We have seen previously that the linker is invoked in three modes i.e. they are command line mode, prompt mode and data file mode. In the Prompt mode the linker is run by simply typing the ‘LINK85’. Responding to the prompt the linker requests the user for the file name. In the given example, responding with ‘MULT.OBJ’ is there in fact MULT is enough.The linker prompts for the address of the offset. The value of the offset which is input by means of the user is finally added to the value of any ORG statements in the file. In response ... Read More

The file MULT.ASM which is created by using an editor is simply a text file. We cannot execute this file directly. At first we have to assemble the file, and then we have to link it. The step of assembly the translation of the program of assembly language to machine code requires the generation of a .OBJ file.Now a .HEX file is generated by using a linker. We have seen previously that how the naming of the input and output file is done.Input filenameThe source name of the file must be responded by the user. In the given example the ... Read More

Generation of .OBJ file by using a cross-assembler:The file assembly language program file, e.g. MULT.ASM which is created by using an editor is simply a text file. We cannot execute this file directly. At first we have to assemble the file, and then we have to link it. The step of assembly the translation of the program of assembly language to machine code requires the generation of a .OBJ file.We have used some examples in the text of ‘2500 A.D i.e. the 8085 cross-assembler of version 4.01.In the prompt mode the translation is:We type ‘X8085’ just for performing the translation ... Read More

The program written in assembly language is entered by us by the help of any editor, we prefer a screen editor because when we enter the program it is of no importance for the particular exact column label of instructions and the starting of the program. But the point to be noted that the instruction must start after column 1. More over the aligned program must be lucidly written for reading if its various fields should be assigned properly. There are many features which the editor provides like many help features which the user gets to be more familiar by ... Read More

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to multiply two numbers using shift and add methods.Problem Statement:Write 8085 Assembly language program to multiply two 8-bit numbers using shift and add method.Discussion:The shift and add method is an efficient process. In this program, we are taking the numbers from memory location 8000H and 8001H. The 16 bit results are storing into location 8050H onwards.In this method we are putting the first number into DE register pair. The actual number is placed at E register, and D is holding 00H. The second ... Read More

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to add the contents of N word locations.Problem Statement:Write 8085 Assembly language program to add N 16-bit numbers stored into memoryDiscussion:The 16-bit numbers are stored into memory location 8001H onwards. The value of N is stored at location 8000H. After addition, the result will be stored at location 8050H onwards.In 8085 we have few number of registers. So we are storing the count into memory, when we need to update it, we will fetch it from memory, update it and then again store ... Read More

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to sort a sequence of numbers in reverse order using selection sort.Problem Statement:Write 8085 Assembly language program to sort a given sequence using selection sort in descending order. The numbers are stored at 8001H onwards. 8000H is holding the block size.Discussion:In the selection sorting technique, we will choose the minimum or the maximum term from a set of numbers. In this case we are considering the sorting in descending order, so we are choosing the maximum number. By taking the maximum number, we ... Read More

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to sort a sequence of numbers using selection sort.Problem Statement:Write 8085 Assembly language program to sort a given sequence using selection sort in ascending order. The numbers are stored at 8001H onwards. 8000H is holding the block size.Discussion:In the selection sorting technique, we will choose the minimum or the maximum term from a set of numbers. In this case we are considering the sorting in ascending order, so we are choosing the minimum number. By taking the minimum number, we are swapping it ... Read More

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to perform bubble sort in based on choice.Problem Statement:Write 8085 Assembly language program to perform bubble sorting operation on a set of data, and arrange them into ascending or descending order based on choice.Discussion:In this program we are arranging some numbers into ascending or descending order based on some choice. We are storing the choice at location A000H. If the choice value is 00H, then data will be sorted in ascending order, otherwise it will be sorted in descending order. The 8000H is ... Read More

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to do different operations on BCD numbers based on choice.Problem Statement:Write 8085 Assembly language program to perform some operations on two 8-bit BCD numbers base on choice.Discussion:In this program we are taking a choice. The choice value is stored at memory location 8000H (named as X). And the BCD numbers are stored at location 8001H and 8002H. We are storing the result at location 8050H and 8051H.Here if the choice is 00H, then it will perform addition, for 01H, it will perform subtraction, ... Read More