TDT4255/theory1.org

• Theory question EX1
  • Question 1
    • Part 1
      • Part 1½
      • Part 1¼
    • Part 2
  • Question 2
  • Question 3

Theory question EX1

Keep in mind that your design and your implementation are separate entities, thus you should answer these questions based on your ideal design, not your finished implementation. Consequently I will not consider your implementation when grading these questions, thus even with no implementation at all you should still be able to score 100% on the theory questions.

All questions can be answered in a few sentences. Remember that brevity is wit, and also the key to getting a good score. You should easily be able to fit your entire answer on a single screen.

Question 1

2 points.

Part 1

Part 1½

½ points. When decoding the BNE branch instruction in the above assembly program

bne x6, x2, "loop",

In your design, what is the value of each of the control signals below?

• regWrite
• memRead
• memWrite
• branch
• jump

Part 1¼

½ points. When decoding the LW instruction in the above assembly program

jal x1, 0x10(x1)

In your design, what is the value of each of the control signals below?

• regWrite
• memRead
• memWrite
• branch
• jump

Keep in mind that your design and your implementation are separate entities, thus you should answer this question based on your ideal design, not your finished implementation.

Part 2

During execution, at some arbitrary cycle the control signals are:

• regWrite = 1
• memRead = 0
• memWrite = 0
• branch = 0
• jump = 1

In your design, which intruction(s) could be executing?

Keep in mind that your design and your implementation are separate entities, thus you should answer this question based on your ideal design, not your finished implementation.

Question 2

4 points. NO PARTIAL CREDITS Since you can test your solution with the testing framework I will not offer any points for a near correct solution to this problem.

Reading the binary of a RISC-V program you get the following:

0x0:  0x00a00293   --   0000 0000 1010 0000 0000 0010 1001 0011
0x4:  0x01400313   --   0000 0001 0100 0000 0000 0011 0001 0011
0x8:  0xfff30313   --   1111 1111 1111 0011 0000 0011 0001 0011
0xc:  0x00628463   --   0000 0000 0110 0010 1000 0100 0110 0011
0x10: 0xff9ff06f   --   1111 1111 1001 1111 1111 0000 0110 1111

For each instruction describe the format and name and corresponding RISC-V source To give you an idea on how decoding would work, here is the decoding of 0x40635293:

0x40635293   --   0100 0000 0110 0011 0101 0010 1001 0011

Opcode: 0010011 => Format is IType, thus the funct3 field is used to decode further
funct3: 101     => Instruction is of type SRAI, the instruction looks like ~srai rd, rx, imm~

rs1:    00110   => x6
rd:     00101   => x5
shamt:  000110  => 6

Resulting in ~srai x5, x6, 6~

Your answer should be in the form of a simple asm program.

• hint 1: the original asm program had a label, you need to infer where that label was
• hint 2: Verify your conclusion by assembling your answer. To do this, make an asm program, place it with the rest of the tests and set printBinary to true in singleTestOptions in Manifest.scala which will print the full binary of your program. As long as your program generates the same binary as the supplied your program is correct.

Question 3

4 points. NO PARTIAL CREDITS Since you can test your solution with the testing framework I will not offer any points for a near correct solution to this problem.

In order to load a large number LUI and ADDI are used. consider the following program

li x0, 0xFF
li x1, 0x600
li x2, 0x8EE
li x3, 0xBABEFACE
li x4, 0xBABE07CE

1. Which of these instructions will be split into ADDI LUI pairs?
2. Explain in 3 sentences or less how the two last ops are handled differently and why.

• hint 1: The parser and assembler in the test suite can help you answer the first part of this question (a). Create an asm file, put it with the rest of the tests and run it, setting the correct test options in singleTestOptions defined in Manifest.scala and observe the output.

• hint 2: While it's probably easier to solve this problem using the internet, however you can also figure out what is happening by browsing the assembler source code which will hopefully give you a deeper insight into what is going on here.

  Look at Parser.scala, specifically what happens when an li instruction is parsed. When parsing an instruction the parser first attempts to apply the singleInstruction rule, however this only succeeds if the immediate value obeys certain restrictions (nBits <= 12), if not it fails.

  If the singleInstruction rule fails the parser then attempts to apply the multipleInstructions rule instead which expands operations into a list of real ops. When this happens the resulting operations are defined as the following:

  stringWs("li") ~> (reg <~ sep, (hex | int).map(_.splitHiLo(20))).mapN{ case(rd, (hi, lo)) => {
    List(
    ArithImm.add(rd, rd, lo),
    LUI(rd, if(lo > 0) hi else hi+1),
  )}}.map(_.widen[Op]),

  This is quite a lot to unpack, but you can focus on the line where the LUI is constructed. hi and lo are the results of splitHiLo which splits a 32 bit word into a 12 bit and a 20 bit. Try this for yourself on paper; what happens when lo ends up being a negative number? What is the interplay between incrementing hi with 1 and adding a lo that is represented as a negative value?