MICROPROCESSORS & APPLICATIONS LAB MANUAL(Second half))

6(A). LARGEST ELEMENT IN AN ARRAY

AIM:

            To find the largest element in an array.

ALGORITHM:

1. Place all the elements of an array in the consecutive memory locations.
2. Fetch the first element from the memory location and load it in the accumulator.
3. Initialize a counter (register) with the total number of elements in an array.
4. Decrement the counter by 1.
5. Increment the memory pointer to point to the next element.
6. Compare the accumulator content with the memory content (next                             element).
7. If the accumulator content is smaller, then move the memory content                            (largest element) to the accumulator. Else continue.
8. Decrement the counter by 1.
9. Repeat steps 5 to 8  until the counter reaches zero
10. Store the result (accumulator content) in the specified memory location.

RESULT:

            Thus the largest number in the given array is found out.

PROGRAM:

ADDRE SS	OPCO DE	LABEL	MNEM ONICS	OPER AND	COMMENTS
8001			LXI	H,8100	Initialize HL reg. to 8100H
8002
8003
8004			MVI	B,04	Initialize B reg with no. of comparisons(n-1)
8005
8006			MOV	A,M	Transfer first data to acc.
8007		LOOP1	INX	H	Increment HL reg.  to point next memory location
8008			CMP	M	Compare M & A
8009			JNC	LOOP	If A is greater than M then go to loop
800A
800B
800C			MOV	A,M	Transfer data from M to A reg
800D		LOOP	DCR	B	Decrement B reg
800E			JNZ	LOOP1	If B is not Zero go to loop1
800F
8010
8011			STA	8105	Store the result in a memory location.
8012
8013
8014			HLT		Stop the program

OBSERVATION:

INPUT		OUTPUT
ADDRESS	DATA	ADDRESS	DATA
8100		8105
8101
8102
8103
8104

6(B). SMALLEST ELEMENT IN AN ARRAY

AIM:

            To find the smallest element in an array.

ALGORITHM:

1. Place all the elements of an array in the consecutive memory locations.
2. Fetch the first element from the memory location and load it in the accumulator.
3. Initialize a counter (register) with the total number of elements in an array.
4. Decrement the counter by 1.
5. Increment the memory pointer to point to the next element.
6. Compare the accumulator content with the memory content (next                             element).
7. If the accumulator content is smaller, then move the memory content                            (largest element) to the accumulator. Else continue.
8. Decrement the counter by 1.
9. Repeat steps 5 to 8  until the counter reaches zero
10. Store the result (accumulator content) in the specified memory location.

RESULT:

            Thus the smallest number in the given array is found out.

PROGRAM:

ADDRE SS	OPCO DE	LABEL	MNEM ONICS	OPER AND	COMMENTS
8001			LXI	H,8100	Initialize HL reg. to 8100H
8002
8003
8004			MVI	B,04	Initialize B reg with no. of comparisons(n-1)
8005
8006			MOV	A,M	Transfer first data to acc.
8007		LOOP1	INX	H	Increment HL reg.  to point next memory location
8008			CMP	M	Compare M & A
8009			JC	LOOP	If A is lesser than M then go to loop
800A
800B
800C			MOV	A,M	Transfer data from M to A reg
800D		LOOP	DCR	B	Decrement B reg
800E			JNZ	LOOP1	If B is not Zero go to loop1
800F
8010
8011			STA	8105	Store the result in a memory location.
8012
8013
8014			HLT		Stop the program

OBSERVATION:

INPUT		OUTPUT
ADDRESS	DATA	ADDRESS	DATA
8100		8105
8101
8102
8103
8104

7(A).ASCENDING ORDER

AIM:

            To sort the given number in the ascending order using 8085 microprocessor.

ALGORITHM:

 1. Get the numbers to be sorted from the memory locations.

 2. Compare the first two numbers and if the first number is larger than second then   I                       interchange the number.

3. If the first number is smaller, go to step 4

 4. Repeat steps 2 and 3 until the numbers are in required order

RESULT:

            Thus the ascending order program is executed and thus the numbers are arranged in ascending order.

