作业2—lab2指令集流水化实验—计算机体系结构.docx

Lab2 指令集流水化实验报告学号 姓名：崔雪莹Exercise 1:What is hex code for NOP or SLL R0, R0, R0 00 00 00 00 H .Exercise 2:By looking at the binary code of the instruction, discover the following fields: Rs: 00100 , Rt: 00011 , Rd 00101 , 16-bit Imm (Imm16)0010 1000 0010 0000 , Shift amount 00000 , Function Code 100000 , Write MemWrite 0 and RegWrite 1 for this instruction. What is use of 16-bit Immediate field in this instruction? 16位立即数是对branch指令来说的跳转目的地址的偏移地址，add指令没有作用。If the same instruction were a Branch instruction, the immediate field would have represented a Branch Offset. What is the potential Target Address 43092 立即数十进制表示为10272，目的地址为10272*4+2004=43092 Exercise 3:LW Instruction Rs: 01000 , Rt 00111 , Rd 32（00000） , 16-bit Imm (Imm16) 1111 1111 0011 1000 , Shift amount 29（11101）, Function Code 100011 , Write MemWrite 0 and RegWrite 1 for this instruction. What is use of Function filed and Shift Amount in this instruction? 操作数表明操作为LW读操作，是指令操作的唯一标记，移位量是相对与移位指令来说移几位，在LW指令没有作用。If the same instruction were a Branch instruction, the immediate field would have represented a Branch Offset. What is the potential Target Address 1204 ，（-200*4）+2004=1204 。Exercise 4: Branch Target Found Adress 1204 , Is It same as EX.3是? B input to ALU -200 , Why? 因为对于I型指令只有rs，rt和立即数，没有rd，是rs和立即数Imm运算的结果进行操作，所以B最终送ALU的是立即数而不是rt, ALUOut800 , ALUOP = ADD , Why 因为立即数是偏移量，所以运算器是对R8和-200求和，得到的为目的地址. LMD 525252, Rdest 7 why this? 因为没有rd，但是是取800地址的内容，回送到R7里。RegWrite 1 ,MemWrite 0 , What are two values at the input of very last MUX in the WB stage 525252 , 800 , Which value is selected 525252 , why因为最终写回的是525252的值，而不是ALU运算的结果800 . Final Register File Values. R7 , 525252 , R8 , 1000 .Exercise 5:SW instruction Rs: 01000 , Rt 00111 , Rd 32（00000） , 16-bit Imm (Imm16) 1111 1111 0011 1000 ,, Shift amount 28（11100）, Function Code 101011 , Write MemWrite 1 and RegWrite 0 for this instruction. What