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8085 program to count total even numbers in series of 10 numbers

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Program – Write an assembly language program in 8085 microprocessor to count even numbers in a series of 10 numbers. 

Example –


Assumption – Ten 8-bit numbers are stored from starting memory location 2050. Value of count is stored at memory location 3050. 

Algorithm –

  1. Initialize register H with 20 and register L with 4F so that indirect memory points to memory location 204F.
  2. Initialize register C with 00 and register D with 0A.
  3. Increment indirect memory by 1.
  4. Store value of M in accumulator A.
  5. Check whether the content in A is even or odd by performing AND operation of A with 01.
  6. If content of A is 00 after AND operation then the number scanned was even, If so then increment C by 01 else if content of A is 01 after AND operation then number scanned was odd, If so then decrements D by 01.
  7. Check if zero flag is not set i.e. ZF = 0 then jump to step 3 otherwise store value of C at memory location 3050.

Program (Using ANI Instruction) –

2000LXI H 204FH <- 20, L <- 4F
2003MVI C, 00C <- 00
2005MVI D, 0AD <- 0A
2007INX HM <- M + 01
2008MOV A, MA <- M
2009ANI 01A <- A (AND) 01
200BJNZ 200FJump if ZF = 0
200EINR CC <- C + 01
200FDCR DD <- D – 01
2010JNZ 2007Jump if ZF = 0
2013MOV A, CA <- C
2014STA 3050M[3050] <- A

Explanation – Registers A, B, C, D, H, L are used for general purpose.

  1. LXI H 204F: assign 20 to H and 4F to L.
  2. MVI C, 00: assign 00 to C.
  3. MVI D, 0A: assign 0A to D.
  4. INX H: increment indirect memory location M by 01.
  5. MOV A, M: move content of M to A.
  6. ANI 01: perform AND operation of A with 01 and store the result in A.
  7. JNZ 200F: jump if ZF = 0 to memory location 200F.
  8. INR C: increment C by 01.
  9. DCR D: decrements D by 01.
  10. JNZ 2007: jump if ZF = 0 to memory location 2007.
  11. MOV A, C: moves the content of C to A.
  12. STA 3050: store the content of A to memory location 3050.
  13. HLT: stops executing the program and halts any further execution.

Alternative Approach (Using Rotate Instructions):

For a given set of numbers, If the Least Significant Bit(LSB) of each number’s Binary representation is 1, then the given number is assured to be an Odd Number.

But if the Least Significant Bit (LSB) of each number’s binary representation is 0, then the given number is assured to be an Even Number.

Hence, here we will use Rotate Instructions of 8085 Microprocessor. i.e. RAR Instruction (Rotate Accumulator Right involving Carry flag in Rotation).

If we rotate the given binary number in the right direction by 1-bit then we will get the carry a flag as Zero or One.

If we get Carry Flag = 1, then the given number is ODD, Otherwise EVEN.



2000LXI H,204FHLoad HL Pair with Previous location of 2050H , H <- 20, L <- 4F
2003MVI C,00HFor storing count of Even Numbers [C]=0H
2005MVI D,0AHFor series of 10 numbers we have stored array size as [D]=10
2007INX HIncrement memory pointer pointed by HL pair
2008MOV A,MMove Content of memory location 2050H into Accumulator A. {[A] <- [M]}
2009RARRotate Accumulator Right with Carry .
200BJC 200FAfter rotating if Carry Flag=1 then we can say that given number is ODD, So we just need to move to next element nothing to do extra here.
200EINR Cif we found that Carry Flag=0 then we can say that given Number is EVEN, So we need to increment our count by 1.Hence INR C will happen for all subsequent conditions.
200FDCR DNow if we have found out whether given number is ODD/EVEN then we need to decrement array size because here size is 10 and we need to decrement till 0.
2010JNZ 2007Until [D]=0 we have to decrement D and move to next memory location and take input of 10 numbers (This is looping Structure.)
2013MOV A,C(After whole looping is over)Transfer/Copy Contents of register C which contains total number of even numbers into Accumulator.
2014STA 3050HStore Accumulator Content which contains count of Even Numbers to Memory Location 3050H as required in Question
2017HLTEnd of a Program Execution 
Last Updated : 30 May, 2022
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