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Machine Level Programs
In this section, a few machine levels programming instance, rather then, instruction sequences are presented for comparing the 8086 programming with that of 8085. These programs are ii the form of instruction sequences as 8085 programs. These can even be hand-coded entered byte by byte and executed on an 8086 based system but due to the complicated instruction set of 8086 and its tedious opcode conversion procedure, mostly programmers prefer to use assemblers. However, we will deeply discuss the hand- coding,
Example :
Write a program to add data byte situated at offset 0500H in 2000H segment to another data byte available at 0600H in the similar segment and the result is store at 0700H in the similar segment.
Solution :
The flow chart for this problem might be drawn as given figure
The above instruction is quite straight-forward. As the immediate data can't be loaded into a segment register, the data is transferred to one general purpose resistors AX. And then the register general purpose registers AX, and then the register content is moved to the segment registers DS. Thus the data segment register DS have 2000H. The instruction MOV AX,[500H] signifies that the contents of the specific location, whose offset is indicated in the brackets having the segment pointed to by DS segment register, is to be moved to register AX. The MOV [0700], AX instruction moves the contents of the AX to an offset 0700H in DS (DS = 2000H). Make a point that the code segment register CS gets automatically loaded by the code segment address of the program whenever it is executed. In actual it is the monitor program that accepts the CS:IP address of the program and passes it to the equivalent registers on the time of execution. Hence no instructions are needed for loading the CS register like SS or DS.
move a byte string ,16 bytes long from the offset 0200H to 0300H in the segment 7000H
DW : Define Word:- The DW directive serves the same purposes as the DB directive, but now it makes the assembler which reserves thenumber ofmemory words (16-bit) instead of by
program to accept 23 students name using while loop let your variable control the value negative 4
Write a program to mask bits D3D2D1D0 and to set bits D5D4 and to invert bits D7D6 of ax register
CBW: Convert Signed Byte to Word: This instruction converts a signed byte to a signed word. In other terms, it copies the sign bit of a byte to be converted to all of the bits in
AND: Logical AND: This instruction bit by bit ANDs the source operand that might be an immediate, or a memory location or register to the destination operand that might be a memor
Program: Write a program to perform addition of a series of 8-bit numbers. The series have 100 (numbers). Solution : In the first program, we have been implemented the add
ADD: Add :- This instruction adds an immediate contents of a memory location specified in the a register ( source ) or instruction to the contents of another register (destinat
MyLocation SDWORD 14 TheTest SDWORD 8 mov eax,MyLocation mov ebx,TheTest neg eax,ebx sub eax,ebx Show exactly what lives in eax after executi
Memory Segmentation : The memory in an 8086/8088 based system is organized as segmented memory. In this scheme, the whole physically available memory can be divided into a n
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