PROGRAM:

ADDRE SS	OPCO DE	LABEL	MNEM ONICS	OPER AND	COMMENTS
8000			MVI	B,04	Initialize B reg with number of comparisons (n-1)
8001
8002		LOOP 3	LXI	H,8100	Initialize HL reg. to 8100H
8003
8004
8005			MVI	C,04	Initialize C reg with no. of comparisons(n-1)
8006
8007		LOOP2	MOV	A,M	Transfer first data to acc.
8008			INX	H	Increment HL reg.  to point next memory location
8009			CMP	M	Compare M & A
800A			JC	LOOP1	If A is less than M then go to loop1
800B
800C
800D			MOV	D,M	Transfer data from M to D reg
800E			MOV	M,A	Transfer data from acc to M
800F			DCX	H	Decrement HL pair
8010			MOV	M,D	Transfer data from D to M
8011			INX	H	Increment HL pair
8012		LOOP1	DCR	C	Decrement C reg
8013			JNZ	LOOP2	If C is not zero go to loop2
8014
8015
8016			DCR	B	Decrement B reg
8017			JNZ	LOOP3	If B is not Zero go to loop3
8018
8019
801A			HLT		Stop the program

OBSERVATION:

INPUT		OUTPUT
MEMORY LOCATION	DATA	MEMORY LOCATION	DATA
8100		8100
8101		8101
8102		8102
8103		8103
8104		8104

7(B). DESCENDING ORDER

AIM:

            To sort the given number in the descending order using 8085 microprocessor.

ALGORITHM:

 1. Get the numbers to be sorted from the memory locations.

 2. Compare the first two numbers and if the first number is smaller than second then   I                       interchange the number.

3. If the first number is larger, go to step 4

 4. Repeat steps 2 and 3 until the numbers are in required order

RESULT:

            Thus the descending order program is executed and thus the numbers are arranged in descending order.

PROGRAM:

ADDRE SS		OPCO DE	LABEL	MNEM ONICS	OPER AND	COMMENTS
8000				MVI	B,04	Initialize B reg with number of comparisons (n-1)
8001
8002			LOOP 3	LXI	H,8100	Initialize HL reg. to 8100H
8003
8004
8005				MVI	C,04	Initialize C reg with no. of comparisons(n-1)
8006
8007			LOOP2	MOV	A,M	Transfer first data to acc.
8008				INX	H	Increment HL reg.  to point next memory location
8009				CMP	M	Compare M & A
800A				JNC	LOOP1	If A is greater than M then go to loop1
800B
800C
800D				MOV	D,M	Transfer data from M to D reg
800E				MOV	M,A	Transfer data from acc to M
800F				DCX	H	Decrement HL pair
8010				MOV	M,D	Transfer data from D to M
8011				INX	H	Increment HL pair
8012			LOOP1	DCR	C	Decrement C reg
8013				JNZ	LOOP2	If C is not zero go to loop2
8014
8015
8016				DCR	B	Decrement B reg
8017				JNZ	LOOP3	If B is not Zero go to loop3
8018
8019
801A				HLT		Stop the program

OBSERVATION:

INPUT		OUTPUT
MEMORY LOCATION	DATA	MEMORY LOCATION	DATA
8100		8100
8101		8101
8102		8102
8103		8103
8104		8104

8(A). CODE CONVERSION –DECIMAL TO HEX

AIM:

            To convert a given decimal number to hexadecimal.

ALGORITHM:

1.      Initialize the memory location to the data pointer.

2.      Increment B register.

3.      Increment accumulator by 1 and adjust it to decimal every time.

4.      Compare the given decimal number with accumulator value.

5.      When both matches, the equivalent hexadecimal value is in B register.

6.      Store the resultant in memory location.

RESULT:

Thus an ALP program for conversion of decimal to hexadecimal was written and executed.

PROGRAM:

ADDRE SS	OPCO DE	LABEL	MNEM ONICS	OPER AND	COMMENTS
8000			LXI	H,8100	Initialize HL reg. to 8100H
8001
8002
8003			MVI	A,00	Initialize A register.
8004
8005			MVI	B,00	Initialize B register..
8006
8007		LOOP	INR	B	Increment B reg.
8008			ADI	01	Increment A reg
8009
800A			DAA		Decimal Adjust Accumulator
800B			CMP	M	Compare M & A
800C			JNZ	LOOP	If acc and given number are not equal, then go to LOOP
800D
800E
800F			MOV	A,B	Transfer B reg to acc.
8010			STA	8101	Store the result in a memory location.
8011
8012
8013			HLT		Stop the program

RESULT:

INPUT		OUTPUT
ADDRESS	DATA	ADDRESS	DATA
8100		8101

8(B). CODE CONVERSION –HEXADECIMAL TO DECIMAL

AIM:

            To convert a given hexadecimal number to decimal.

ALGORITHM:

1.      Initialize the memory location to the data pointer.

2.      Increment B register.

3.      Increment accumulator by 1 and adjust it to decimal every time.

4.      Compare the given hexadecimal number with B register value.

5.      When both match, the equivalent decimal value is in A register.

6.      Store the resultant in memory location.

RESULT:

Thus an ALP program for conversion of hexadecimal to decimal was written and executed.

PROGRAM:

ADDRE SS	OPCO DE	LABEL	MNEM ONICS	OPER AND	COMMENTS
8000			LXI	H,8100	Initialize HL reg. to 8100H
8001
8002
8003			MVI	A,00	Initialize A register.
8004
8005			MVI	B,00	Initialize B register.
8006
8007			MVI	C,00	Initialize C register for carry.
8008
8009		LOOP	INR	B	Increment B reg.
800A			ADI	01	Increment A reg
800B
800C			DAA		Decimal Adjust Accumulator
800D			JNC	NEXT	If there is no carry go to NEXT.
800E
800F
8010			INR	C	Increment c register.
8011		NEXT	MOV	D,A	Transfer A to D
8012			MOV	A,B	Transfer B to A
8013			CMP	M	Compare M & A
8014			MOV	A,D	Transfer D to A
8015			JNZ	LOOP	If acc and given number are not equal, then go to LOOP
8016
8017
8018			STA	8101	Store the result in a memory location.
8019
801A
801B			MOV	A,C	Transfer C to A
801C			STA	8102	Store the carry in another memory location.
801D
801E
801F			HLT		Stop the program

RESULT:

INPUT		OUTPUT
ADDRESS	DATA	ADDRESS	DATA
8100		8101
		8102

9(A) BCD ADDITION

AIM:

             To add two 8 bit BCD numbers stored at consecutive memory locations.

ALGORITHM:

1.      Initialize memory pointer to data location.

2.      Get the first number from memory in accumulator.

3.      Get the second number and add it to the accumulator

4.      Adjust the accumulator value to the proper BCD value using DAA instruction.

5.      Store the answer at another memory location.

RESULT:

Thus the 8 bit BCD numbers stored at 4500 &4501 are added and the result stored at 4502 & 4503.

PROGRAM:

ADDRESS	OPCODE	LABEL	MNEMONICS	OPERAND	COMMENT
4100		START	MVI	C, 00	Clear C reg.
4101
4102			LXI	H, 4500	Initialize HL reg. to 4500
4103
4104
4105			MOV	A, M	Transfer first data to accumulator
4106			INX	H	Increment HL reg.  to point next memory    Location.
4107			ADD	M	Add first number to acc. Content.
4108			DAA		Decimal adjust accumulator
4109			JNC	L1	Jump to location if result does not yield carry.
410A
410B
410C			INR	C	Increment C reg.
410D		L1	INX	H	Increment HL reg.  to point next memory Location.
410E			MOV	M, A	Transfer the result from acc. to memory.
410F			INX	H	Increment HL reg. to point next memory Location.
4110			MOV	M, C	Move carry to memory
4111			HLT		Stop the program

OBSERVATION:

INPUT		OUTPUT
4500		4502
4501		4503

                                                             

9(B). BCD SUBTRACTION

AIM:

             To Subtract two 8 bit BCD numbers stored at consecutive memory locations.

ALGORITHM:

1.      Load the minuend and subtrahend in two registers.

2.      Initialize Borrow register to 0.

3.      Take the 100’s complement of the subtrahend.

4.      Add the result with the minuend  which yields the result.

5.      Adjust the accumulator value to the proper BCD value using DAA instruction. If there is a carry ignore it.

6.      If there is no carry, increment the carry register by 1

7.      Store the content of the accumulator (result)and borrow register in the specified memory location

RESULT:

Thus the 8 bit BCD numbers stored at 4500 &4501 are subtracted and the result stored at 4502 & 4503.

