Chapter 4

Chapter 4

The Processor

CprE 381 Computer Organization and Assembly Level Programming, Fall 2013

Zhao ZhangIowa State UniversityRevised from original slides provided by MKP

Week 8 Overview CPU design overview Datapath and Control Control Unit ALU Control Unit

Chapter 1 — Computer Abstractions and Technology — 2

Announcements Mini-project B starts in week 9

Mini-projects B and C will be revised The grading scale will be discussed by

Friday (week 8)

Chapter 1 — Computer Abstractions and Technology — 3

Chapter 4 — The Processor — 4

Introduction CPU performance factors

Instruction count Determined by ISA and compiler

CPI and Cycle time Determined by CPU hardware

We will examine two MIPS implementations A simplified, single-cycle version A more realistic, pipelined version

§4.1 Introduction

Nine-Instruction MIPS We will first use a MIPS subset of nine

instructions, then extend the subset It’s enough to illustrate the most aspects of CPU

design, particularly datapath and control design

Memory reference: LW and SW

Arithmetic/logic: ADD, SUB, AND, OR, SLT

Branch: BEQ, BNE

Chapter 1 — Computer Abstractions and Technology — 5

Chapter 4 — The Processor — 6

Instruction Execution PC instruction memory, Fetch instruction Register numbers register file, Read registers

Then, depending on instruction class Execute: Use ALU to calculate

Arithmetic result Memory address for load/store Branch target address

Memory access: Access data memory for load/store Register writeback: Write data back to registers

PC update (for all): PC target address or PC + 4

Chapter 4 — The Processor — 7

CPU Overview

Next Sequential PC= PC + 4

Branch Target= (PC+4)+offset

A Sketchy view

An instruction may change1. PC (all instructions)2. Some register (arithmetic/logic, load)3. Some memory word/halfword/byte (store)

Chapter 4 — The Processor — 8

Multiplexers Can’t just join

wires together Use multiplexers

What would happen if you just join signals in VHDL?

Chapter 4 — The Processor — 9

Control

Control signals: mux select, read/write enable, ALU opcode, etc.

Chapter 4 — The Processor — 10

Logic Design Basics§4.2 Logic D

esign Conventions

Combinational element Operate on data Output is a function of input

State (sequential) elements Store information Output is a function of internal state and

input

Chapter 4 — The Processor — 11

Combinational Elements

AND-gate Y = A & B

AB

Y

I0I1

YMux

S

Multiplexer Y = S ? I1 : I0

A

B

Y+

A

B

YALU

F

Adder Y = A + B

Arithmetic/Logic Unit Y = F(A, B)

Chapter 4 — The Processor — 12

Sequential Elements Register: stores data in a circuit

Uses a clock signal to determine when to update the stored value

Edge-triggered: update when Clk changes from 0 to 1

Data output Q is stable for a clock cycle

D

Clk

QClk

D

Q

Chapter 4 — The Processor — 13

Sequential Elements Register with write control

Only updates on clock edge when write control input is 1

VHDL: rising_edge(Clk) AND Write Used when stored value is required later

D

Clk

Q

Write

Write

D

Q

Clk

Chapter 4 — The Processor — 14

Clocking Methodology Combinational logic transforms data during

clock cycles Input from state elements Output must stabilize within one cycle

Longest delay determines clock period Output to state element at the next rising edge

Clocking Methodology Processor is a big state machine

Works like a Moore machine in non-I/O phase Output is a function of the state

States include PC, all registers and memory contents

Chapter 1 — Computer Abstractions and Technology — 15

Chapter 4 — The Processor — 16

Building a Datapath Datapath elements

Elements that process data and addressesin the CPU

Registers, ALUs, mux’s, memories, …

We will build a MIPS datapath incrementally Refining the overview design

§4.3 Building a D

atapath

Chapter 4 — The Processor — 17

Instruction Fetch

32-bit register

Increment by 4 for next instruction

Datapath elements: PC register, instruction memory, 32-bit adder

Chapter 4 — The Processor — 18

R-Format Instructions Read two register operands Perform arithmetic/logical operation Write register result

Datapath elements: Register file, ALU

Chapter 4 — The Processor — 19

Load/Store Instructions Read register operands Calculate address using 16-bit offset

Use ALU, but sign-extend offset Load: Read memory and update register Store: Write register value to memory

