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

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

General terms for cache-microprocessor, General terms for Cache : Cac...

General terms for Cache : Cache Hits : When the cache consisted the information requested, the transaction is said to be a cache hit. Cache Miss : When the cache does n

Assembler directives and operators-microprocessor, Assembler Directives and...

Assembler Directives and Operators The major advantage of machine language programming is directly that the memory control is in the hands of the programmer, so that, he can be

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

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

Rol-logical instruction-microprocessor, ROL : Rotate Left without Carry: T...

ROL : Rotate Left without Carry: This instruction rotates the content of the destination operand to the left by the specified count bit-wise excluding the carry. The most signific

General bus operation-microprocessor, General Bus Operation The 8086 ha...

General Bus Operation The 8086 has a joined data and address bus commonly referred to as a time multiplexed address and data bus. The major reason behind  multiplexing address

Cbw-cwd-arithmetic instruction-microprocessor, CBW: Convert Signed Byte to...

CBW: Convert Signed Byte to Word: This instruction converts a signed byte to a signed word. In other terms, it copies the sign bit of a byte to be converted to all of the bits in

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

Write a mips program that reads a string from user input, Description Wr...

Description Write a MIPS program that reads a string from user input, reverse each word (defined as a sequence of English alphabetic letters or numeric digits without any punctu

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