PROGRAM:

ADDRESS	OPCODE	LABEL	MNEMONICS	OPERAND	COMMENT
4100		START	MVI	D, 00	Clear D reg.
4101
4102			LXI	H, 4500	Initialize HL reg. to 4500
4103
4104
4105			MOV	B, M	Transfer first data to accumulator
4106			INX	H	Increment HL reg. to point next mem. Location.
4107			MOV	C, M	Move second no. to B reg.
4108			MVI	A, 99	Move 99 to the Accumulator
4109
410A			SUB	C	Subtract [C] from  acc. Content.
410B			INR	A	Increment  A  register
410C			ADD	B	Add [B] with [A]
410D			DAA		Adjust Accumulator value for Decimal digits
410E			JC	LOOP	Jump on carry to loop
410F
4110
4111			INR	D	Increment D reg.
4112		LOOP	INX	H	Increment HL register pair
4113			MOV	M , A	Move the Acc.content to the memory location
4114			INX	H	Increment HL reg. to point next mem. Location.
4115			MOV	M, D	Transfer D register content to memory.
4116			HLT		Stop the program

OBSERVATION:

INPUT		OUTPUT
4500		4502
4501		4503

10. 2 X 2 MATRIX MULTIPLICATION

AIM:

            To perform the 2 x 2 matrix multiplication.

ALGORITHM:

 

1. Load the 2 input matrices in the separate address and initialize the HL and the DE register pair with the starting address respectively.
2. Call a subroutine for performing the multiplication of one element of a matrix with the other element of the other matrix.
3. Call a subroutine to store the resultant values in a separate matrix.

RESULT:

Thus the 2 x 2  matrix multiplication is performed and the result is stored at 4700,4701 , 4702 & 4703.

PROGRAM:

ADDRESS	OPCODE	LABEL	MNEMONICS	OPERAND	COMMENT
8100			MVI	C, 00	Clear C reg.
8101
8102			LXI	H, 8500	Initialize HL reg. to 4500
8103
8104
8105		LOOP2	LXI	D, 8600	Load DE register pair
8106
8107
8108			CALL	MUL	Call subroutine MUL
8109
810A
810B			MOV	B,A	Move A to B reg.
810C			INX	H	Increment HL register pair .
810D			INX	D	Increment DE register pair
810E			INX	D	Increment DE register pair
810F			CALL	MUL	Call subroutine MUL
8110
8111
8112			ADD	B	Add [B] with [A]
8113			CALL	STORE	Call subroutine STORE
8114
8115
8116			DCX	H	Decrement HL register pair
8117			DCX	D	Decrement DE register pair
8118			CALL	MUL	Call subroutine MUL
8119
811A
811B			MOV	B,A	Transfer A reg content  to B reg.
811C			INX	H	Increment HL register pair
811D			INX	D	Increment DE register pair
811E			INX	D	Increment DE register pair
811F			CALL	MUL	Call subroutine MUL
8120
8121
8122			ADD	B	Add A with B
8123			CALL	STORE	Call subroutine MUL
8124
8125
8126			MOV	A,C	Transfer C register content to Acc.
8127			CPI	04	Compare with 04 to check whether all elements are multiplied.
8128
8129			JZ	LOOP1	If completed, go to loop1
812A
812B
812C			INX	H	Increment HL register Pair.
812D			JMP	LOOP2	Jump to LOOP2.
812E
812F
8130		LOOP1	HLT		Stop the program.
8131		MUL	LDAX	D	Load acc from the memory location pointed by DE pair.
8132			MOV	D,A	Transfer acc content to D register.
8133			MOV	H,M	Transfer from memory to H register.
8134			DCR	H	Decrement H register.
8135			JZ	LOOP3	If H is zero go to LOOP3.
8136
8137
8138		LOOP4	ADD	D	Add Acc with D reg
8139			DCR	H	Decrement H register.
813A			JNZ	LOOP4	If H is not zero go to LOOP4.
813B
813C
813D		LOOP3	MVI	H,85	Transfer 85 TO H register.
813E
813F			MVI	D,86	Transfer 86 to D register.
8140
8141			RET		Return to main program.
8142		STORE	MVI	B,87	Transfer 87 to B register.
8143
8144			STAX	B	Load A from memory location pointed by BC pair.
8145			INR	C	Increment C register.
8146			RET		Return to main program.

