# 8085 Program to check for two out of five code

Now let us see a program of Intel 8085 Microprocessor. This program will help us to check a given value is a valid 2 out of 5 code or not.

Problem Statement:

Write 8085 Assembly language program to check whether a given number is 2 out of five code or not. The number is stored at location 8000H.

Discussion:

The checking of two out of five code is simple. At first we have to check the higher order three bits are 0 or not. If they are 0, then we will check next five bits. If there are exactly two 1s in these 5-bits, then it is a valid 2 out of 5 code.

Here at first we are ANDing the number by (1110 0000), if the first three bits are 0, then the result will be 0, after that we are checking the number of 1s by using rotating operation. Using rotation operation if the carry flag is enabled, then the count will be increased. Thus the total count of 1s will be reflected.

This program will store FFH into location 8050H if the number is valid 2 out of 5 code, otherwise it will store 00H into 8050H.

Input:

First Input

Data
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8000
12
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.

Second Input

Data
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8000
13
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.
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.

Third Input

Data
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8000
03
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.

Flow Diagram:

Program:

HEX Codes
Labels
Mnemonics
F000
3A, 00, 80

LDA 8000H
F003
67

MOV H,A
F004
2E, 00

MVI L,00H
Clear register L
F006
E6, E0

ANI E0H
AND Acc and 11100000b
F008
C2, 21, F0

JNZ DONE
If Z = 0, go to Done
F00B
7C

MOV A,H
F00C
0E, 05

MVI C,05H
Load C with 05H for counting
F00E
16, 00

MVI D,00H
Clear register D
F010
0F
LOOP
RRC
Rotate acc content to the right
F011
D2, 15, F0

JNC SKIP
F014
14

INR D
Increase D by 1
F015
0D
SKIP
DCR C
Decrease C by 1
F016
10, F0

JNZ LOOP
F019
3E, 02

MVI A,02H
Initialize A with 02H
F01B
BA

CMP D
Compare D with A
F01C
C2, 21, F0

JNZ DONE
if Z = 0, go to DONE
F01F
2E, FF

MVI L,FFH
F021
7D
DONE
MOV A,L
Take result from L to A
F022
32, 50, 80

STA 8050H
Sore result at 8050H
F025
76

HLT
Terminate the program

Output:

First Output

Data
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8050
FF
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Second Output

Data
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8050
00
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Third Output

Data
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8050
FF
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