Direct video practice and Keyboard Operations

1
Direct video practice and Keyboard Operations
2
Midterm

• Avg 182/195 (93)

3
Midterm

• 1.1. For the data bits 01001001, the parity bit
  is
• (a) 0 (b) 1
• (c) irrelevant (d) unknown

4
Midterm

• 1.2. The data segment contains
• (a) read-only memory,
• (b) the machine instructions that are to
  execute,
• (c) a program's defined data, constants, and
  work areas,
• (d) values that a program needs to save
  temporarily.

5
Midterm

• 1.3. The binary representation for decimal 117
  is
• a. 01110101 c. 01111011
• b. 01110011 d. 01100101

6
Midterm

• 1.5 Which of the following is an invalid name?
• a. _hello b. hello
• c. hello d. hello

7
Midterm

• 1.6. Which one of the following statements is
  invalid?
• (a) MOV AL,035AH (b) MOV CX,035AH
• (c) MOV AX,035AH (d) MOV DI,035AH

8
Midterm

• 1.7. INT 21H function 09H for displaying
  requires defining a display string in the data
  area, immediately followed by this character
• (a) _at_ (b)
• (c) (d) _

9
Midterm

• 1.8 Given DATSG is the name of the data segment.
  The instruction MOV DS,DATSG is a correct
  instruction to initialize the address of the data
  segment in the DS register.
• a. True
• b. False

10
Midterm

• 1.9 Identify the following colors in the
  attribute byte 00000001
• (a) Blue foreground and black background
• (b) Blue foreground and white background
• (c) White foreground and blue background
• (d) Black foreground and blue background

11
Midterm

• 1.10 In what significant way do the following MOV
  instructions differ in execution
• MOV DX,AC24H
• MOV DX,AC24H
• a. The first MOV moves immediate value AC24H to
  DX while the second MOV moves AC24H to location
  specified by DSDX
• b. The first MOV moves immediate value AC24H to
  DX while the second MOV moves the contents of
  locations AC24H and AC25H to DX.
• c. The first MOV moves immediate value AC24H to
  location specified by DSDX while the second MOV
  moves the content of locations AC24H and AC25H
  to DX
• d. The first MOV moves immediate value AC24H to
  DX while the second MOV moves the content of
  locations AC24H to DX.

12
Midterm

• 1. The _code____ segment contains the machine
  instructions that are to execute. The _data____
  segment contains a program's defined data,
  constants, and work areas. The _stack_____
  contains any data and addresses that you need to
  save temporarily.
• 2. The _IP__ register contains the offset address
  of the instruction that is to execute next and is
  associated with the _CS__ register that addresses
  the code segment.

13
Midterm

• 3. The LOOP instruction requires an initial value
  in the _CX__ register. For each iteration, LOOP
  deducts 1 from this value if it is __zero____,
  control drops through to the following
  instruction, and if it is __non zero_____,
  control jumps to the operand address.
• 4. For INT 21H function 0AH, the area for
  keyboard input requires definition of a
  _parameter list__. The first byte contains
  maximum number of character entered_. The second
  byte is for the operation to store the _actual
  number of character entered___. The third byte
  begins a field that is to contain _entered
  characters______________________.

14
Midterm

• 5. A RET instruction that returns from a near
  procedure pops the original value from the stack
  into the IP___ register and increments the _SP__
  register by 2.
• 6. For screen clearing or scrolling, use BIOS INT
  _10H__ and insert function _06H__ in AH

15
Midterm

• 7. The first byte of an instruction is the
  _operation_ the other bytes are _operands__,
  which reference an immediate value, register, or
  memory location.
• 8. On loading an .EXE program into memory for
  execution, the loader stores the address of the
  PSP in the _DS___ and _ES___ registers, the
  address of the stack in the __SS__ register, and
  the size of the stack in the ___SP_ register.

16
Midterm

• 9. The BIOS Data Area begins in low memory at
  segment address __400 H______
• 10. The statement DB 0 defines one _byte___
  initialized with a _zero___ value.

17
Midterm

• III. 1.
• SS 2A600H
• 100H
• 2A700H -gt SS 2A70H
• CS 2A600H
• 100H
• 30H
• 40H
• 2A770H -gt CS 2A77H

18
Midterm

• III. 1.
• c. ES 2A60H
• d. SP 0030H

19
Midterm

• III. 2
• a.
• b. 3412H 011AH 352CH

BL
BH
34H
12H
BL
BH
35H
2CH
20
Midterm

• III. 2
• c.

