : $‘
CR,LF , 'IN UPPER CASE IT IS : ‘
?,'$'
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Our program begins by displaying the
first message and reading the character:
LEA
MOV
INT
MOV
INT
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DX,MSG1
AH,9
21H
AH,1
21H
; get first message
; display string function
;display first message
; read character function
; 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
; input a character and
.DATA
convert to upper case
CR
EQU
0DH
MOV AH,1
LF
EQU
0AH
INT 21H
MSG1 DB 'ENTER A LOWER CASE LETTER : $'
SUB AL,20H
MSG2 DB CR,LF,'IN UPPER CASE IT IS : '
MOV CHAR,AL
CHAR DB ?,'$'
; display on the next line
.CODE
LEA DX,MSG2
MAIN
PROC
MOV AH,9
; initialize DS
INT 21H
MOV AX,@DATA
; return TO DOS
MOV DS,AX
MOV AH,4CH
INT 21H
;print user prompt
MAIN ENDP
LEA DX,MSG1
END MAIN
MOV AH,9
INT 21H
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