OBSERVATION:

INPUT				OUTPUT
4500		4600		4700
4501		4601		4701
4502		4602		4702
4503		4603		4703

11. BIOS/DOS CALLS – DISPLAY

AIM:

      To display a message on the CRT screen of a microcomputer using DOS calls.

ALGORITHM:

1. Initialize the data segment and the message to be displayed.
2. Set function value for display.
3. Point to the message and run the interrupt to display the message in the CRT.

PROGRAM:

ASSUME CS: CODE, DS: DATA

DATA SEGMENT

MSG DB 0DH, 0AH, “GOOD MORNING” , ODH, OAH, “$”

DATA ENDS

CODE SEGMENT

START:     MOV AX, DATA

                  MOV DS, AX

                  MOV AH, 09H

                  MOV DX, OFFSET MSG

                  INT 21H

                  MOV AH, 4CH

                  INT 21H

CODE ENDS

END START

RESULT:

A message is displayed on the CRT screen of a microcomputer using DOS calls
12. BIOS/DOS CALLS – FILE MANIPULATION

AIM:

To open a file using DOS calls.

ALGORITHM:

1. Initialize the data segment, file name and the message to be displayed.
2. Set the file attribute to create a file using a DOS call.
3. If the file is unable t o create a file display the message 

PROGRAM:

ASSUME CS: CODE, DS: DATA

DATA SEGMENT

FILENAME DB “SAMPLE.DAT”, “$”

MSG DB 0DH, 0AH, “FILE NOT CREATED”, ODH, OAH, “$”

DATA ENDS

CODE SEGMENT

START:           MOV AX, DATA

                        MOV DS, AX

                        MOV DX, OFFSET FILENAME

                        MOV CX, 00H

                        MOV AH, 3CH

                        INT 21H

                        JNC LOOP1

                        MOV AX, DATA

                        MOV DS, AX

                        MOV DX, OFFSET MSG

                        MOV AH, 09H

                        INT 21H

LOOP1           MOV AH, 4CH

                        INT 21H

CODE ENDS

END START

RESULT:

A file is opened using DOS calls.
13. BIOS/DOS CALLS – DISK INFORMATION

AIM:

      To display the disk information.

ALGORITHM:

1. Initialize the data segment and the message to be displayed.
2. Set function value for disk information.
3. Point to the message and run the interrupt to display the message in the CRT.

PROGRAM:

ASSUME CS: CODE, DS: DATA

DATA SEGMENT

MSG DB 0DH, 0AH, “GOOD MORNING” , ODH, OAH, “$”

DATA ENDS

CODE SEGMENT

START:     MOV AX, DATA

                  MOV DS, AX

                  MOV AH, 36H

                  MOV DX, OFFSET MSG

                  INT 21H

                  MOV AH, 4CH

                  INT 21H

CODE ENDS

END START

RESULT:

      The disk information is displayed.

1.8086 STRING MANIPULATION – SEARCH A WORD

AIM:

To search a word from a string.

ALGORITHM:

1. Load the source and destination index register with starting and the ending address respectively.
2. Initialize the counter with the total number of words to be copied.
3. Clear the direction flag for auto incrementing mode of transfer.
4. Use the string manipulation instruction SCASW with the prefix REP to search a word from string.
5. If a match is found (z=1), display 01 in destination address. Otherwise, display 00 in destination address.

RESULT:

      A word is searched and the count of number of appearances is displayed.

PROGRAM:

ASSUME CS: CODE, DS: DATA

DATA SEGMENT

LIST DW 53H, 15H, 19H, 02H

DEST EQU 3000H

COUNT EQU 05H

DATA ENDS

CODE SEGMENT

START:           MOV AX, DATA

                        MOV DS, AX

                        MOV AX, 15H

                        MOV SI, OFFSET LIST

                        MOV DI, DEST

                        MOV CX, COUNT

                        MOV AX, 00

                        CLD

REP                 SCASW

            JZ LOOP

            MOV AX, 01

LOOP             MOV [DI], AX

                        MOV AH, 4CH

                        INT 21H

CODE ENDS

END START

INPUT:                     

LIST: 53H, 15H, 19H, 02H

OUTPUT:

      3000    01