8085 program to count the number of ones in contents of register B

In this program, we will see how to count the number of 1’s in an 8-bit number stored in register B.

Problem Statement

Write 8085 Assembly language program to count the number of 1s in 8-bit number stored in register B.

Discussion

In this program, we are using the rotate operation to count the number of 1’s. As there are 8 different bits in 8-bit number, then we are rotating the number eight times. we can use RRC or RLC. Here we have used the RRC instruction. This instruction sends the LSb to MSb also to carry flag. So after each iteration, we can check the carry status to get the count of 1s.

If the number is DA (1101 1010) then the answer will be 5, as there are five 1s in the number.

Input

Register	Data
B	DA

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	06, DA		MVI B, DA	Take number 1101 1010
F002	37		STC	Set Carry
F003	3F		CMC	Complement Carry
F004	78		MOV A,B	Load the number into A from B
F005	0E, 08		MVI C,08H	Initialize the counter to 08H
F007	06, 00		MVI B,00H	Clear B register
F009	0F	LOOP	RRC	Rotate right
F00A	D2, 0E, F0		JNC SKIP	If CY = 0, skip the next step
F00D	04		INR B	If CY = 1, increase B
F00E	0D	SKIP	DCR C	Decrease C by 1
F01F	C2, 09, F0		JNZ LOOP	If Z = 0, jump to loop
F012	78		MOV A,B	Load the count to A
F013	32, 50, 80		STA 8050H	Store the result at 8050H
F016	76		HLT	Terminate the program