0135
0134
2CH
35H
21
Midterm

• 3.
• WORD3 1030 300
• 4. CX 5 AX 10
• CX 4 AX 20
• CX 3 AX 30
• CX 2 AX 40
• CX 1 AX 50

22
Direct Video Display

• Fastest way to display text or graphic characters
  is to transfer them directly to video display
  area
• The address of page 0 for mode 3 is B8000H.
  Each character require two bytes (one for
  representing the character itself and one for its
  attribute)
• Moving a pair of characterattribute into video
  area of the active page will cause the character
  appear immediately on screen

23
Direct Video Display
80 columns (0-79) 25 rows (0-24)
Hint Similar to the use one dimension array to
represent this area. Each element put in this
area takes 2 bytes. We start this array from
0 Example ES00 48H ES01 1FH
(0,0)
MOV ES00, 48H MOV ES01, 1FH
24
Direct Video Display
(0,39)
ES78 48H ES79 0FH
MOV ES78, 48H MOV ES79, 0FH
(1,0)
(1,39)
MOV ES160, 48H MOV ES161, 0FH MOV ES238,
48H MOV ES239, 0FH
ES160 48H ES161 0FH ES238
48H ES239 0FH
25
Example of direct video display

• MOV AX, 0B800h
• MOV ES, AX
• print message using ES
• as segment
• MOV ES00h, 48h H
• MOV ES02H, 45h E
• MOV ES04H, 4ch L
• MOV ES06H, 4ch L
• MOV ES08H, 4fh O

26
Example of direct video display
27
Example of direct video display
MOV AX, 0B800H MOV ES, AX MOV AX, 0003H Set
the current video mode to text mode INT 10H
INT 10H, function 00H (mode03H is standard
text) MOV AX, 0500H Set the active page
00 INT 10H CALL DISPLAY
28
Example of direct video display
DISPLAY PROC NEAR mov cx,25 vertical line
mov di,0 again1 mov esdi,48h mov
esdi1,4eh add di,80 loop again1 DISPLAY
ENDP
29
Direct Video Display
30
Direct video display Practice
1. Open your browser and open this
page C\emu8086\documentation\8086_instruction_se
t.html And C\emu8086\documentation\8086_and_dos_i
nterrupts.html 2. Open your emu8086 software 3.
Cut and paste (or type) the following code (as
shown in the next page) and save as
directvideo.asm
31
page 60,132 TITLE VideoPractice DirectDisplay
--------------------------------------------- STA
CK SEGMENT PARA STACK 'Stack' DW 32
DUP(0) STACK ENDS ------------------------------
---------------- DATASEG SEGMENT PARA 'Data'
Please insert your data declaration here
DATASEG ENDS CODESEG SEGMENT PARA
'Code' MAIN PROC FAR MOV AX, 0B800H
MOV ES, AX Please enter
your code here MOV AX,4C00H
exit procedure INT 21H MAIN
ENDP CODESEG ENDS END MAIN End of program
32
Direct video display Practice

• 4. Modify your code so that it performs the
  following tasks
• Draw horizontal line (40 character E) on the
  screen using direct video display
• 5. Compile and run your code. It should look
  something like

33
Coding

• MOV AX, 0B800h
• MOV ES, AX
• mov cx,40 horizontal line
• mov di,0
• again2
• mov esdi,45h
• mov esdi1,5eh
• add di,2
• loop again2

34
Lesson plan

• Brief go over keyboard operations
• Focus on String operation

35
Keyboard Operations

• Keyboard provides three basic types of keys
• Standard characters A-Z, a-z,0-9,,,.
• Extended function key
• Function keys F1,F2..Shift-F1..
• Numeric keypad keys Home,End, Arrows, Page Up,
  Page Down.
• Special keys Alt, Ctrl, and Shift

36
INT 21H Functions

• Direct keyboard input no echo
• if there is no character in the keyboard buffer,
  the function waits until any key is pressed.
• Example
• example
• MOV AH, 7
• INT 21H

37
INT 21H

• Function 0BH Check Keyboard status
• Return FFH in AL if an input character is
  available in the keyboard buffer and
• Return 00H if no character is available
• MOV AH, 7
• INT 21H
• MOV AH, 0BH
• INT 21H
• CMP AL, 00H
• JE PRINTLABEL
• JMP CONT
• PRINTLABEL
• ..
• CONT

38
INT 21H

• Function 0Ch - flush keyboard buffer and read
  standard input.Example
• MOV AH, 0CH
• MOV AL, 07H
• INT 21H

39
Keyboard buffer

• BIOS Data Area starts at 401EH
• INT 09H is automatically executed when you press
  a key. This interrupt points to an
  interrupt-handling routines in ROM BIOS.
• The BIOS routine reads the scan code and compares
  it with entries in a scan code table for the
  associate ASCII character

40
BIOS INT 09H and the keyboard buffer

• When we press a key, the keyboard processor
  automatically generates its scan code and issue
  INT 09H
• When we release the key with one-half second, it
  generates a second scan code to tell the
  interrupt routine that you have released the key.

41
Address

• 41AH current head of the buffer, the next
  position for INT 16H is to get a character
  entered from the keyboard
• 41CH current tail of the buffer. The next
  position for INT 09H to store a character entered
  from the keyboard
• 41EH beginning of the keyboard buffer itself.

