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

8086 microprocessors, program to find negative and positive integers from g...

program to find negative and positive integers from given signed numbers with output and explanation of every instructions.

Read architecture:look through-microprocessor, Read Architecture: Look Thro...

Read Architecture: Look Through Main memory that located is conflicting the system interface. The least concerning feature of this cache unit is that it remain between the proc

Zero flag, Zero flag: The next line compares the value in register. A ...

Zero flag: The next line compares the value in register. A with the value 1. If they are equivalent, the Zero flag is set (to 1). The next line then jumps to start: only if th

Second generation microprocessor, IInd Generation Microprocessor : The ...

IInd Generation Microprocessor : The second generation microprocessor by using n MOS technology seemed in the market in 1973.   The Intel  8080, of nMOS  technology

Program to find out positive and negative numbers-assembly, Program : W...

Program : Write an assembly program to find out the number of positive numbers and negative numbers from a given series of signed numbers. Solution : Take the i th num

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

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

Aad-arithmetic instruction-microprocessor, AAD: ASCII Adjust for Division ...

AAD: ASCII Adjust for Division though the names of these 2 instructions (AAM and AAD) seem to be same, there is many difference between their functions. The AAD instruction conver

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

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