Machine coding the programs-microprocessor, Assembly Language

Assignment Help:

Machine Coding the Programs

So far we have describe five programs which were  written  for hand coding  by a programmer. In this, we will now have a deep look at how these programs may be  translate to machine codes. In Appendix, the instruction set along with the Appendix is presented. This Appendix is self-explanatory to hand code mostly of the instructions. The V,S W, D, MOD, REG  and R/M  fields are appropriate decided depending upon the data types, addressing mode and the registers  used. The table shows the details about how to select these fields.

Most of the instructions either have particular opcodes or they may be decided only by setting the V,S, W, D, REG, MOD and R/M fields suitably but the critical point is  the calculation of jump addresses for intra segment branch instructions. Before beginning the coding of call or jump instructions, we will see some simpler coding examples.

Example :

MOV BL, CL

For hand coding this instruction, first we will have to note down the following features.

(i) It sets in the register/memory to/from register format.

(ii) It is an 8-bit operation.

(iii) BL is the destination register and CL is a source register.

Now from the feature (i) by using the Appendix, the op code format is given below.

1485_mcp.jpg

If d =1, then transformation of data is to the register shown by the REG field, for example the destination is a register (REG). If d = 0, the source is a register shown by the REG field. It is an 8-bit operation, therefore w bit is 0. If it had been a 16-bit operation, the w bit would have been 1.From referring to given table to search the REG to REG addressing in it, for example the last column with MOD 11. According to the Appendix when MOD is 11, the R/M field is treated as a REG field. The REG field which is used for source register and the R/M field are used for the destination register, if d is 0. If d =1, the REG field is utilized for destination and the R/M field is used to indicate source. the complete machine code of this instruction comes out to be now.

code    dw       MOD   REG    R/M

MOV BL, CL 1 0 0 0 1 0 0 0     1   1   001    0 1 1= 88 CB


Related Discussions:- Machine coding the programs-microprocessor

Comparison between 8086 and 8088, Comparison between 8086 and 8088 All ...

Comparison between 8086 and 8088 All the changes in 8088 above 8086 are indirectly or directly related to the 8-bit, 8085 compatible data and control bus interface. 1) The p

Program to add 8-bit series numbers-assembly language, Program: Write a pr...

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

Pc bus and interrupt system-microprocessor, PC Bus and Interrupt System ...

PC Bus and Interrupt System The PC Bus utilized a bus controller, address latches, and data transceivers (bidirectional data buffers). 1) Bus controller : ( Intel 8288 Bus

Space don''t come in ASCII characters while printing?, .MODEL SMALL .STACK...

.MODEL SMALL .STACK 100H .DATA PROMPT DB \''The 256 ASCII Characters are : $\'' .CODE MAIN PROC MOV AX, @DATA ; initialize DS MOV DS, AX

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

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

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

Instructions, Difference between div and idiv

Difference between div and idiv

Encrypting, write an assembly language program that has two subroutines : o...

write an assembly language program that has two subroutines : one for encrypting alphabates of a string and second fo decrypting the encoded string . in encryption simply converta

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

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