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

In this program we will see how to search an element in an array of bytes using 8085.Problem StatementWrite 8085 Assembly language program to search a key value in an array of data using linear search technique.DiscussionIn this program the data are stored at location 8002H to 8007H. The 8000H is containing the size of the block, and 8001H is holding the key value to search. After executing this program, it will return the address of the data where the item is found and store the address at location 9000H and9001H. If the item is not found, it will return ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will convert binary or hexadecimal number to ASCII values.Problem StatementWrite 8085 Assembly language program to convert binary or Hexadecimal characters to ASCII values. DiscussionWe know that the ASCII of number 00H is 30H (48D), and ASCII of 09H is39H (57D). So all other numbers are in the range 30H to 39H. TheASCII value of 0AH is 41H (65D) and ASCII of 0FH is 46H (70D), so all other alphabets (B, C, D, E, F) are in the range 41H to 46H.Here we are providing hexadecimal digit at memory location ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will convert 8-bit BCDnumbers to two digit ASCII values.Problem StatementWrite 8085 Assembly language program where an 8-bit BCD number is stored in memory location 8050H. Separate each BCD digit 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 BCD digit(nibble) to its equivalent ASCII values. As the 8-bit BCD number contains two nibbles, so we can execute this subroutine to find ASCIIvalues of them. We can get the lower nibble ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will convert gray code to binary code.Problem StatementWrite an assembly language program for 8085 to convert gray code to binary code. The data is stored at address 8200H & store the result at memory location 8201H.DiscussionHere we are loading the number from memory and in each step we are performing right shift, and XOR the intermediate result with the previous one. Thus we are getting the result. In the following demonstration you can get the logic.C       1110 1011    (A) (EBH) 07H     0111 ... Read More

In this program we will see how to find the gray code from an 8-bit number.Problem StatementWrite 8085 Assembly language program to convert an 8-bit number stored at 8000H to its equivalent gray code. The result will be stored at 8050H.DiscussionIn this program we are converting binary to gray code. The procedure is simple. At first we have to shift the content to the right, then perform XOR operation with the sifted content and the actual content. Thus we will get the gray code. For an example if the number is ABH, then the binary value will be (1010 1011), ... Read More

In this program we will see how to find the gray code from an 8-bit number.Problem StatementWrite 8085 Assembly language program to convert an 8-bit number stored at 8000H to its equivalent gray code. The result will be stored at 8050H.DiscussionIn this program we are converting binary to gray code. The procedure is simple. At first we have to shift the content to the right, then perform XOR operation with the sifted content and the actual content. Thus we will get the gray code. For an example if the number is ABH, then the binary value will be (1010 1011), ... Read More

In this program we will see how to add a blocks of data using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to add numbers in an array, where the size of the array is N. The value of N is provided.DiscussionIn this problem we are using location 8000H to hold the length of the block. The main block is stored from address 8010H. We are storing the result at location 9000H and 9001H. The 9000H holding the lower byte, and 9001H is holding the upper byte.Repeatedly we are taking the number from the memory, then adding it with accumulator and ... Read More