In this program we will see how to convert binary numbers to its BCD equivalent.Problem StatementA binary number is store dat location 800H. Convert the number into its BCD equivalent and store it to the memory location 8050H.DiscussionHere we are taking a number from the memory, and initializing it as a counter. Now in each step of this counter we are incrementing the number by 1, and adjust the decimal value. By this process we are finding the BCD value of binary number or hexadecimal number.We can use INRinstruction to increment the counter in this case but this instruction will ... Read More

In this program we will see how to convert BCD numbers to binary equivalent.Problem StatementA BCD number is stored at location 802BH. Convert the number into its binary equivalent and store it to the memory location 802CH.DiscussionIn this problem we are taking a BCD number from the memory and converting it to its binary equivalent. At first we are cutting each nibble of the input. So if the input is 52 (0101 0010) then we can simply cut it by masking the number by 0FH and F0H. When the Higher order nibble is cut, then rotate it to the left ... Read More

In this program we will see how to sort a block of bytes in descending order using bubble sorting technique.Problem StatementWrite8085 Assembly language program to sort numbers in descending order where n number of numbers are stored in consecutive memory locations starting from 8041H and the value of n is available in memory location 8040H (Using BUBBLE sort).DiscussionIn this program we will arrange the numbers in bubble sorting technique. In this sorting technique, it will be executed in different pass. In each pass the smallest number is stored at the end of the list. Here we are taking the numbers ... Read More

In this program we will see how to sort a block of bytes in ascending order using bubble sorting technique.Problem StatementWrite8085 Assembly language program to sort numbers in ascending order where n number of numbers are stored in consecutive memory locations starting from 8041H and the value of n is available in memory location 8040H (Using BUBBLE sort).DiscussionIn this program we will arrange the numbers in bubble sorting technique. In this sorting technique, it will be executed in different pass. In each pass the largest number is stored at the end of the list. Here we are taking the numbers ... Read More

In this program we will see how to sort a block of bytes using bubble sorting technique.Problem StatementWrite8085 Assembly language program to sort numbers in ascending order where n number of numbers are stored in consecutive memory locations starting from 8041H and the value of n is available in memory location 8040H (Using BUBBLE sort).DiscussionIn this program we will arrange the numbers in bubble sorting technique. In this sorting technique, it will be executed in different pass. In each pass the largest number is stored at the end of the list. Here we are taking the numbers from location 8041H ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will convert 8-bit numbers to two digit ASCII values.Problem StatementWrite 8085 Assembly language program where an 8-bit binary number is stored in memory location 8050H. Separate each nibbles and convert it to corresponding ASCII code and store it to the memory location 8060H and 8061H.DiscussionIn this problem we are using a subroutine to convert one hexa-decimal digit (nibble) to its equivalent ASCII values. As the 8-bit number contains two nibbles, so we can execute this subroutine to find ASCII values of them. We can get the lower nibble ... Read More

In this program we will see how to generate Fibonacci sequence.Problem StatementWrite 8085 Assembly language program to generate the first ten elements of the Fibonacci sequence using registers only and store them in memory locations 8050H to 8059H. DiscussionThis program will generate the Fibonacci numbers. The Fibonacci numbers follows this relation F(i) = F(i - 1) + F(i - 2) for all i >2 with F(1) = 0, F(2) = 1.InputIn this case we are not providing any input, this program will generate ten Fibonacci numbers.Flow DiagramProgramAddressHEX CodesLabelsMnemonicsComments800021, 50, 80STARTLXI H 8050H Pointer to the OUT-BUFFER8003AFXRA A Clear accumulator and reg. B800447MOV B, ... Read More

In this program we will see how to exchange a block of bytes using 8085.Problem StatementWrite 8085 Assembly language program to exchange a block of data, where block size is given.DiscussionThe data are stored at location 8010H to 8019H and 9010H to 9019H. The location 8000H is holding the number of bytes to exchange.The logic is very simple, The HL and DE register pair is pointing the first and second data block respectively. By taking the data we are just swapping the values of each memory locations. Then repeating this process to swap two blocks completely.InputAddressData......800006......801000801111801222801333801444801555......9010849011639012129013479014489015AD......Flow DiagramProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, 10, 80LXI ... Read More

In this program we will see how to sort a block of bytes using bubble sorting technique.Problem StatementWrite 8085 Assembly language program to sort numbers in ascending order where n number of numbers are stored in consecutive memory locations starting from 8041H and the value of n is available in memory location 8040H (Using BUBBLE sort).DiscussionIn this program we will arrange the numbers in bubble sorting technique. In this sorting technique, it will be executed in different pass. In each pass the largest number is stored at the end of the list. Here we are taking the numbers from location ... Read More

In this program we will see how to find the largest number from a block of bytes using 8085.Problem StatementWrite 8085 Assembly language program to find the largest number from a block of bytes.DiscussionIn this program the data are stored at location 8001H onwards. The 8000H is containing the size of the block. After executing this program, it will return the largest number and store it at location 9000H.Logic is simple, we are taking the first number at register B to start the job. In each iteration we are getting the number from memory and storing it into register A. ... Read More