ASSEMBLY LANGUAGE PROGRAMMING EXAMPLES

ASSEMBLY LANGUAGE PROGRAMMING EXAMPLES

Addition Programs

Example 1: Addition of two 8-bit numbers whose sum is 8-bits.

Explanation: This assembly language program adds two 8-bit numbers stored in two memory locations .The sum of the two numbers is 8-bits only. The necessary algorithm and flow charts are given below.

ALGORITHM:

Step1. : Initialize H-L pair with memory address XX00 (say: 9000).

Step2. : Clear accumulator.

Step3. : Add contents of memory location M to accumulator.

Step4. : Increment memory pointer (i.e. XX01).

Step5. : Add the contents of memory indicated by memory pointer to accumulator.

Step6. : Store the contents of accumulator in 9002.

Step7. : Halt

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,90		LXI	H, 9000	Initialise memory pointer to point the first data location 9000.
8003	3E		MVI	A, 00	Clear accumulator
8004	00
8005	86		ADD	A, M	The first number is added to accumulator [A] = [A] + M
8006	23		INX	H	Increment the memory pointer to next location of the Data.
8007	86		ADD	A, M	The 2nd number is added to contents of accumulator
8008	32,02,90		STA	9002	The contents of accumulator are stored in memory 8009 02 location 9002.
800B	76		HLT		Stop the execution

Ex: Input: Ex: (i) 9000 – 29 H & 9001 – 16 H Ex :(ii) 9000 –49 H   & 9001 –32 H

Result: Ex: (i) 9002 – 3F H Ex :( ii) 9002 – 7B

Example 2: Addition of two 8-bit numbers whose sum is 16 bits.

Explanation: The first 8-bit number is stored in one memory location (say 8500) and the second 8-bit number is stored in the next location (8501).Add these two numbers and check for carry. Store the LSB of the sum in one memory location (8502) and the MSB (carry) in the other location (8503).

ALGORITHM:

Step1. : Initialize H-L pair with memory address X (say: 8500).

Step2. : Clear accumulator.

Step3. : Add contents of memory location M to accumulator.

Step4. : Increment memory pointer (i.e. 8501).

Step5. : Add the contents of memory indicated by memory pointer to accumulator.

Step6. : Check for Carry

Step 7: Store the sum in 8502.

Step 8: Store the Carry in 8503 location

Step 9: Halt

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LXI	H, 8500	Initialise memory pointer to point the first data location 8500.
8003	3E		MVI	A, 00	Clear accumulator
8004	00
8005	86		ADD	A, M	The first number is added to accumulator [A] = [A] + M
8006	0E		MVI	C, 00	Initial value of Carry is 0
8007	00
8008	23		INX	H	Increment the memory pointer to next location of the Data.
8009	86		ADD	A, M	The 2nd number is added to contents of accumulator
800A	32,0E,80		JNC	FWD	Is Carry exists? No, go to the label FWD
800D	0C		INR	C	Make carry =1
800E	32,02,85	FWD	STA	8502H	The sum is stored in memory location 8502.
8011	79		MOV	A, C	A=C
8012	32,03,85		STA	8503 H	Store the carry at 8503 location
8015	76		HLT		Stop the execution

Ex: Input: 8500 – 97 H & 8501 – 98H                      

RESULT: 8502 – 32 H & 8503 - 01 H

Example 3: Decimal addition of two 8-bit numbers whose sum is 16 bits.

Explanation: Decimal addition of two 8-bit numbers is same as that of two 8-bit numbers program. Except that the use of DAA instruction. The first 8-bit number is stored in one memory location (say 8500) and the second 8-bit number is stored in the next location(8501).Add these two numbers and use the DAA instruction to get the result in decimal. Also check for carry. Store the LSB of the sum in one memory location (8502) and the MSB (carry) in the other location (8503).

ALGORITHM:

Step1. : Initialize H-L pair with memory address XXXX (say: 8500).

Step2. : Clear Carry register C.

Step3. : Move contents of memory location M to accumulator.

Step4. : Increment memory pointer (i.e. 8501).

Step5. : Add the contents of memory indicated by memory pointer to accumulator.

Step6. : Apply the instruction DAA (Decimal adjust after addition)

Step7: Check for Carry

Step8: Store the sum in XX02.

