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| * 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 the | |||||
| soul of wit, and also the key to getting a good score. | |||||
| ** Question 1 | |||||
| 2 points. | |||||
| *** Part 1 | |||||
| When decoding the BNE branch instruction in the above assembly program | |||||
| #+begin_src asm | |||||
| bne x6, x2, "loop", | |||||
| #+end_src | |||||
| In your design, what is the value of each of the control signals below? | |||||
| + memToReg | |||||
| + 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: | |||||
| + memToReg = 0 | |||||
| + 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. | |||||
| Reading the binary of a RISC-V program you get the following: | |||||
| #+begin_src text | |||||
| 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 | |||||
| #+end_src | |||||
| 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: | |||||
| #+begin_src text | |||||
| 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~ | |||||
| #+end_src | |||||
| *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) | |||||
| ** Question 3 | |||||
| 4 points. | |||||
| In order to load a large number LUI and ADDI are used. | |||||
| consider the following program | |||||
| #+begin_src asm | |||||
| li x0, 0xFF | |||||
| li x1, 0x600 | |||||
| li x2, 0x8EE | |||||
| li x3, 0xBABEFACE | |||||
| li x4, 0xBABE07CE | |||||
| #+end_src | |||||
| a) Which of these instructions will be split into ADDI LUI pairs? | |||||
| b) Why do the two last instructions need to be handled differently from each other? | |||||
| (hint: The parser and assembler in the test suite can help you answer this question) | |||||