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

Problem: Write an assembly language program to count the number of ones in the contents of register B and store the result at memory location 3050. 

Example:

Algorithm:

• Convert the decimal number in Accumulator to its binary equivalent.
• Rotate the digits of the binary number right without carry.
• Apply a loop till the count is not zero to change the values of the D register and count.
• Copy the value of the D register to the accumulator and store the result.

Program:

MEMORY ADDRESS	MNEMONICS	COMMENTS
2000	MVI B 75	B ← 75
2002	MVI C 08	C ← 75
2004	MVI D 00	D ← 00
2006	MOV A, B	A ← B
2007	RRC	Rotate right without carry
2008	JNC 200C	Jump if Not Carry
200B	INR D	D ← D+1
200C	DCR C	C ← C-1
200D	JNZ 2007	Jump if Not Zero
2010	MOV A, D	A ← D
2011	STA 3050	A → 3050
2014	HLT	Stops execution

Explanation:

1. MVI B 75 moves 75 decimal numbers into the B register
2. MVI C 08 moves 08 decimal number into C register, which is taken as counter as the number is of 8 bytes
3. MVI D 00 moves 00 number into d register
4. MOV A, B moves the contents of B register into A (accumulator) register
5. RRC rotates the contents of A (which is 75 with binary equivalent 01110101) right
6. JNC 200C jumps to 200C address and performs the instructions written there if the carry flag is not zero
7. INR D increases the value of the D register by adding one to its contents
8. DCR C decreases the value of the C register by subtracting one from its contents
9. JNZ 2007 jumps to the 2007 address and performs the instructions written there if the zero flags are not zero
10. MOV A, D moves the contents of the B register into the A register
11. STA 3050 store the contents of A at 3050 memory location
12. HLT stops execution

Approach : 2 [Using Rotate Instruction RLC] :

Here algorithm will be the same as above but here we are moving contents of accumulator bits to 1-bit left and then checking carry flag values and updating the count register accordingly.

If Carry Flag = 1 , then Count = Count + 1

Otherwise: Rotate Accumulator Left again without carrying and repeat the above procedure.

Both Different Instructions do the same task and concept of achieving end-goal is one and the same. Only Rotating Style is Different but these 2 methods can do the same task for us.

MEMORY ADDRESS	MNEMONICS	COMMENTS
2000	MVI B,75H	75H is the value that is to be tested for the Number of 1’s
2002	MVI C,08H	C is storing 8 because of the 8-bit data which we are using and we need to rotate the given number 8 times hence for loop [C]=08
2004	MVI D,00H	This is a counter variable that will increment when 1 will be encountered
2006	MOV A, B	Move 75H into Accumulator for rotating purposes because RLC will work only on Accumulator.
2007	RLC	Rotate Accumulator Left Without Carry
2008	JNC 200C	If Carry Flag = 0 then just decrement counter and again start Rotating Accumulator using RLC.
200B	INR D	If Carry Flag = 1 , then we have to increment Counter Register D =D + 1
200C	DCR C	Decrement value of C register because we have to rotate data 8 times because of 8 bits. So limit of rotation = 8
200D	JNZ 2007	While [C] =0 this looping structure will execute from location 2007 and keep rotating accumulator contents until [C]=0
2010	MOV A, D	Move Contents of register D which contains a total number of 1’s in given data into Accumulator.
2011	STA 3050	Store Accumulator Contents into memory location 3050H as asked in the question.
2014	HLT	Stop execution of the given Program.

Advantages of counting ones using this program:

• It is a simple and efficient way to count the number of ones in the contents of register B using only a few instructions.
  
• It is a low-level operation that provides a deep understanding of the internal workings of the 8085 microprocessor.
  
• It can be easily adapted for use in other programs and applications that require counting the number of ones in binary data.
  

Disadvantages of counting ones using this program:

• It requires specialized knowledge of assembly language programming and the 8085 microprocessor architecture.
  
• It may not be efficient for larger sets of data, as it involves looping through each bit of the register and checking for a carry.
  
• It may not be practical for use in high-level programming languages or applications where speed and efficiency are critical.

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