Cd 4002: Fill & Download for Free

In decimal addition, to add a 4 digit number we perform 4 single digit addition and store the result of each single digit addition. If the result is more than 10 then we use the carry for the next addition. We do not perform all 4 digit simultaneously. We can do only one addition at a time. Once we write (store) the result of the addition on paper (memory) then we forget that result and add the next numbers.Example:1 1 -> Carry2 5 7 8 -> Number 16 8 9 5 -> Number 2--------9 4 7 3 -> Total--------How we do the addition. We first read 8 & 5 . Perform the addition. We get result 13. We write 3 and store 1 as carry for next digit addition. Then we add 7, 9 & 1. So we get 17. We do the same thing for the 4 digits.8085 is a 8 bit microprocessor. It can perform only 8 bit operations. So the 32 bit can be divided into 4 8bits.Example: c c c -> carryNumber 1: 11001101 00110001 11101001 11110001Number 2: 10110011 10001100 01101100 01101101First we need to store these numbers in memory. Let us assume we store number 1 in 2000,2001,2002 and 2003, number 2 in 3000,3001,3002,3004. The is stored in 4000,4001,4002,4003 & 4004. The result has 1 byte extra.Let us first manually add the above two binary numbers.Example: 0 1 1 -> carryNumber 1: 11001101 00110001 11101001 11110001Number 2: 10110011 10001100 01101100 01101101-------------------------------------------------Result: 00000001 10000000 10111110 01010110 01011110Now let us write a 8085 Assembly language program to do this 32 bit addition.------------------------------------------------Store the number in Memory-------------------------------------------------Initially memory will contain some random value. In the following figure 'x' represents 0 or 1.Since writing the binary number is tedious, let us use the short form of binary, the hexa decimal.MVI A, 11110001 is written as****** number 1, byte 0, bits 0-7 are stored in memory 2000 ******MVI A, F1 -> in general it is written as 0F1hST A,2000 -> in general it is written as 2000h****** number 1, byte 1, bits 8-15 are stored in memory 2001 ******MVI A, E9ST A,2001****** number 1, byte 2, bits 16-23 are stored in memory 2002 ******MVI A, 31ST A,2002****** number 1, byte 3, bits 24-31 are stored in memory 2003 ******MVI A, CDST A,2003After executing the above instruction the memory will contain values as shown below.Now let us write a program to store the number 2 in memory****** number 2, byte 0, bits 0-7 are stored in memory 3000 ******MVI A, 6DST A,3000****** number 2, byte 1, bits 8-15 are stored in memory 3001 ******MVI A, 6CST A,3001****** number 2, byte 2, bits 16-23 are stored in memory 3002 ******MVI A, 8CST A,3002****** number 2, byte 3, bits 24-31 are stored in memory 3003 ******MVI A, B3ST A,3003After executing the above instruction the memory will contain values as shown below.Now let us perform the 32 bit addition.LD A,2000LD B,3000ADD BST A,4000LD A,2001LD B,3001ADC BST A,4001LD A,2002LD B,3002ADC BST A,4002LD A,2003LD B,3003ADC BST A,4003MVI A,0MVI B,0ADC BST A,4004After executing the above instructions the memory will contain values as shown below.I hope this helps you to understand the 32 bit addition. If you need any more details write a comment. I will explain.All the best.