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0s and 1s Count in 1 Byte Memory Word

This program will count all the 1s and 0s individualy presents in a 1 byte word. And store the result in individual places in Memory. Suppose, argument byte is stored in memory location 2050H and store the result of 1's count at 2060H memory location and 0's at 2061H

;Initiate 1's Counter
Memory LocationHex CodeLabelMnemonicsComment
3050
3051
3052		3A
50
20		START:LDA 2050H;Load Accumulator with the number stored at location 2050H
3053
3054		06
00		MVI B, 00H;Initiate 0's Counter
3055
3056		0E
00		MVI C, 00H
3057
3058		16
08		MVI D, 08H;Initiate Bit Counter
305907LOOP:RLC;Rotate Accumulator Left with Carry
305A
305B
305C		DA
61
30		JC ONE;If 1 is found count 1
305D04INR B;If 1 is not found it must be 0. So increase 0's Counter
305E
305F
3060		D2
62
30		JNC NEXT;Go to next Bit
30610CONE:INR C;Increase 1's Counter
306215NEXT:DCR D;Decrease the Bit Counter
3063
3064
3065		C2
59
30		JNZ LOOP;If all Bits is not checked execute for next Bit
306679MOV A, C;Move 1's Count Result to Accumulator
3067
3068
3069		32
60
20		STA 2060H;Store the Result for 1's Count
306A78MOV A, B;Move 0's Count Result to Accumulator
306B
306C
306D		32
61
20		STA 2061H;Store the Result for 0's Count
306E76HLT;End of the Program
Above program can be modified as,
  1. Count 1s from the Byte
  2. Store the Result for 1s count
  3. Deduce this result from 8
  4. Store the Result for 0s count (what is work out by deduction)
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