Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

How could each of the two-proposed changes decrease the size of an mips assembly program? On the other hand, how could the proposed change increase the size of an mips assembly program?

Sagot :

Martebi

Based on the question attached, the Number of bits that is needed to address a register is 7 bits.

3.  How could the proposed change increase the size of an mips assembly program?

The two-proposed changes decrease the size of an mips assembly program because a lot of complicated operations will need to be put in place or implemented in one instructions instead of numerous instructions due to the lowered register leaking issue. The program's size will be therefore be lowered as a result of this issue.

The proposed change will increase the size of an mips assembly program because  The size of the instruction word will then be brought up or raised if the required bits are added to the opcode and that of the register fields, which will bring up the size of the programmed.

In  the MIPS register file;

The Number of MIPS registers = 128

The Number of bits needed = log₂128

The  Number of bits needed = 7 bits

So Increasing number bits for opcode will be = (6 + 2)

                                                                       = 8

Therefore the answers to question one:

1. R-type instruction

Op-code = 6 bits

rs = 5 bits

rt = 5 bits

rd = 5 bits

shamt = 5 bits

funct = 6 bits

Hence, the size of the opcode field will be go up by two bits, to eight bits and the size of the rs ,rt, and rd fields is also go up to 7 bits.

For question 2 which is I-type instruction, there is:

Op-code = 6 bits

rs = 5 bits

rt = 5 bits

Immediate = 16 bits

Hence, The size of the opcode field will also go up by two bits, to eight bits and the size of the rs, rt  fields will is also go up to 7 bits.

Therefore, Based on the question attached, the Number of bits that is needed to address a register is 7 bits.

Learn more about program from

https://brainly.com/question/14897427

#SPJ1

See full question below

Assume that we would like to expand the MIPS register file to 128 registers and expand the instruction set to contain four times as many instructions. 1. How this would this affect the size of each of the bit fields in the R-type instructions? 2. How this would this affect the size of each of the bit fields in the I-type instructions? 3. How could each of the two proposed changes decrease the size of an MIPS assembly program? On the other hand, how could the proposed change increase the size of an MIPS assembly program?

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.