Assembly language, Assembly Language

Assignment Help:

Assembly Language:

Inside the 8085, instructions are really stored like binary numbers, not a very good manner to look at them and very difficult to decipher. An assembler is a program that permits you to write instructions in, more or less, English form, much more simply read and understood, and after that converted or assembled into hex numbers and at last into binary numbers.

The program is written with a text editor (NOTEPAD or similar), saved like an ASM file, and then assembled by the assembler (TASM or MASM or similar) program. The end result is an OBJ file you download to the 8085. Here is an instance of the problem of adding 2 plus 2 :

mvi A,2; move 2 into the A register

mvi B,2; move 2 into the B register

add B; add reg. B to reg. A, store result in reg. A

The first line moves a 2 into a register A. The second line moves a 2 into a register B. it is all the data we require for the program. The third line adds up the accumulator with register B and holds the result back into the accumulator, destroying the 2 that were originally in it. The accumulator contains a 4 in it now and B still contains a 2 in it. In the program above all text after the ';' are treated as comments, and not executed. It is a very significant habit to acquire.

Assembly language follows some rules that we shall describe as they come up. With most of the instructions, especially those including data transfer, the instruction is first, followed by at least 1 space, after that the destination followed through a comma, and then the source. The destination is where the result of the instruction shall end up and the source is where the data is coming from.

Next we shall read a switch, and light an LED if the switch is pressed. This happens quite frequently in your lab experiments. Bit 0 of Port 0 shall be the switch. While the switch is closed or pressed, bit 0 will be a 1, and if the switch is open or not pressed, bit 0 will be a 0. Bit 0 of Port l is the LED. If bit 0 is a 0 the LED is off and if bit 0 is a 1, the LED shall be on. All the other bits of reg. A shall be ignored and assumed to be all 0's, for the sake of discussion

Start

IN         0          ; read Port 0 into reg. A

CMP    1          ;compare reg. A with the value 1

JNZ      start     ; jump to start if the comparison does not yield 0

OUT     1          ; send a 1 to Port 1, turning the LED on

JMP     start

 

The first line has something new. It's called as a label. In this case it is start:. A label is a manner of telling the assembler that this line has a name that may be referred to later to get back to it. All of the labels are followed by the symbol:, that tells the assembler that it is a label. In the first line we also read the switch by reading Port and putting it into the accumulator. Register. A is the only register that may read in/send out data using ports or perform compares. Therefore, we need not write 'A' in the command it's implied.


Related Discussions:- Assembly language

Imul-arithmetic instruction-microprocessor, IMUL: Signed Multiplication: T...

IMUL: Signed Multiplication: This instruction multiplies a signed byte by a signed bit in source operand e in the register AL or signed word in source operand by signed word in th

Cryptography, Write a program that allows a user to encrypt/decrypt a mess...

Write a program that allows a user to encrypt/decrypt a message using a shift cipher. The user should specify the key and the message. Your program should be able to handle messa

Code for reading flow & generating led output, Code for Reading Flow & Gene...

Code for Reading Flow & Generating LED Output The code starts with the scanning of the PORT 3, for reading the flow status to check for various flow conditions and compare to

Declare a constant in assembly language, Declare 1 constant. This can be do...

Declare 1 constant. This can be done just below the prototype section. Put the following divider above the constant section. ;************************ Constants ****************

8086 assembly language program, move a byte string ,16 bytes long from the ...

move a byte string ,16 bytes long from the offset 0200H to 0300H in the segment 7000H

Relocate program and data, ) What is the difference between re-locatable pr...

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

Opcode-microprocessor, Opcode : The opcode generally appear in the firs...

Opcode : The opcode generally appear in the first byte.but in a few instructions, a register objective is in the first byte and few other instructions may have their 3-bits of

Intel''s 8237 dma controller-microprocessor, Intel's 8237 DMA controller : ...

Intel's 8237 DMA controller : 1) The 8237 contain 4 independent I/O channels 2) It contains 27 registers, 7 of which are system-wide registers and 5 for each channel. 3)

Cache components-microprocessor, Cache components The cache sub-system ...

Cache components The cache sub-system may be divided into 3 functional blocks: Tag RAM, SRAM and theCache Controller. In real designs, these blocks can be implemented  by multi

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

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