In this program we will see how to transfer a block from one location to another location.Problem StatementWrite 8086 Assembly language program to transfer a block from one memory section to another memory section. The numbers are stored at memory offset 501 onwards. The block size is stored at memory offset 500.DiscussionHere we are initially setting up the source index register with the source of data blocks, then set the destination index register to store into another block. Then set the Data segment register and Extra Segment register to 0000H. By using MOVSB instruction, the entire block is transferred from ... Read More

It is possible to address an I/O port as if it were a memory location. For example, let us say, the chip select pin of an I/O port chip is activated when address = FFF0H, IO/M* = 0, and RD* = 0. This is shown in the following fig.In this case, the I/O port chip is selected when the 8085 is thinking that it is addressing memory location FFF0H for a read operation. Note that 8085 thinks that it is addressing a memory location because it has sent out IO/M* as a logic 0. But in reality, an input port ... Read More

In this program we will see how to transfer a 4-byte block from one location to another location.Problem StatementWrite 8086 Assembly language program to transfer a four-byte block from one memory section to another memory section. The numbers are stored at memory offset 500 – 503.DiscussionHere we are initially setting up the source index register with the source of data blocks, then set the destination index register to store into another block. Then set the Data segment register and Extra Segment register to 0000H. By using MOVSB instruction, the entire block is transferred from one location to another. As the ... Read More

In this program we will see how to sort array elements in ascending order by using selection sort.Problem StatementWrite 8086 Assembly language program to sort the elements in a given array using selection sort technique. The array is started from memory offset 501. The size of the series is stored at memory offset 500.DiscussionIn the selection sort technique, in each phase we are taking the smallest number from the array, swap the smallest element with the first element inside the array. Then move to second position, and check for second highest number form the second position to the end of ... Read More

In this program we will see how to sort array elements in ascending order.Problem StatementWrite 8086 Assembly language program to sort the elements in a given array, which is starts from memory offset 501. The size of the series is stored at memory offset 500.DiscussionHere we are sorting the number in bubble sorting technique. In this sorting technique there will be n passes for n different numbers. In ith pass the ith largest element will be placed at the end. This is comparison based sort. We taking two consecutive numbers, compare them, and then swap them if the numbers are ... Read More

In this program we will see how to find the minimum number in a given array.Problem StatementWrite 8086 Assembly language program to find the minimum number in a given array, which is starts from memory offset 501. The size of the series is stored at memory offset 500. Store the minimum number at memory offset 600.DiscussionAt first we are taking the size of the array from memory offset 500. Then using that size, we are initializing the counter to read and check all the numbers. We are taking the first number into AL, then check each number and compare it ... Read More

In this program we will see how to find the largest number in a given array.Problem StatementWrite 8086 Assembly language program to find the largest number in a given array, which is starts from memory offset 501. The size of the series is stored at memory offset 500. Store the largest number at memory offset 600.DiscussionAt first we are taking the size of the array from memory offset 500. Then using that size, we are initializing the counter to read and check all the numbers. We are taking the first number into AL, then check each number and compare it ... Read More

The Central Processing Unit and the main memory are always very accurate and fast as compared with electromechanical input or output devices like printers, etc. Here in this case it is essential for us that the data lines of the computer are not engaged for a long time during communication process with input/output devices. Else the overall speed of the computer system drastically falls below. So the Input Output devices gets connected to a computer through the Input Output ports. Two Instructions are fetched which are called as IN and OUT where there are 256 Input Ports and 256 Output ... Read More

In this program we will see how to find the equivalent BCD number from an ASCII value.Problem StatementWrite 8086 Assembly language program to find the equivalent BCD number from an ASCII value. The number is stored at memory location 2050 and store the result at memory location 3050.DiscussionThis program can change ASCII value of a number to its BCD (Decimal) form. The ASCII values of the numbers are like below:ASCII (in Hex)30313233343536373839BCD00010203040506070809 From this table we can easily find that the last nibble of the ASCII value is actually the BCD equivalent. So to take the last nibble we have masked ... Read More

In this program we will see how to find the equivalent hexadecimal number from a BCD number.Problem StatementWrite 8086 Assembly language program to find the equivalent hexadecimal number from a BCD number. The number is stored at memory offset 500 and store the result at memory offset 600.DiscussionTo convert BCD to hexadecimal at first we have to cut the BCD digits. The most significant digit will be multiplied with 0AH (10D), and then least significant digit will be added with the result of multiplication. Thus the BCD will be converted to its equivalent hexadecimal form.InputAddressData……50059…… Flow Diagram ProgramOutputAddressData……6003B……