Presentation on microprocessor programming

8/13/2019 Presentation on microprocessor programming

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UNIVERSITY OF CALCUTTA.DEPARTMENT OF APPLIED PHYSICS.

SAIBAL KUMAR MITRA.INSTRUMENTATION ENGINEERING.

ROLL NO: 24.


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Some unsigned single byte numbers are storedat consecutive memory locations starting fromX. Write a programme to add those numbers

taking into account of any possible overflow ateach step. Store the result as a double precisionnumber at two consecutive memory locations Y

and Y+1. Number of data to be added is givenat the memory location X-1 .


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Step 1. START.Step 2. Initialize H-L pair as pointer to memory location X-1.Step 3. Move data from memory location X-1 , to register C, which will be used asa counter.Step 4. Clear register D, and register E.Step 5. Increment the pointer.Step 6. Move data from memory location currently pointed to by H-L pair, toaccumulator.Step 6. Add content of register E to that of accumulator.Step 7. Move the result to register E.Step 8. Clear the accumulator.Step 9. Add the content of register D, and accumulator A with carry.Step 10. Move data from accumulator to register D.Step 11. Decrement the count in register C.Step 12. Jump to step 5 if the count is not zero.Step 13. Exchange contents of D-E pair with H-L pair.Step 14. Store contents of H-L pair to memory location Y and Y+1.Step 15. STOP.


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Address Mnemonic Opcode/ data Comment

8000 LXI H 84FF 21 H-L pair initializedto 1 st address 84FF.

8001 FF

8002 84

8003 MOV C, M 4E C is loaded with thecontent of memory

location.

8004 MVI D, 00 16 D is made empty by loading 00

8005 00

8006 MOV E, D 5A E is also madeempty by loading

00

8007 INX H 23 Content of H-L pairis incremented by

1.


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8008 MOV A, M 7E Content of thememory location

currently pointed to

by H-L pair, istransferred toaccumulator.

8009 ADD E 83 Content of E isadded to that of aand the result is

stored in A.

800A MOV E, A 5F Content of A istransferred back toE. E contains theleast significcant

byte of the double precision number.

800B MVI A, 00 3E A is made empty byloading 00

800C 00


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address Mnemonic Opcode/ data Comment

800D ADC D 8A Content of D andthe carry bit of flag

register(from theoperation ADD E)

are added to thecontent of A, and

the result stored inA.

800E MOV D, A 57 Now the content ofA is transferred

back to D.

800F DCR C 0D Counter isdecremented by 1after each addition

process


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8010 JNZ 8007 C2 Jump to memorylocation 8007 if

Z=08011 07

8012 808013 XCHG EB Content of D-E pair

and H-L pair areexchanged.

Hcontains the

higher byte, while Lcontains the lower byte.

8014 SHLD 9500 22 Content of L goesto the address given

in the instructionwhile content of H

goes to the nextaddress.

8015 00

8016 95

8017 HLT 76 End of programme.


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Memory location Data

84FF 05

8500 F6

8501 E88502 AD

8503 C9

8504 BC

OUTPUT


9500 10

9501 04


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Ten 8-bit numbers are stored starting from

memory location 3100 H. Find thegreatest of the ten numbers and store it atmemory location 3500 H.


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1. START2. Store 0A in the register D.3. Point the HL pair to the starting

address of the data block.4. Move the content of memory

location to accumulator.5. Decrement register D.6. Check if the D register is 0. if

yes go to step 10.7. Increment pointer.8. Compare content of memory

location with accumulator.9. Check for CY Flag. If CY=1 go

to step 4, else go to 5.10.Store accumulator at the

memory location 9050H.

11.STOP


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8000 MVI D, 0A 16 D register is loadedwith 0A.

8001 0A

8002 LXI H 9000 21 H-L pair initializedto 1 st address 9000.

8003 00

8004 90

8005 MOV A, M 7E Content of thememory location

currently pointed to by H-L pair, is

transferred toaccumulator.


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8006 DCR D 15 Counter is

decremented by 1after each addition process

8007 JZ 8012 CA Jump to memorylocation 8012 if Z=1

8008 12

8009 80

800A INX H 23 Content of H-L pair is

incremented by 1.

800B CMP M Content of thememory location is

compared with that ofthe accumulator.


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800C JC 8005 Jump to memorylocation 8005 if

CY=1800D 05

800E 80

800F JMP 8006 Jumpunconditionally to

address 80068010 06

8011 80

8012 STA 9050H 32 Store accumulator

in the memorylocation 90508013 50

8014 90

8015 HLT 76 End of programme.


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9000 109001 20

9002 30

9003 40

9004 509005 60

9006 70

9007 80

9008 90

9009 A0

INPUT


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Memory location Data9050 A0

OUTPUT


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Write a programme that reads the gray code bits

from a memory location, find the binary codecorresponding to the gray code and store it inanother location.


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The following algorithm converts an 8 bit graycode g 7, g7, g1, g0 into an 8 bit binary number

b7, b6 , b 1, b0 b7= g 7 b i = g i gi+1 0 i 7


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Step 1. START.Step 2. load the accumulator with the data taken from a

specified locationStep 3. copy that data to register B from Accumulator.Step 3. Do AND operation between 80 and the content ofaccumulator.

Step 4. copy data from accumulator to register B.Step 5. rotate content of accumulator to right.Step 6. do XOR operation between content of accumulatorand that of register B.Step 7. do OR operation between content of register C andthat of accumulator.Step 8. store result in a specified memory location.Step 9. STOP.


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A000 LDA A200 3A A is loaded withthe content of the

memory locationA200.

A001 00

A002 A2

A003 MOV B, A 47 B Is loaded with

the content of A to preserve the inputdata.

A004 ANI 80 E6 Content of A isANDed with 80.

This is done tocreate a numberwhose msb is same

as that of theoriginal input data,

with other data being 0.

A005 80


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A006 MOV C,A 4F This newlycreated number is

stored in C

A007 MOV A,B 78 Original inputdata is again

taken into B.A008 RRC 0F Input data in A is

roatated right.

A009 XRA B A8 Original input

data is XORedwith the number

created byshifting the inputdata by one place

to the right.


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Address Mnemonic Opcode/ data CommentA00A ORA C B1 The result of XORing

is ORed with thecontent of C, so thatthe msb of the outputdata is same as that of

input data

A00B STA A300 32 Content of a is stored

at memory locationA300 as final output.

A00C 00

A00D A3

A00E HLT 76 End of programme.


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Memory location DataA200 28H

Input


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Memory location DataA300 3CH

OUTPUT


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Here28H = 0010 1000 2

And its converted gray code is

0011 1100

which is equal to 3CH


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THANK YOU.

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Presentation on microprocessor programming