Step9: Store the Carry in XX03 location

Step10: Halt

PROGRAM

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LXI	H, 8500	Initialise memory pointer to point the first data location 8500.
8003	3E		MVI	C, 00	Clear Carry register C.
8004	00
8005	7E		MOV	A, M	Move contents of memory location M to accumulator.
8006	23		INX	H	Increment memory pointer (i.e. 8501).
8007	86		ADD	A, M	The first number is added to accumulator [A] = [A]+M
8008	27		DAA		Apply the instruction DAA(Decimal adjust after addition)
8009	D2,OD,80		JNC	FWD	Is Carry exists? No, go to the label FWD
800C	OC		INR	C	Make carry =1
800D	32,02,85	FWD	STA	8502 H	The contents of accumulator are stored in memory location 8502.
8010	79		MOV	A,C	Carry is moved to accumulator
8011	32,03,85		STA	8503 H	A Carry is stored in the location 8503
8014	76		HLT		Stop the execution

Ex: Input: 8500 – 67 D & 8501 – 85 D                        

RESULT: 8502 – 52 D & 8503 – 01 (Carry)

Example 4: Addition of two 16-bit numbers whose sum is 16 bits or more

Explanation: First 16-bit number is stored in two consecutive locations (Ex 8500 &8501) because in each location we can store only one 8-bit number. Store the second 16-bit number in the next two consecutive locations (For Ex: 8502 &8503).Add the LSB of the first number to the LSB of the second number and the MSB of the first number to the MSB of the second number using the DAD instruction. Store the sum in the next two locations and the carry (if any) in the third location

ALGORITHM:

Step1: First 16 bit number is in locations 8500 & 8501 respectively

Step2: Second 16-bit number is in locations 8502 & 8503

Step3: Add the two 16-bit numbers using DAD Instruction.

Step4: Sum is stored in locations 8504 & 8505.

Step5: Carry (if any) is stored in the location 8506.

Step6: Halt

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	2A,00,85		LHLD	8500 H	First 16-bit number in H-L pair
8003	EB		XCHB		Exchange first number to D-E Pair
8004	2A,02,85		LHLD	8502
8007	0E,00		MVI	00	MSB of the sum is initially 00
8009	19		DAD	D	Add two 16 –bit numbers
800A	D2,0E,80		JNC	FWD	FWD Is Carry? If yes go to the next line .Else go to the 800E Location
800D	OC		INR	C	Increment carry
800E	22,04,85	FWD	SHLD	8504 H	Store the LSB of the Sum in 8504 & MSB in 8505 locations
8011	79		MOV	A,C	MSBs of the sum is in Accumulator
8012	32,06,85		STA	8506 H	Store the MSB (Carry) of the result in 8506 location
8015	76		HLT		Stop the execution

Ex: INPUT:

·         8500- 12 H LSB of the 1^st Number

·         8501- 13 H MSB of the 1^st Number

·         8502 -13 H LSB of the 2^nd Number

·         8503 -12H MSB of the 2^nd number

RESULT: 8504 - 25H LSB of the Sum8505 – 25H MSB of the Sum 8506 -- 00 Carry.

Subtraction Programs:

Example 5: Subtraction of two 8-bit numbers without borrows.

Explanation: It’s a simple program similar to addition of two 8- bit numbers, except that we use the instruction SUB instead of ADD. The first 8-bit number is stored in XX00 memory location and the second 8-bit number is stored in the XX01 location .Use the SUB instruction and store the result in the XX02 location.

ALGORITHM:

Step1. : Initialise H-L pair with the address of minuend.

Step2. : Move the minuend into accumulator

Step3. : Increment H-L pair

Step4. : Subtract the subtrahend in memory location M from the minuend.

Step5. : Store the result in XX02.

Step6. : Stop the execution

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LXI	H, 8500	Initialise H-L pair and get the First number in to 8500 location
8003	7E		MOV	A,M	[A] = [M]
8004	23		INX	H	Increment the memory pointer to next location of the Data.
8005	96		SUB	M	The first number is Subtract  to accumulator [A] = [A] - M
8006	23		INX	H	Increment the memory pointer to next location of the Data.
8007	77		MOV	M,A	Store the result in the location 8502
8008	76		HLT		Stop the execution

INPUT: Ex: 8500- 59H & 8501- 30H

Result: 8502 – 29H

Example 6: Subtraction of two 8-bit Decimal numbers.

