Section 4.3: Sequential Y86 Implementation - Tracing an Register-Memory Move Instruction

1. Instruction Memory is combinational logic.
   1. We assume it has the following instructions stored starting at location 0:
      

      irmovl $11, %eax
      irmovl $7, %ecx
      rmmovl %eax, 8(%ecx)
      

      
      
   2. In binary the instuction memory contains (all values in hex)

      Address	Contents	Comments
      00	30	irmovl
      01	F0	rB=%eax
      02	0B	11=0x0B (little endian)
      03	00	second byte of 11 is 0 (little endian)
      04	00	third byte of 11 is 0 (little endian)
      05	00	fourth byte of 11 is 0 (little endian)
      06	30	irmovl
      07	F1	rB=%ecx=1
      08	07	7=0x07 (little endian)
      09	00	second byte of 7 is 0 (little endian)
      0A	00	third byte of 7 is 0 (little endian)
      0B	00	fourth byte of 7 is 0 (little endian)
      0C	40	rmmovl
      0D	01	rA=%eax=0, rB=%ecx=1
      0E	08	8=0x08 (little endian)
      0F	00	second byte of 8 is 0 (little endian)
      10	00	thrid byte of 8 is 0 (little endian)
      11	00	fourth byte of 8 is 0 (little endian)

2. The register file acts like combinational logic when reading values.
3. Each block of combinational logic has a fixed propgation delay of .1 (in some units)
4. We start our trace with the rmmovl instruction, so we have:
   
   PC = 0C
   %eax = 0B
   %ecx = 07

1. In Figure 4.27, coming out of the PC is 0C.
   1. Byte 0 = 40, Byte 1 = 01, Byte 2 = 08, Bytes 3-5 are 0
   2. icode = 4, ifun = 0
   3. Need regids = 1, Need valC = 1
   4. valP = 12 = 0x0C + 6
   5. rA = 0, rB = 1
   6. valC = 00000008
   7. Instr valid = 1
2. Transfer these results to Figure 4.23 and 4.28
3. In Figure 4.28:
   1. have rA=0, rB=1, icode=4
   2. srcA=rA= 0, srcB=rB=1
   3. valA = 0B, valB = 07
4. Transfer these values to Figures 4.23 and 4.29
5. In Figure 4.29
   1. valC = 00000008, valB = 07, icode=4, ifun=0
   2. ALUfun = 0 (0=add, 1=sub, 2=and, 3=xor)
   3. ALU-A output = valC = 00000008, ALU-B output = valB = 07
   4. SetCC output is 0
   5. valE = 0F (0x08 + 0x07 = 11 + 7 = 15 = 0x0F)
6. Transfer these values to Figures 4.23 and 4.30
7. In Figure 4.23, New PC output is valP = 12
8. In Figure 4.30, icode=4, valE = 0F, valA=0B.
   Since icode=4, Mem addr out is valE, and data in is valA, Mem read is 0, and Mem write is 1.
   On the next clock cycle, valA=0B will be stored in addres valE=0F.
9. Also on the next clock cycle, newPC=12 will be stored in the PC