8085 program to sum of two 8 bit numbers without carry
Problem – Write an assembly language program to sum two 8 bit numbers without using carry operation in 8085 microprocessor.
Assumption:
- The starting address of the program is 2000.
- Memory address of the first number is 2050.
- Memory address of the second number is 2051.
- Memory address of result is 2052.
Example:
Input: 2050: 03
: 2-51: 04
Output: 2052: 07
Algorithm:
- Load the first number to the accumulator through memory address 2050.
- Move the content of accumulator to the register B.
- Load the second number to the accumulator through memory address 2051.
- Add the content of accumulator and register B and result will be stored at the accumulator.
- Store the result from the accumulator to the memory address 2052.
- Terminate the program.
Program:
Memory Address |
MNEMONICS |
Comment |
2000 |
LDA 2050 |
A<-[2050] |
2003 |
MOV B, A |
B<-A |
2004 |
LDA 2051 |
A<-[2051] |
2007 |
ADD B |
A<-A+B |
2008 |
STA 2052 |
[2052]<-A |
200B |
HLT |
Terminate |
Explanation:
- LDA 2050: This instruction will load the number from memory to the accumulator.
- MOV B, A: This instruction will move the content of accumulator to the register B.
- LDA 2051: This instruction will load the number from memory to the accumulator.
- ADD B: This instruction will sum the content of the accumulator with the content of the register B.
- STA 2052: This instruction will store the content of accumulator to the memory address 2052.
- HLT: This instruction will terminate the program.
Hence we successfully sum the two 8 bit numbers without carry using 8085 microprocessor.
Last Updated :
30 Jan, 2019
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