Processing String Data and Binary Data

(continue)

Code for last practice exercise

DATASEG SEGMENT PARA 'Data'

; Please insert your data declaration here

DES_STR DB 50 DUP ('$')

ParaList label byte

maxlen DB 50

actlen DB ?

SRC_STR DB 50 DUP ('$')

DATASEG ENDS

Code for last practice exercise; Code for reading input from keyboard

MOV AH, 0AH LEA DX, ParaList INT 21H ; Code for reversing the string LEA SI, SRC_STR LEA DI, DES_STR MOV CH,0 MOV CL, ACTLEN DEC CX ADD DI, CX INC CX L10: LODSB MOV [DI], AL DEC DI LOOP L10

CMPSB, CMPSW Compare String Instruction

Compare bytes: ES:[DI] from DS:[SI].

DS:[SI] - ES:[DI]

set flags according to result:OF, SF, ZF, AF, PF, CF

if DF = 0 then SI = SI + 1 DI = DI + 1

else SI = SI - 1 DI = DI - 1

Example

STRING1 DB ‘COMPUTER’STRING2 DB ‘COMPUTER’

MOV CX, 8LEA DI, STRING2LEA SI, STRING1REPE CMPSBJNE exit

exit: …

REPE (repeat if equal)

Repeat following CMPSB, CMPSW, SCASB, SCASW instructions while ZF = 1 (result is Equal), maximum CX times.

check_cx:

if CX <> 0 then do following chain instruction

CX = CX - 1 if ZF = 1 then:

go back to check_cx else

exit from REPE cycle else

exit from REPE cycle

SCASB: SCAN STRING INSTRUCTION

Scan a string for a specified value. Continue to Compare bytes: AL from ES:[DI] while the comparison is not equal or until CX is 0

(Similar to indexOf in java or strstr in C)ES:[DI] - AL

set flags according to result:OF, SF, ZF, AF, PF, CF

if DF = 0 then DI = DI + 1

else DI = DI - 1

Example:

STRING1 DB ‘COMPUTER’

…

MOV AL, ‘r’

MOV CX, 8

LEA DI, STRING1

REPNE SCASB

REPNE

Repeat following CMPSB, CMPSW, SCASB, SCASW instructions while ZF = 0 (result is Not Equal), maximum CX times.

check_cx:

if CX <> 0 then do following chain instruction

CX = CX - 1 if ZF = 0 then:

go back to check_cx else

exit from REPNE cycle else

exit from REPNE cycle

Replicating a pattern

PATTERN DB ‘|----|’

RESULT DB 42 DUP (‘ ‘)

MOV CX, 21

LEA DI, RESULT

LEA SI, PATTERN

REP MOVSW

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.html

2. Open your emu8086 software

3. Cut and paste (or type) the following code (as shown in the next page) and save move.asm

page 60,132TITLEMovePractice Move; ---------------------------------------------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 insert your code here MOV AX,4C00H ;exit procedure INT 21HMAIN ENDP

CODESEG ENDSEND MAIN ;End of program

Practice

4. Modify your code so that:- Define: BUS_TITLE DB ‘COMPUTER WIZARDS’- WORK_SPACE DB 16 DUP(20H)

a. - Move BUS_TITLE to WORK_SPACE from left to right (CLD)

b. - MOV BUS_TITLE to WORK_SPACE from right to left (STD)

c. - COMPARE BUS_TITLE with WORK_SPACE, print “equal” if equal, print “not equal” if not5. Compile and run

Code

; Data segment declaration

EQUALPROMPT DB 'Equal', '$'

NOTEQUALPROMPT DB 'Not Equal', '$'

BUS_TITLE DB 'COMPUTER WIZARDS','$'

WORK_SPACE DB 17 DUP(20H)

Code ; Moving from left to right

CLD LEA SI, BUS_TITLE LEA DI, WORK_SPACE MOV CX, 17 REP MOVSB

; Moving from right to left STD

LEA SI, BUS_TITLE+16 LEA DI, WORK_SPACE+16 MOV CX, 17 REP MOVSB

Code ; Print equal if equal, print not equal if not MOV CX, 17 LEA DI, BUS_TITLE LEA SI, WORK_SPACE REPE CMPSB JE EQUALPRINT MOV AH, 09H LEA DX, NOTEQUALPROMPT INT 21H JMP exitEQUALPRINT: MOV AH, 09H LEA DX, EQUALPROMPT INT 21H

Binary Data

ADC CBW NEG SBB

ADC

Add with Carry.

Algorithm:

operand1 = operand1 + operand2 + CF

Example

STC ; set CF =1

MOV AL, 5 ; AL = 5

ADC AL, 1 ;

AL =?

CBW

Convert byte into word.

Algorithm:

if high bit of AL = 1 then: AH = 255 (0FFh)

else AH = 0

Example

MOV AX, 0 ; AH = 0, AL = 0

MOV AL, -5 ; AX = 000FBh (251)

CBW ; AX = 0FFFBh (-5)

NEG

Negate. Makes operand negative (two's complement).

Algorithm:

Invert all bits of the operand

Add 1 to inverted operand

Example

MOV AL, 5 ; AL = 05h

NEG AL ; AL = 0FBh (-5)

NEG AL ; AL = 05h (5)

SBB

Subtract with Borrow.

Algorithm:

operand1 = operand1 - operand2 - CF

Example

STC ; Set CF=1

MOV AL, 5

SBB AL, 3

AL =?

Distinguish between a carry and an overflow

Arithmetic carry (CF)

An arithmetic operation transfers the resulting sign bit (0 or 1) to CF.

When the carry occurs with unsigned data, the result is invalid. When carry occurs with signed data, the result is valid

Distinguish between a carry and an overflow

• Arithmetic overflow (OF)

Example: unsigned signed decimal

11110110 246 -10

+ 10001001 37 -119

(1)01111111 127 (invalid) -129 (invalid)

CF = 1

OF = 1

Note about DIV/ IDIV

If divisor is a byte, a value must be greater than the left byte of the dividend (AH)

If divisor is a word, its value must be greater than left word of the dividend (DX)

Define doubleword using words

Assume, we have 13290147H, we can define this as:

DDVAR DW 0147H

DW 1329H

Practice 1. Open your browser and open this page:C:\emu8086\documentation\

8086_instruction_set.htmlAndC:\emu8086\documentation\

8086_and_dos_interrupts.html2. Open your emu8086 software3. Cut and paste (or type) the following code

(as shown in the next page) and save add.asm

page 60,132TITLE AddPractice Add; ---------------------------------------------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 insert your code here MOV AX,4C00H ;exit procedure INT 21HMAIN ENDPCODESEG ENDSEND MAIN ;End of program

Practice 4. Define data:

BIN1 DW 0147H DW 1329H

BIN2 DW 02B3H DW 0241H

SUM DW 0 DW 0

Write a program to add the doublewords beginning at BIN1 and BIN2 and store the result to doubleword beginning at SUM

SUM= 156A03FAH