Datapath elements: Data memory, sign extender

Chapter 4 — The Processor — 20

Branch Instructions Read register operands Compare operands

Use ALU, subtract and check Zero output Calculate target address

Sign-extend displacement Shift left 2 places (word displacement) Add to PC + 4

Already calculated by instruction fetch

Chapter 4 — The Processor — 21

Branch Instructions

Justre-routes wires

Sign-bit wire replicatedNew: Shifter, 2nd 32-bit Adder

Chapter 4 — The Processor — 22

Composing the Elements First-cut data path does an instruction in

one clock cycle Each datapath element can only do one

function at a time Hence, we need separate instruction and data

memories Use multiplexers where alternate data

sources are used for different instructions

Chapter 4 — The Processor — 23

R-Type/Load/Store Datapath

Chapter 4 — The Processor — 24

Full Datapath

Chapter 4 — The Processor — 25

Performance Issues Longest delay determines clock period

Critical path: load instruction Instruction memory register file ALU data

memory register file Not every instruction requires the same time

Chapter 4 — The Processor — 26

Performance Issues Some instructions may take substantially longer

time, e.g. multiply/division Not feasible to vary clock cycle for different

instructions Must use the worst-case delay as the clock cycle Violates design principle making the common

case fast We will improve performance by pipelining

Chapter 4 — The Processor — 27

ALU Control ALU used for

Load/Store: F = add Branch: F = subtract R-type: F depends on funct field

§4.4 A S

imple Im

plementation S

cheme

ALU control Function

0000 AND

0001 OR

0010 add

0110 subtract

0111 set-on-less-than

1100 NOR

Chapter 4 — The Processor — 28

ALU Control Assume 2-bit ALUOp derived from opcode

Combinational logic derives ALU control

opcode ALUOp Operation funct ALU function ALU control

lw 00 load word XXXXXX add 0010

sw 00 store word XXXXXX add 0010

beq 01 branch equal XXXXXX subtract 0110

R-type 10 add 100000 add 0010

subtract 100010 subtract 0110

AND 100100 AND 0000

OR 100101 OR 0001

set-on-less-than 101010 set-on-less-than 0111

VHDL Notes How to program the ALU control? -- Behavior style process (alu_op, funct) begin case alu_op is when ‘00’ => alu_code <= ‘0010’; when ’01’ => … end case; end process;

Chapter 1 — Computer Abstractions and Technology — 29

Chapter 4 — The Processor — 30

The Main Control Unit Control signals derived from instruction

0 rs rt rd shamt funct

31:26 5:025:21 20:16 15:11 10:6

35 or 43 rs rt address

31:26 25:21 20:16 15:0

4 rs rt address

31:26 25:21 20:16 15:0

R-type

Load/Store

Branch

opcode always read

read, except for load

write for R-type and load

sign-extend and add

Chapter 4 — The Processor — 31

Datapath With Control

Summary of Control Signals RegDst: Write to register rt or rd? ALUSrc: Immediate to ALU? MemtoReg: Write memory or ALU output? RegWrite: Write to regfile at all? MemRead: Read from Data Memory? MemWrite: Write to the Data Memory? Branch: Is it a branch intruction? ALUOp[1:0]: ALU control field

Chapter 1 — Computer Abstractions and Technology — 32

Chapter 4 — The Processor — 33

R-Type Instruction

R-Type: Control Signals

RegDst ALUSrc MemtoRegRegWriteMemReadMemWriteBranchALUOp[1:0]

1 (write to rd)0 (No immediate)0 (wrote not from memory)1 (does write regfile)

0 (no memory read)

0 (no memory write)

0 (does write regfile)

10 (R-type ALU op)

Chapter 1 — Computer Abstractions and Technology — 34

Chapter 4 — The Processor — 35

Load Instruction

Load: Control Signals

RegDst ALUSrc MemtoRegRegWriteMemReadMemWriteBranchALUOp[1:0]

011110000

Chapter 1 — Computer Abstractions and Technology — 36

Store: Control Signals

RegDst ALUSrc MemtoRegRegWriteMemReadMemWriteBranchALUOp[1:0]

X1X001000

Chapter 1 — Computer Abstractions and Technology — 37

Chapter 4 — The Processor — 38

Branch-on-Equal Instruction

BEQ: Control Signals

RegDst ALUSrc MemtoRegRegWriteMemReadMemWriteBranchALUOp[1:0]

X0X000101

Chapter 1 — Computer Abstractions and Technology — 39

Control Signal Setting What’re the control signal values for each

instruction or instruction type?

Inst Reg-Dst

ALU-Src

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp1

ALUOp0

R- 1 0 0 1 0 0 0 1 0

lw 0 1 1 1 1 0 0 0 0

sw X 1 X 0 0 1 0 0 0

beq X 0 X 0 0 0 1 0 1

Chapter 1 — Computer Abstractions and Technology — 40

Note: “R-” means R-format

VHDL Notes How to program the control?

entity control is port (op_code : in m32_6bits; reg_dst : out m32_1bit; alu_src : out m32_1bit; mem_to_reg : out m32_1bit; reg_write : out m32_1bit; mem_read : out m32_1bit; mem_write : out m32_1bit; branch : out m32_1bit; alu_op : out m32_2bits);end control;

