Found 60 Articles for 8086

In this program we will see how to find the squares of n numbers stored in an array.Problem StatementWrite 8086 Assembly language program to calculate square of each numbers stored in an array of size n. The array size is stored at location offset 600, and Numbers are stored at 601 onwards.DiscussionTo solve this problem, we are taking the size of the array into the CL register, and make CH = 00H for counting. Now from each location take the number into accumulator, to make square, we have to multiply it two times. so we are multiplying AL with AL. ... Read More

Here we will see how to find product of two array elements and store result into memory.Problem StatementWrite 8086 Assembly language program to find product of two arrays stored at 501 onwards and 601 onwards. The size of array is stored at location 500. After calculating product store result at 501 onwards.DiscussionTo solve this problem, we are taking elements from first array using source register SI, and second array using destination register DI. Repeatedly take elements from SI to AL, then multiply with the content of DI, and store again into SI address. Thus it is solved.InputAddressData……500055012C5020B5037D5042550521……6010460212603026040460505…… Flow Diagram ProgramOutputAddressData……501B0502C6503FA504B9505A5……    Read More

In this program we will see how to find the cubes of n numbers stored in an array.Problem StatementWrite 8086 Assembly language program to calculate cubes of each numbers stored in an array of size n. The array size is stored at location offset 600, and Numbers are stored at 601 onwards.DiscussionTo solve this problem, we are taking the size of the array into the CL register, and make CH = 00H for counting. Now from each location take the number into accumulator, to make cube, we have to multiply it three times. so we are storing the number temporarily ... Read More

In this program we will see how to perform modulus of the first array corresponding to the next array.Problem StatementWrite 8086 Assembly language program perform modulus of the first array corresponding to the next array.DiscussionIn this example there are two different arrays. The arrays are stored at location 501 onwards and 601 onwards. The size of these two arrays are stored at offset location 500. We are taking the array size to initialize the counter, then by using loops we are getting the modulus of the elements one by oneInputAddressData……500045010F5020B5030550408……601046020A6030260403……Flow DiagramProgram    MOV SI, 500     ;Point Source index ... Read More

In this program we will see how to subtract the contents of two different arrays.Problem StatementWrite 8086 Assembly language program to subtract the contents to corresponding elements which are stored in two different arraysDiscussionIn this example there are two different arrays. The arrays are stored at location 501 onwards and 601 onwards. The size of these two arrays are stored at offset location 500. We are taking the array size to initialize the counter, then by using loops we are subtracting the elements one by oneInputAddressData……5000450109502035030850406……60104602016030260403……Flow DiagramProgram    MOV SI,  500     ;Point Source index to 500     MOV CL,  [SI]    ;Load the ... Read More

We call mode 1 as the strobed Input Output or handshake Input Output. We use this mode when the data is supplied by the input device to the microprocessor at irregular interval of time. A port which is functioned to program in mode uses three handshake signals. These handshake signals are provided by Port C. Only port A and B works in mode 1. The pins PC2, PC1, and PC0 provides handshake signals for port B when we configure it for Input port or Output port. Moreover, the pins PC7, PC6, and PC3 provides handshake signals for port A. The ... Read More

We interface the I/O device in a very segmented manner and is carried systematically. In the interfacing of seven segment display to 8085 microcontroller it is found that An output device which is very common is, especially in the kit of 8085 microprocessor and it is the Light Emitting Diode consisting of seven segments. Moreover, we have eight segments in a LED display consisting of 7 segments which includes ‘.’, consisting of character 8 and having a decimal point just next to it. We denote the segments as ‘a, b, c, d, e, f, g, and dp’ where dp signifies ... Read More

Before having a discussion regarding the demerits of I/O mapped I/O and merits of memory-mapped I/O, let us have a generic discussion regarding the difference between I/O mapped I/O and memory mapped I/O.In Memory Mapped Input Output −We allocate a memory address to an Input-Output device.Any instructions related to memory can be accessed by this Input-Output device.The Input-Output device data are also given to the Arithmetic Logical Unit.Input-Output Mapped Input Output −We give an Input-Output address to an Input-Output deviceOnly IN and OUT instructions are accessed by such devices.The ALU operations are not directly applicable to such Input-Output data.So as ... Read More

In Memory Mapped Input Output −We allocate a memory address to an Input-Output device.Any instructions related to memory can be accessed by this Input-Output device.The Input-Output device data are also given to the Arithmetic Logical Unit.Input-Output Mapped Input Output −We give an Input-Output address to an Input-Output device.Only IN and OUT instructions are accessed by such devices.The ALU operations are not directly applicable to such Input-Output data.So as a summary we can mention that −I/O is any general-purpose port used by processor/controller to handle peripherals connected to it.I/O mapped I/Os have a separate address space from the memory. So, total ... Read More

In this program we will see how to find a number n from a string (an array of numbers).Problem StatementWrite 8086 Assembly language program to find a number in a string (an array of numbers). The numbers are stored at memory offset 600 onwards.DiscussionIn this program we are taking only 5 numbers. We are searching the number 25. after successful search the DX register will hold the offset address, and BX register will hold the index of that number.We are taking each number from that array and then compare it with 25. If the numbers are same, then we will ... Read More