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

Assignment Help:

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.

 


Related Discussions:- Program to move contents in memory-machine level programs

Interrupt system based on 8259 a-microprocessor, Interrupt System Based on ...

Interrupt System Based on Single 8259 A The 8259A is contained in a 28-pin dual-in-line package that need only a + 5-V supply voltage.  Its organization is shown in given figur

Write a mips assembly language program, Write a MIPS/SPIM assembly language...

Write a MIPS/SPIM assembly language program that prints the smallest and largest values found in a non-empty table of N word-sized integers. The address of the first entry in your

End-endp-assemblers directive-microprocessor, END : END of Program:- Th...

END : END of Program:- The END directive marks the ending of the assembly language program. When the assembler comes across this END directive, it avoided the source lines avai

Ddition, Write a program that performs the addition, subtraction, multiplic...

Write a program that performs the addition, subtraction, multiplications, division of the given operands. Perform BCD operation for addition and subtraction.

Any project ideas plz, can any one help me in my project by using assembly ...

can any one help me in my project by using assembly language

Rics/cisc architecture-microprocessor, RICS/CISC Architecture An essent...

RICS/CISC Architecture An essential aspect of computer architecture is the design of the instruction set for the processor.  The instruction set selected for a specific compute

Machine level programs-microprocessor, Machine Level Programs In this s...

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

Physical memory mapped and port input output-microprocessor, Physical Memor...

Physical Memory Mapped I/O and Port I/O : CPU controlled I/O comes in 2 ways. Simply the difference is whether we utilize the normal memory addresses for I/O, this is mention

8237 modes-microprocessor, 8237 modes : Intel 8237 can be set to four d...

8237 modes : Intel 8237 can be set to four different type of style of transfer: 1) Single - One transfer at a time,  it allow processor access to the bus between transfers

Motorola 68000 series, Motorola 68000 Series : 68000microprocessor is a...

Motorola 68000 Series : 68000microprocessor is a 16 bit processor that has addressing space of 65536 locations, each of which holds a 64-bits word; In order to address those lo

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd