This is the website for a prior semester's offering of CS 3330.

Your task

  1. Start with your HCL2 solution that implements irmovq, rrmovq, halt, and unconditional jmp. (If you don’t have a working HCL2 solution, fix it first.)

  2. Copy your HCL2 solution to new HCL file called seqlab.hcl.

  3. Add OPq, cmovXX, and rmmovq to the single-cycle processor in that HCL file. For OPq and cmovXX, you only need to implement the SF and ZF` condition codes. (We don’t care about overflow.)

  4. Test your solution with make test-seqlab.

  5. Submit your seqlab.hcl to archimedes.

Advice/Hints

Implementing OPq

The textbook’s Figure 4.18 (page 387) notes the following semantics for OPq:

Stage OPq rA, rB
Fetch icode:ifun ← M1[PC]
rA:rB ← M1[PC + 1]
valP ← PC + 2
Decode valA ← R[rA]
valB ← R[rB]
Execute valE ← valB OP valA
Set CC
Memory  
Writeback R[rB] ← valE
PC Update PC ← valP

All the interesting stuff is in the execute phase, including:

The ALU operation

  1. The ALU operation is essentially a MUX based on ifun with lines inside it like icode == OPQ && ifun == XORQ : reg_outputA ^ reg_outputB;.

The condition codes

  1. Create a register to store the condition codes inside of, like:

     register cC {
     SF:1 = 0;
     ZF:1 = 1;
 }

    (ZF defaulting to 1 is consistent with yis, but we won’t test what you choose as the initial value of the condition codes.)

  2. Record if the (signed) value of valE is <, =, or > 0 (using unsigned comparison operators) in the condition codes.

     c_ZF = (valE == 0);
 c_SF = (valE >= 0x8000000000000000);

  3. You must only update condition codes during an OPq instruction; other operations should not update them. Register banks like cC have a special input stall_C which, if 1, causes the registers to ignore inputs and keep their current value. So, we can use this to avoid updating the condition codes unless there’s an OPq:

     stall_C = (icode != OPQ);

Testing OPq

  1. If you run your simulator on y86/opq.yo, which is an assembled version of

    irmovq   $7, %rdx
irmovq   $3, %rcx
addq %rcx, %rbx # b = 3
subq %rdx, %rcx # c = -3 
andq %rdx, %rbx # b = 2
xorq %rcx, %rdx # d = -6
andq %rdx, %rsi

    you should see (without the -q flag, shown with some lines remove for brevity)

    +------------------- between cycles    0 and    1 ----------------------+
| register cC(N) { SF=0 ZF=1 }                                          |
    +------------------- between cycles    1 and    2 ----------------------+
| RAX:                0   RCX:                0   RDX:                7 |
| register cC(S) { SF=0 ZF=1 }                                          |
    +------------------- between cycles    2 and    3 ----------------------+
| RAX:                0   RCX:                3   RDX:                7 |
| register cC(S) { SF=0 ZF=0 }                                          |
    +------------------- between cycles    3 and    4 ----------------------+
| RAX:                0   RCX:                3   RDX:                7 |
| RBX:                3   RSP:                0   RBP:                0 |
| register cC(N) { SF=0 ZF=0 }                                          |
    +------------------- between cycles    4 and    5 ----------------------+
| RAX:                0   RCX: fffffffffffffffc   RDX:                7 |
| RBX:                3   RSP:                0   RBP:                0 |
| register cC(N) { SF=1 ZF=0 }                                          |
    +------------------- between cycles    5 and    6 ----------------------+
| RAX:                0   RCX: fffffffffffffffc   RDX:                7 |
| RBX:                3   RSP:                0   RBP:                0 |
| register cC(N) { SF=0 ZF=0 }                                          |
    +------------------- between cycles    6 and    7 ----------------------+
| RAX:                0   RCX: fffffffffffffffc   RDX: fffffffffffffffb |
| RBX:                3   RSP:                0   RBP:                0 |
| register cC(N) { SF=1 ZF=0 }                                          |
    +------------------- between cycles    7 and    8 ----------------------+
| RAX:                0   RCX: fffffffffffffffc   RDX: fffffffffffffffb |
| RBX:                3   RSP:                0   RBP:                0 |
| register cC(N) { SF=0 ZF=1 }                                          |
    +----------------------- halted in state: ------------------------------+
| RAX:                0   RCX: fffffffffffffffc   RDX: fffffffffffffffb |
| RBX:                3   RSP:                0   RBP:                0 |
| register cC(S) { SF=0 ZF=1 }                                          |

  2. You should also now be able to get the same results using your simulator as you get from tools/yis when running y86/prog1.yo through y86/prog4.yo (and y86/prog8.yo should still work too).

    We have supplied traces of the output for all the .yo files in testdata/seq-traces.

Implementing cmovXX

  1. cmovXX has the same icode as rrmovq, but non-zero ifuns, and will share much of the same logic with rrmovq.
  2. We suggest creating a wire called conditionsMet, and setting it using a MUX with entries like ifun == LE : C_SF || C_ZF; (C_SF are the outputs of the condition code registers from above).

  3. Once you have conditionsMet wire, in the writeback stage of cmovXX, make the reg_dstE (or reg_dstM, depending on how you implemented rrmovq) REG_NONE if conditionsMet is false.

    Recall that muxes execute only the first true case, so adding something like !conditionsMet && icode == CMOVXX : REG_NONE; before other cases when setting the dst_ should suffice.

    Remember that CMOVXX == RRMOVQ, so your conditionsMet wire should handle ifun == ALWAYS (i.e. ifun == 0). (rrmovq should just become a special case of cmovXX.)

Testing cmovXX

If you run your simulator on y86/cmovXX.yo, which is an assembled version of

irmovq $2766, %rbx  # 0xace → b
irmovq    $1, %rax  # 1 → a
andq    %rax, %rax  # set flags based on a (>)
cmovg   %rbx, %rcx  # move if g  (which it is): b → c  (c now 0xace)
cmovne  %rbx, %rdx  # move if ne (which it is): b → d  (d now 0xace)
irmovq   $-1, %rax  # -1 → a
andq    %rax, %rax  # set flags based on a (<)
cmovl   %rbx, %rsp  # move if l  (which it is): b → sp  (sp now 0xace)
cmovle  %rbx, %rbp  # move if le (which it is): b → bp  (bp now 0xace)
xorq    %rax, %rax  # a ^= a means 0 → a, and sets flags (=)
cmove   %rbx, %rsi  # move if e  (which it is): b → si  (si now 0xace)
cmovge  %rbx, %rdi  # move if ge (which it is): b → di  (di now 0xace)
irmovq $2989, %rbx  # 0xbad → b
irmovq    $1, %rax  # 1 → a
andq    %rax, %rax  # set flags based on a (>)
cmovl   %rbx, %rcx  # move if l  (which it is not): b → c  (not moved)
cmove   %rbx, %rdx  # move if e  (which it is not): b → d  (not moved)
irmovq   $-1, %rax  # -1 → a
andq    %rax, %rax  # set flags based on a (<)
cmovge  %rbx, %rsp  # move if ge (which it is not): b → sp  (not moved)
cmovg   %rbx, %rbp  # move if g  (which it is not): b → bp  (not moved)
xorq    %rax, %rax  # a ^= a means 0 → a, and sets flags (=)
cmovl   %rbx, %rsi  # move if l  (which it is not): b → si  (not moved)
cmovne  %rbx, %rdi  # move if ne (which it is not): b → di  (not moved)
irmovq    $0, %rbx  # 0 → b

you should end with registers

| RAX:                0   RCX:              ace   RDX:              ace |
| RBX:                0   RSP:              ace   RBP:              ace |
| RSI:              ace   RDI:              ace   R8:                 0 |

There is a trace of the expected cycle-by-cycle output in testdata/seq-traces/cmovXX.txt.

Implementing rmmovq

  1. The textbook’s Figure 4.19 (page 389) notes the following semantics for rmmovq:

    Stage rmmovq rA, D(rB)
    Fetch icode:ifun ← M1[PC]
    rA:rB ← M1[PC + 1]
    valC ← M8[PC+2]
    valP ← PC + 10
    Decode valA ← R[rA]
    valB ← R[rB]
    Execute valE ← valB + valC
    Memory M8[valE] ← valA
    Writeback  
    PC Update PC ← valP
  2. Memory is accessed by setting mem_addr to the memory address in question and either
    • setting mem_readbit to 0, mem_writebit to 1, and mem_input to the value to write to memory, which will cause the memory system to write a 8-byte value to memory; or
    • setting mem_readbit to 1 and mem_writebit to 0, which will cause the memory system to read a 8-byte value from memory into mem_output.
  3. You will need to compute the memory address as reg_outputB + valC (the book suggests you do this in the ALU, meaning the same mux you used for OPq’s adding and subtracting).

Testing rmmovq

If both rmmovq is implemented correctly, the test case y86/rmmovq-trivial.yo should result in address 0xa2 containing byte 0x08. This test case is an assembled version of:

irmovq $2, %rax
irmovq $8, %rbx
rmmovq %rbx, 160(%rax)

Additional Testing

You can run make test-seqlab to test your processor over a variety of files. This will compare the output on the list of .yo files in testdata/seqlab-tests.txt to the reference outputs in testdata/seq-reference.