8085 program to divide two 8 bit numbers

Problem – Write 8085 program to divide two 8 bit numbers. 

Example –

Algorithm –

1. Start the program by loading the HL pair registers with address of memory location. 
2. Move the data to B Register. 
3. Load the second data into accumulator. 
4. Compare the two numbers to check carry. 
5. Subtract two numbers. 
6. Increment the value of carry. 
7. Check whether the repeated subtraction is over. 
8. Then store the results(quotient and remainder) in given memory location. 
9. Terminate the program. 
   

Program –

ADDRESS	MNEMONICS	COMMENT
2000	LXI H, 2050
2003	MOV B, M	B<-M
2004	MVI C, 00	C<-00H
2006	INX H
2007	MOV A, M	A<-M
2008	CMP B
2009	JC 2011	check for carry
200C	SUB B	A<-A-B
200D	INR C	C<-C+1
200E	JMP 2008
2011	STA 3050	3050<-A
2014	MOV A, C	A<-C
2015	STA 3051	3051<-A
2018	HLT	terminate the program

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

1. LXI H, 2050 will load the HL pair register with the address 2050 of memory location. 
   
2. MOV B, M copies the content of memory into register B. 
   
3. MVI C, 00 assign 00 to C. 
   
4. INX H increment register pair HL. 
   
5. MOV A, M copies the content of memory into accumulator. 
   
6. CMP B compares the content of accumulator and register B. 
   
7. JC 2011 jump to address 2011 if carry flag is set. 
   
8. SUB B subtract the content of accumulator with register B and store the result in accumulator. 
   
9. INR C increment the register C. 
   
10. JMP 2008 control will shift to memory address 2008. 
    
11. STA 3050 stores the remainder at memory location 3050. 
    
12. MOV A, C copies the content of register into accumulator. 
    
13. STA 3051 stores the quotient at memory location 3051. 
    
14. HLT stops executing the program and halts any further execution. 

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