Chapter 1 — Computer Abstractions and Technology — 41

VHDL Notesarchitecture rom of control is subtype code_t is m32_vector(8 downto 0); type rom_t is array (0 to 63) of code_t; -- The ROM content for control signals signal rom : rom_t := ( 00 => "100100010", -- R-type 35 => "011110000", -- LW … -- More for other instructions others=>"000000000");

begin (reg_dst, alu_src, mem_to_reg, reg_write, mem_read, mem_write, branch, alu_op(1), alu_op(0)) <= rom(to_integer(unsigned(op_code)));end rom;

Chapter 1 — Computer Abstractions and Technology — 42

Chapter 4 — The Processor — 43

Implementing Jumps

Jump uses word address Update PC with concatenation of

Top 4 bits of old PC 26-bit jump address 00

Need an extra control signal decoded from opcode

2 address

31:26 25:0

Jump

Chapter 4 — The Processor — 44

Datapath With Jumps Added

Grading Scale Tentative grading scale

A: 90, A-: 87

B+: 84, B: 80, B-: 75

C+: 70, C: 65, C-: 60

D: 50 There will be a bonus in lab projects

Chapter 1 — Computer Abstractions and Technology — 45

Mini-Project B, Tentative

Implement single-cycle processor (SCP). There will be three parts

1. Part 1, SCPv1: Implement the nine-instruction ISA plus the J instruction

2. Part 2, SCPv2a: Support all the instructions needed to run bubble sorting

3. Part 3, SCPv2b: Detailed modeling of data elements

Chapter 1 — Computer Abstractions and Technology — 46

Mini-Project B Bonus part, SCPv3: Support all integer

instructions on the green sheet, due in the last lab

Some support files will be provided High-level modeling of Register File, ALU,

Adder, to be used in Parts 1 and 2

Partial sample VHDL code will be provided

Chapter 1 — Computer Abstractions and Technology — 47

Mini-Project B The CPU composition must be strongly

structural Parts 1 and 2 may use behavior/dataflow

modeling for data elements Part 3 must use detailed modeling for data

elements – Reuse your VHDL code in the labs

Chapter 1 — Computer Abstractions and Technology — 48

Extend Single-Cycle MIPS

Consider the following instructions addi: add immediate sll: Shift left logic by a constant bne: branch if not equal jal: Jump and link jr: Jump register

Chapter 1 — Computer Abstractions and Technology — 49

Chapter 4 — The Processor — 50

SCPv0: R-Format, LW/SW, BEQ

Chapter 4 — The Processor — 51

SCPv1: R-Format, LW/SW, BEQ, J

SCPv1: Control Signals What’re the control signal values for each

instruction or instruction type?

Inst Reg-Dst

ALU-Src

Mem-toReg

Reg-Write

MemRead

MemWrite

Branch

ALUOp1

ALUOp0

Jump

R- 1 0 0 1 0 0 0 1 0 0

lw 0 1 1 1 1 0 0 0 0 0

sw X 1 X 0 0 1 0 0 0 0

beq X 0 X 0 0 0 1 0 1 0

j X X X 0 0 0 0 X X 1

Chapter 1 — Computer Abstractions and Technology — 52

Note: “R-” means R-format

Extend the Single-Cycle Processor

For each instruction, do we need

1. Any new or revised datapath element(s)?

2. Any new control signal(s)?

Then revise, if necessary,

3. Datapath: Add new elements or revise existing ones, add new connections

4. Control Unit: Add/extend control signals, extend the truth table

5. ALU Control: Extend the truth tableChapter 1 — Computer Abstractions and Technology — 53

SCPv0 + ADDI

addi rs, rt, immediate

R[rt] = R[rs]+SignExtImm Read register operands (only one is used) Sign extend the immediate (in parallel) Perform arithmetic/logical operation Write register result

Chapter 1 — Computer Abstractions and Technology — 54

001000 rs rt immediate

31:26 25:21 20:16 15:0

SCPv0 + ADDI

Chapter 1 — Computer Abstractions and Technology — 55

What changes to this baseline?

Chapter 4 — The Processor — 56

SCPv0 + ADDI

Do we need new or revised datapath elements?

SCPv0 + ADDI Do we need new or revised datapath

elements? Do we need new control signal(s)?

