Machine coding the programs-microprocessor, Assembly Language

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

Posted Date: 10/12/2012 6:47:59 AM | Location : United States







Related Discussions:- Machine coding the programs-microprocessor, Assignment Help, Ask Question on Machine coding the programs-microprocessor, Get Answer, Expert's Help, Machine coding the programs-microprocessor Discussions

Write discussion on Machine coding the programs-microprocessor
Your posts are moderated
Related Questions
take an integer and its base and the base in which you want to convert the number from user and perform conversion.

Memory Segmentation : The  memory in an 8086/8088  based system is organized as segmented memory. In this scheme, the whole physically available memory can be divided into a n

what will be the value of EAX after following instructions execute? mov bx, 0FFFFh and bx, 6Bh

Difference between div and idiv

Segment Registers The 8086 addresses a segmented memory unlike 8085. The complete 1 megabyte memory, which 8086 is capable to address is divided into 16 logical segments.Thusea

Basic Microprocessor Architecture and Interface : Introduction: Intel launches its first 4-bit microprocessor 4004 in the year 1971 and 8-bit microprocessor 8008 in the y

describes vertical and horizontal web services protocols. Next, identify the similarities and differences between vertical and horizontal web services protocols. Finally, explain w

ADC: Add with Carry:- This instruction performs the similar operation a like ADD instruction, but adds the carry flag bit (which might be set as a result of the previous calculatio

Please let me know if you can do an assignment in the next 12 hours