Explanation: In this program we can’t use the DAA instruction after SUB or SBB instruction because it is decimal adjust after addition only. So, for decimal subtraction the number which is to be subtracted is converted to 10’s complement and then DAA is applied.

ALGORITHM:

Step1. : Initialise H-L pair with the address of second number (XX01).

Step2. : Find its ten’s complement

Step3. : Decrement the H-L pair for the first number (XX00)

Step4. : Add the first number to the 10’s complement of second number.

Step5. : Store the result in XX02.

Step6. : Stop the execution

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LXI	H, 8500	Initialise H-L pair and get the Second number in to 8500 location
8003	3E		MVI	A, 99	[A] = 99
8005	96		SUB	M	9’s complement of second number
8006	3C		INR	A	10’s complement of second number
8007	2B		DCX	H	Increment the memory pointer to next location of the Data.
8008	86		ADD	M	Add first number to 10’s complement of second number
8009	27		DAA
800A	32,02,85		STA	8502	Store the result in the location 8502
800D	76		HLT		Stop the execution

Ex: Input: 8500 -76 D & 8501- 35 D

Result: 8502 - 41 D

Example 6: Subtraction of two 16 –bit numbers.

Explanation: It is very similar to the addition of two 16-bit numbers. Here we use SUB &SBB instructions to get the result .The first 16-bit number is stored in two consecutive locations and the second 16-bit number is stored in the next two consecutive locations. The lsbs are subtracted using SUB instruction and the MSBs a are subtracted using SBB instruction. The result is stored in different locations.

ALGORITHM:

Step1. : Store the first number in the locations 8500 & 8501.

Step2. : Store the second number in the locations 8502 &8503.

Step4. : Subtract the second number from the first number with borrow.

Step5. : Store the result in locations 8504 & 8505.

Step6. : Store the borrow in location 8506

Step 7: Stop the execution

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	2A,00,85		LHL	D,8500H	First 16-bit number in H-L pair
8003	EB		XCHG		Exchange first number to D-E Pair
8004	2A,02,85		LHLD	8502 H	Get the second 16-bit number in H-L pair
8007	7B		MOV	A,E	Get the lower byte of the First number in to Accumulator
8008	95		SUB	L	Subtract the lower byte of the second number
8009	6F		MOV	L,A	Store the result in L- register
800A			MOV	A,D	Get higher byte of the first number
800A	9C		SBB	H	Subtract higher byte of second number with borrow
800B	67		MOV	H,A
800C	22,04,85		SHLD	8504	Store the result in memory locations with LSB in 8504 & MSB in 8505
800F	76		HLT		Stop the execution

Ex: INPUT:

·         8500- FF H LSB of the 1^st  Number

·         8501 - FF H MSB of the 1^st  Number

·         8502 -EE H LSB of the 2^nd  Number

·         8503 –EE H MSB of the 2^nd  number

RESULT: 8504 - 11H LSB & 8505 – 11 H MSB

Multiplication Programs

Example 7: Multiplication of two 8-bit numbers. Product is 16-bits.

Explanation: The multiplication of two binary numbers is done by successive addition. When multiplicand is multiplied by 1 the product is equal to the multiplicand, but when it is multiplied by zero, the product is zero. So, each bit of the multiplier is taken one by one and checked whether it is 1 or 0 .If the bit of the multiplier is 1 the multiplicand is added to the product and the product is shifted to left by one bit. If the bit of the multiplier is 0, the product is simply shifted left by one bit. This process is done for all the 8-bits of the multiplier.

ALGORITHM:

Step 1: Initialise H-L pair with the address of multiplicand.(say 8500)

Step 2: Exchange the H-L pair by D-E pair. so that multiplicand is in D-E pair.

Step 3: Load the multiplier in Accumulator.

Step 4: Shift the multiplier left by one bit.

Step 5: If there is carry add multiplicand to product.

Step 6: Decrement the count.

Step 7: If count ¹ 0; Go to step 4

Step 8: Store the product i.e. result in memory location.

