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
#Write a function to calculate the following arithmetic operation and return the result. A = 2 + (3x)3 + y/2n (x, y and n are arguments of the function where x is an integer in the

Architecture Of 8088 The register set of 8088 is accurately the same as in to 8086. The architecture of 8088 is also same to 8086 except for 2 changes; a) 8088 has 4-byte instr

I need a text conversion program written in assembly language

write a program that calculates the fibonacci series: except for the first two numbers in the sequence

) What is the difference between re-locatable program and re-locatable data?

how we can multiply two 8 bit number with rotation

I need a division subroutine. Asks for two inputs, then displays the inputs and shows the answer with a remainder. Mine isnt displaying the inputs correctly.

Ask 2. Exchange higher byte of AX and higher byte of BX registers by using memory location 0160 in between the transfer. Then stores AX and BX registers onto memory location 0174 o

Register Organization of 8086 8086  has  a great  set  of registers  containing  special  purpose and general  purpose  registers.  All the 8086 resisters are 16-bit registers.

Addressing mode of 8086 : Addressing mode specify a way of locating operands or data. Depending on the data types used the memory  addressing  modes and in the instruction  ,