42
INT 16H operations for keyboard

• INT 16H is the basic BIOS keyboard operation used
  extensively by software developers

43
INT 16H operations for keyboard

• INT 16h / AH 00h - get keystroke from keyboard
  (no echo).returnAH BIOS scan code.AL
  ASCII character.
• MOV AH, 00H
• INT 16H

44
Example

• MOV AH, 00H
• INT 16H
• CMP AL,'c'
• JE PRINTPROMPT
• JMP CONT
• PRINTPROMPT
• CONT

45
INT 16H

• INT 16h / AH 01h - check for keystroke in the
  keyboard buffer.
• returnZF 1 if keystroke is not available.ZF
  0 if keystroke available.AH BIOS scan
  code.AL ASCII character.(if a keystroke is
  present, it is not removed from the keyboard
  buffer).

46
String operations
47
Processing String Data

• Two ways to code string instructions
• 1. Address of Operands are implied
• E.g MOVS BYTE1, BYTE2
• Need to have
• BYTE1 is stored at ESDI
• BYTE2 is stored at ESSI

48
Processing String Data

• Two ways to code string instructions
• 2. Move the addresses of operands in DI and SI
  and use the instruction without OPERANDS
• E.g
• LEA DI, BYTE2 BYTE2 at ESDI
• LEA SI, BYTE1 BYTE1 at DSSI
• MOVSB MOVE BYTE2 to BYTE1

49
String operations

• Dont forget to initialize the DS and ES by
• MOV AX, DATASEG
• MOV DS, AX
• MOV ES, AX

50
MOVSB

• Copy byte at DSSI to ESDI. Update SI and
  DI.
• if DF 0 then
• SI SI 1
• DI DI 1
• else
• SI SI - 1
• DI DI - 1

51
MOVSB

• Example
• a1 DB 5 DUP('')
• a2 DB 5 DUP(' ')
• LEA SI, a1
• LEA DI, a2
• MOV CX, 5
• REP MOVSB

52
Repeat String Prefix (REP)

• Repeat following MOVSB, MOVSW, LODSB, LODSW,
  STOSB, STOSW instructions CX times.
  check_cxif CX ltgt 0 then
• do following chain instruction
• Decrease CX CX - 1
• go back to check_cx
• else exit

53
MOVSW

• Copy word at DSSI to ESDI. Update SI and
  DI.ESDI DSSI
• if DF 0 then
• SI SI 2
• DI DI 2
• else
• SI SI - 2
• DI DI - 2

54
Example

• a1 DW 5 DUP('')
• a2 DW 5 DUP(' ')
• LEA SI, a1
• LEA DI, a2
• MOV CX, 5
• REP MOVSW

55
LODSB

• Load byte at DSSI into AL. Update SI.
• AL DSSI
• if DF 0 then SI SI 1
• else SI SI - 1

56
Example

• String reverse
• SRC_STR DB 'NEWYORK'
• DES_STR DB 7 DUP (' ')
• LEA SI, SRC_STR
• LEA DI, DES_STR6
• MOV CX, 7
• L10
• LODSB
• MOV DI, AL
• DEC DI
• LOOP L10

57
STOB

• Store byte in AL into ESDI. Update
  DI.Algorithm
• ESDI AL
• if DF 0 then
• DI DI 1
• else
• DI DI - 1

58
Example

• LEA DI, a1
• MOV AL, 12h
• MOV CX, 5
• REP STOSB

59
COMPSB

• Compare bytes ESDI from DSSI. DSSI -
  ESDI
• set flags according to resultOF, SF, ZF, AF,
  PF, CF
• if DF 0 then
• SI SI 1
• DI DI 1
• else
• SI SI - 1
• DI DI - 1

60
Lab/Practice

• 1. Open your browser and open this page
• C\emu8086\documentation\8086_instruction_set.html
  
• And
• C\emu8086\documentation\8086_and_dos_interrupts.h
  tml
• 2. Open your emu8086 software
• 3. Cut and paste (or type) the following code (as
  shown in the next page) and save reverse.asm

61

• page 60,132
• TITLE StringPractice StringReverse
• ---------------------------------------------
• STACK SEGMENT PARA STACK 'Stack'
• DW 32 DUP(0)
• STACK ENDS
• ----------------------------------------------
• DATASEG SEGMENT PARA 'Data'
• Please insert your data declaration here
• DATASEG ENDS
• CODESEG SEGMENT PARA 'Code'
• MAIN PROC FAR
• MOV AX, DATASEG
• MOV DS, AX
• MOV ES, AX
• Please enter your code here
• MOV AX,4C00H exit procedure

62
Lab/Practice

• 4. Insert the code to
• - Read a string from keyboard (INT 21H, 0AH).
  Max typed characters 50
• - Reverse that string using the string
  operations learned today. (see code for reversing
  a string)
• 5. Compile and run the code