Step 9: Stop the execution

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	2A,00,85		LHLD	H,8500	Load the multiplicand in to H-L pair
8003	EB		XCHG		Exchange the multiplicand in to D-E pair
8004	3A,02,85		LDA	8502	Multiplier in Accumulator
8007	21,00,00		LXI	H,0000	Initial value in H-L pair is 00
800A	0E,08		MVI	C,08	Count =08.
800C	29	LOOP	DAD	H	Shift the partial product left by one bit
800D	17		RAL		Rotate multiplier left by one bit
800E	D2,12,80		JNC	FWD	Is Multiplier bit =1? No go to label FWD
8011	19		DAD	D	Product =Product +Multiplicand
8012	0D	FWD	DCR	C	COUNT=COUNT-1
8013	C2		JNZ	LOOP
8016	22,03,85		SHLD	8503	Store the result in the locations 8503 & 8504
8019	76		HLT		Stop the execution

INPUT:

·         8500 8AH – LSB of Multiplicand

·         8501 00 H – MSB of Multiplicand

·         8502 52 H - Multiplier

Result: 8503 34 H – LSB of Product & 8504 2C H – MSB of Product

Division Programs

Example 7: Division of a 16- bit number by a 8-bit number.

Explanation: The division of a 16/8-bit number by a 8-bit number follows the successive subtraction method. The divisor is subtracted from the MSBs of the dividend .If a borrow occurs, the bit of the quotient is set to 1 else 0.For correct subtraction process the dividend is shifted left by one bit before each subtraction. The dividend and quotient are in a pair of register H-L. The vacancy raised due to shifting is occupied by the quotient .In the present example the dividend is a 16-bit number and the divisor is an 8-bit number. The dividend is in locations 8500 &8501.Similarly the divisor is in the location 8502.The quotient is stored at 8503 and the remainder is stored at 8504 locations.

ALGORTHM:

STEP1. : Initialise H-L pair with address of dividend.

STEP2. : Get the divisor from 8502 to register A & then to Reg. B

STEP3. : Make count C=08

STEP4. : Shift dividend and divisor left by one bit

STEP 5: Subtract divisor from dividend.

STEP6. : If carry = 1: go to step 8 else step7.

STEP7. : Increment quotient register.

STEP8. : Decrement count in C

STEP9. : If count not equal to zero go to step 4

STEP10: Store the quotient in 8503

STEP11: Store the remainder in 8504

STEP12: Stop execution.

PROGRAM:

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LHLD	H, 8500	Initialize the H-L pair for dividend
8003	3A,02,85		LDA	8502 H	Load the divisor from location 8502 to accumulator
8006	47		MOV	B,A	Move Divisor to Reg. B from A
8007	0E,08		MVI	C,08	Count =08
8009	29	BACK	DAD	H	Shift dividend and quotient eft by one bit Ex: Input & Result
800A	7C		MOV	A,H	MSB of dividend in to accumulator
800B	90		SUB	B	Subtract divisor from MSB bits of divisor
800C	DA,11,80		JC	FWD	Is MSB part of dividend > divisor? No, go to label FWD
800F	67		MOV	H,A	MSB of the dividend in Reg. H
8010	2C		INR	L	Increment quotient
8011	0D	FWD	DCR	C	Decrement count
8012	C2,09,80		JNZ	BACK	If count is not zero jump to8009 location
8015	22,03,85		SHLD	8503 H	Store quotient in 8503 and remainder in 8504 locations
8018	76		HLT		Stop the execution

Ex: Input & Result

·         8500 64 _ LSB of Dividend

·         8501 00 _ MSB of Dividend

·         8502 07 _ Divisor

·         8503 0E _ Quotient

·         8504 02 _ Remainder

Largest & Smallest numbers in an Array

Example 8: To find the largest number in a data array

Explanation: To find the largest number in a data array of N numbers (say)first the count is placed in memory location (8500H) and the data are stored in consecutive locations.(8501….onwards).The first number is copied to Accumulator and it is compared with the second number in the memory location. The larger of the two is stored in Accumulator. Now the third number in the memory location is again compared with the accumulator. And the largest number is kept in the accumulator. Using the count, this process is completed, until all the numbers are compared .Finally the accumulator stores the smallest number and this number is stored in the memory location.85XX.

ALGORTHM:

Step1: Store the count in the Memory location pointed by H-L register.

Step2: Move the I st number of the data array in to accumulator

Step3: Compare this with the second number in Memory location.

Step4: The larger in the two is placed in Accumulator

Step5: The number in Accumulator is compared with the next number in memory .

Step 6: The larger number is stored in Accumulator.

Step 7; The process is repeated until the count is zero.

Step 8: Final result is stored in memory location.