Inst Reg-Dst

ALU-Src

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp1

ALUOp0

R- 1 0 0 1 0 0 0 1 0

lw 0 1 1 1 1 0 0 0 0

sw X 1 X 0 0 1 0 0 0

beq X 0 X 0 0 0 1 0 1

addi

Chapter 1 — Computer Abstractions and Technology — 57

SCPv0 + ADDI Like LW

I-format instruction Write to register[rt] Use add operation

Inst Reg-Dst

ALU-Src

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp1

ALUOp0

R- 1 0 0 1 0 0 0 1 0

lw 0 1 1 1 1 0 0 0 0

sw X 1 X 0 0 1 0 0 0

beq X 0 X 0 0 0 1 0 1

addi 0 1 0 1 0 0 0 0 0

Chapter 1 — Computer Abstractions and Technology — 58

Like R-format arithmetic Write ALU result to

register file

SCPv0 + SLL

sll rd, rs, shamt

R[rd] = R[rt]<<shamt Read register operands (only one is used) Perform shift operation Write register result

Note: sllv rd, rt, rs for shift left logic variable

Chapter 1 — Computer Abstractions and Technology — 59

000000 rs rt rd shamt 000000

31:26 5:025:21 20:16 15:11 10:6

SCPv0 + SLL

Chapter 1 — Computer Abstractions and Technology — 60

What changes to the datapath elements?

SCPv0 + SLL

Chapter 1 — Computer Abstractions and Technology — 61

ALU needs to do the shift operation

ALU 1st input needs anothersource: shamt extended to 32-bit

SCPv0 + SLL Add another source to the 1st

input of ALU Shamt: Instruction[10-6]

Add a Mux and ALUSrc1 control line 0: R[rs] 1: Shamt (sign-extended)

Rename ALUSrc to ALUSrc2 Extend ALU control

Add an ALU control code for SLL

Chapter 1 — Computer Abstractions and Technology — 62

Chapter 4 — The Processor — 63

SCPv0 + SLL Extend ALU control: Choose a code of

your choice (kkkk shown in the table)

opcode ALUOp Operation funct ALU function ALU control

lw 00 load word XXXXXX add 0010

sw 00 store word XXXXXX add 0010

beq 01 branch equal XXXXXX subtract 0110

R-type 10 add 100000 add 0010

subtract 100010 subtract 0110

AND 100100 AND 0000

OR 100101 OR 0001

set-on-less-than 101010 set-on-less-than 0111

shift-left-logic 000000 shift-left-logic kkkk

SCPv0 + SLLInst Reg-

DstALU-Src

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp

R- 1 0 0 1 0 0 0 1 0

lw 0 1 1 1 1 0 0 0 0

sw X 1 X 0 0 1 0 0 0

beq X 0 X 0 0 0 1 0 1

sll

Chapter 1 — Computer Abstractions and Technology — 64

Inst Reg-Dst

ALUSrc1

ALU-Src2

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp

R- 1 0 0 0 1 0 0 0 1 0

lw 0 0 1 1 1 1 0 0 0 0

sw X 0 1 X 0 0 1 0 0 0

beq X 0 0 X 0 0 0 1 0 1

sll 1 1 0 0 1 0 0 0 1 0

SCPv0 + BNEbne rs, rt, label

PC = (R[Rs]!=R[rt]) ? PC+4+(SignExtImm<<2) : PC+4

Read register operands Compare operands

Use ALU, subtract and check Zero output Calculate target address

Sign-extend displacement Shift left 2 places (word displacement) Add to PC + 4

Already calculated by instruction fetchChapter 1 — Computer Abstractions and Technology — 65

000101 rs rt offset

31:26 25:21 20:16 15:0

Chapter 4 — The Processor — 66

SCPv0 + BNE

Make what changes tothe datapath?

SCPv0 + BNE Extend Branch to two bits

10: Branch-Equal 11: Branch-Not-Equal

Replace the AND gate with the following logic

Can use a different truth table

Chapter 1 — Computer Abstractions and Technology — 67

Branch Zero Branch taken?

1 0 1 1

1 1 0 1

otherwise 0

SCPv0 + BNEInst Reg-

DstALU-Src

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp

R- 1 0 0 1 0 0 0 1 0

lw 0 1 1 1 1 0 0 0 0

sw X 1 X 0 0 1 0 0 0

beq X 0 X 0 0 0 1 0 1

bne

Chapter 1 — Computer Abstractions and Technology — 68

Inst Reg-Dst

ALU-Src

Mem-toReg

Reg-Write

Mem-Read

Mem-Write

Branch

ALUOp

R- 1 0 0 1 0 0 0 0 1 0

lw 0 1 1 1 1 0 0 0 0 0

sw X 1 X 0 0 1 0 0 0 0

beq X 0 X 0 0 0 1 0 0 1

bne X 0 X 0 0 0 1 1 0 1

SCPv1 + JAL

jal target

PC = JumpAddrR[31] = PC+4

Jump uses word address Update PC with JumpAddr: concatenation

of top 4 bits of old PC, 26-bit jump address, and 00 (called pseudo-direct)

Save PC+4 to $raChapter 1 — Computer Abstractions and Technology — 69

000011 address

31:26 25:0

Chapter 4 — The Processor — 70

SCPv1 + JAL

Make what changes tothe datapath?