Program to move contents in memory-machine level programs, Assembly Language

Example : Write a program to move the contents of the memory location 0500H to BX and also to register CX. Add immediate byte 05H to the data residing in memory location, whose address is computed by using DS=2000H and offset=0600H. Store up the result of the addition in 0700H. Consider that the data is located in the segment specified by the data segment register which contain 2000H.

Solution :

The flow chart for the program is shown in given figure.

After initializing the data segment register the content of the location 0500H are moved to the BX register by using MOV instruction. The similar data is moved also to the CX register. For this data transfer, there can be 2 options as shown.

1739_example1.jpg

826_example2.jpg

(a) MOV CX,       BX      ; As the contents of BX register will be similar as 0500H after execution

                                   ; of MOV BX,[0500H].

(b) MOV CX, [050OH]   ; Move directly from 0500H to CX

 

In the first option the opcode is just of 2 bytes, whereas the second option will have 4 bytes of opcode. Thus the second option will need execution time and more memory. Due to these reasons, the first option is preferable.

The immediate data byte 05H is added to content of 0600H by using the ADD instruction. The result will be goes in destination operand 0600H. It is next stored at the location 0700H. In particular case of the 8086/8088 instruction set, there is no instruction for direct transfer of data from the memory source operand to the memory destination operand rather than the string instructions. So the result of addition which is present at 0600H should be moved to any one registration amongst the general purpose registers, except BX and CX, or else the contents of BX and CX will be changed. We have chosen DX (we could have selected register AX also, because once DS is initialized to 2000H the contents of register AX are no longer useful for this purpose. therefore the transfer of result from 0600H to 0700H is accomplished in two stages by using successive MOV instructions, for an instance,

Firstly, the content of 0600H is DX register and then the content of DX register is moved to 0700H. The program terminated with the instruction HLT.

 

Posted Date: 10/12/2012 6:32:47 AM | Location : United States







Related Discussions:- Program to move contents in memory-machine level programs, Assignment Help, Ask Question on Program to move contents in memory-machine level programs, Get Answer, Expert's Help, Program to move contents in memory-machine level programs Discussions

Write discussion on Program to move contents in memory-machine level programs
Your posts are moderated
Related Questions
$NOMOD51 $NOSYMBOLS ;***************************************************************************** ; Spring 2013 Project ; ; FILE NAME : Project.ASM ; DATE : 3/30/20

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

Assembly Code for Reading Flow & Generating Serial Output The timer is timer 1 is set for the baud rate 9600, as the crystal used is of 11.0592 Hz.  Then the timer 1 is starte

Interrupt Table Each interrupt level has a booked memory location, called an interrupt vector.  All these vectors (or pointers) are stored in the interrupt table. Table lies at

PLEASE MAY YOU ASSIST ME WITH SAMPLE CODES FOR PROGRAMING A FIRE ALARM MINI PROJECT

Write an assembly language program that defines symbolic constants for all seven days of the week

Write a program to separate out positive and negative numbers from a given series of 16-bit hexadecimal numbers.

Format of Control Register The format for the control register is given in Figure. Bit 0 of this register might be one before data may be output  and  bit  two  might be  one

how to code

Internal Hardware-Interrupts Internal hardware-interrupts are the outcome of sure situations that occur during the execution of a program, for example. Divide by 0. The interru