Step 9: Stop the execution

PROGRAM

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LXI	H, 8500	Initialise H-L PAIR
8003	7E		MOV	C,M	Count in the C register
8004	23		INX	H	First number in H-L pair
8005	4E		MOV	A, M	Move first number in to Accumulator
8006	0D		DCR	C	Decrement the count
8007	91	LOOP1	INX	H	Get the next number
8008	BE		CMP	M	Compare the next number with previous number
8009	D2,0D,80		JNC	LOOP2	Is next number > previous maximum? No, go to the loop2
800C	7E		MOV	A,M	If, yes move the large number in to Accumulator
800D	OD	LOOP2	DCR	C	Decrement the count
800E	C2,07,80		JNZ	LOOP1	If count not equal to zero, repeat
8012	78
8013	32,XX,85		STA	85XX	Store the largest number in the location 85XX
8016	76		HLT		Stop the execution

Ex: Input:

·         8500- N(Say N=7 )

·         8501-05

·         8502-0A

·         8503-08

·         8504-14

·         8505 -7F

·         8506-25

·         8507-2D

Result: 8508 - 7F

Example 9: To find the smallest number in a data array.

Explanation: To find the smallest number in a data array of N numbers (say)first the count is placed in memory location (8500H) and the data are stored in consecutive locations.(8501….onwards).The first number is copied to Accumulator and it is compared with the second number in the memory location. The smaller of the two is stored in Accumulator. Now the third number in the memory location is again compared with the accumulator. and the smallest number is kept in the accumulator. Using the count, this process is completed until all the numbers are compared .Finally the accumulator stores the smallest number and this number is stored in the memory location.85XX.

ALGORTHM:

Step1: Store the count in the Memory location pointed by H-L register.

Step2: Move the 1st number of the data array in to accumulator

Step3: Compare this with the second number in Memory location.

Step4: The smaller in the two is placed in Accumulator

Step5: The number in Accumulator is compared with the next number in memory .

Step 6: The smaller number is stored in Accumulator.

Step 7; The process is repeated until the count is zero.

Step 8: Final result is stored in memory location.

Step 9: Stop the execution

PROGRAM

Address of the memory location	Hex code	Label	Mnemonics		Comments
			Op-code	Operand
8000	21,00,85		LXI	H, 8500	Initialise the H-L pair.
8003	7E		MOV	C,M	Count in the C register
8004	23		INX	H	First number in H-L pair
8005	4E		MOV	A, M	Move first number in to Accumulator
8006	0D		DCR	C	Decrement the count
8007	91	LOOP1	INX	H	Get the next number
8008	BE		CMP	M	Compare the next number with previous number
8009	D2,0D,80		JC	LOOP2	Is next number <previous smallest ?If yes go to the loop2
800C	7E		MOV	A,M	No, move the smaller number in to Accumulator
800D	0D	LOOP2	DCR	C	Decrement the count
800E	C2,07,80		JNZ	LOOP1	If count not equal to zero, repeat
8012	78
8013	32,XX,85		STA	85XX	Store the smallest number in the location 85XX
8016	76		HLT		Stop the execution

Ex: Input:

·         8500 - N((Say N=7)

·         8501-09

·         8502-0A

·         8503-08

·         8504-14

·         8505 -7F

·         8506-04

·         8507-2D

Result: 8508 – 04

REFERENCES

1.      R. S. Gaonkar, Microprocessor Architecture, Programming, and Applications with the 8085, Fifth Edition, Penram International Publishing (India) Private Limited.

2.      S Ghoshal, Microprocessor Based System Design, Macmillan India Limited, 1996

3.      M. Mano, Digital Logic and Computer Design, Prentice – Hall India

4.      B. Ram - Fundamentals of Microprocessor and Microcontrollers

5.      “Microprocessors: Principles and Applications” by A Pal

6.      “Microprocessors and Microcontrollers : Architecture, Programming and Interfacing Using 8085, 8086 and 8051” by Soumitra Kumar Mandal

7.      “Introduction to Microprocessors and Microcontrollers” by Crisp John Crisp

8.      “Microprocessors And Microcontrollers” by A Nagoor Kani

9.      “Microprocessors And Microcontrollers : Architecture, Programming and System Design 8085, 8086, 8051, 8096” by KRISHNA KANT

10.  “8 - Bit Microprocessor” by Vibhute