8085 program to find the sum of first n natural numbers
Problem – Write an assembly language program for calculating the sum of first n natural numbers using 8085 microprocessor.
Input : 04H Output : 0AH as 01+02+03+04 = 10 in decimal => 0AH
The formula for calculating the sum of first n natural numbers is .
- With n as the input, increment it to obtain n+1.
- Multiply n with n+1.
- Divide the product obtained by 2.
In 8085 microprocessor, no direct instruction exists to multiply two numbers, so multiplication is done by repeated addition as 4×5 is equivalent to 4+4+4+4+4 (i.e., 5 times).
Add 04H 5 times
Similarly, in 8085 microprocessor, no direct instruction exists to divide two numbers, so division is done by repeated subtraction.
Keep subtracting 2 from the input till it reduces to 0.
Since subtraction has to be performed 1010 times before 14H becomes 0, the quotient is 1010 => 0AH.
- Load the data from the memory location (201BH, arbitrary choice) into the accumulator
- Move this data into B
- Increment the value in the accumulator by one and move it to the register C
- Initialise the accumulator with 0
- Multiplication: Keep adding B to accumulator. The number of times B has to be added is equal to the value of C
- Initialise B with 00H. B will store the quotient of the division
- Initialise C with 02H. This is the divisor for the division
- Division: Keep subtracting C from A till A becomes 0. For each subtraction, increment B by one
- The final answer is in B. Move it to A. Then store the value of A in 201CH (arbitrary choice again)
201CH contains the final answer.
|2003H||MOV B, A|
|2005H||MOV C, A|
|2006H||MVI A, 00H|
|200DH||MVI C, 02H|
|200FH||MVI B, 00H|
|2016H||MOV A, B|
Store the value of n in 201BH. The sum can be found at 201CH.