Building a Computer I wonder where this goes? MIPS Kit ALU 1

Slides:



Advertisements
Similar presentations
1 Lecture 3: MIPS Instruction Set Today’s topic:  More MIPS instructions  Procedure call/return Reminder: Assignment 1 is on the class web-page (due.
Advertisements

Lecture 5: MIPS Instruction Set
ELEN 468 Advanced Logic Design
COMP541 Datapath & Single-Cycle MIPS
COMP541 Datapath & Single-Cycle MIPS
L18 – Pipeline Issues 1 Comp 411 – Spring /03/08 CPU Pipelining Issues Finishing up Chapter 6 This pipe stuff makes my head hurt! What have you.
L17 – Pipeline Issues 1 Comp 411 – Fall /1308 CPU Pipelining Issues Finishing up Chapter 6 This pipe stuff makes my head hurt! What have you been.
The Processor 2 Andreas Klappenecker CPSC321 Computer Architecture.
Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved. Digital Architectures1 Machine instructions execution steps (1) FETCH = Read the instruction.
L16- Building a Computer 1 Comp 411 – Spring /27/08 Building a Computer I wonder where this goes? Instruction Memory A D 0 1 MIPS Kit ALU AB.
Fall EE 333 Lillevik 333f06-l7 University of Portland School of Engineering Computer Organization Lecture 7 ALU design MIPS data path.
COSC 3430 L08 Basic MIPS Architecture.1 COSC 3430 Computer Architecture Lecture 08 Processors Single cycle Datapath PH 3: Sections
6.004 – Fall /29/0L15 – Building a Beta 1 Building the Beta.
COMP541 Datapaths II & Single-Cycle MIPS
Computer Organization and Design Building a Computer!
CDA 3101 Fall 2013 Introduction to Computer Organization
CS2100 Computer Organisation The Processor: Datapath (AY2015/6) Semester 1.
How Computers Work Lecture 3 Page 1 How Computers Work Lecture 3 A Direct Execution RISC Processor: The Unpipelined BETA.
EE 3755 Datapath Presented by Dr. Alexander Skavantzos.
Computer Organization and Design Pipelining Montek Singh Mon, Dec 2, 2013 Lecture 16.
Computer Organization and Design Pipelining Montek Singh Dec 2, 2015 Lecture 16 (SELF STUDY – not covered on the final exam)
COM181 Computer Hardware Lecture 6: The MIPs CPU.
Computer Organization and Design Building a Computer! Montek Singh Nov 18-20, 2015 Lecture 14.
L17 – Pipeline Issues 1 Comp 411 – Fall /23/09 CPU Pipelining Issues Read Chapter This pipe stuff makes my head hurt! What have you been.
MIPS Processor.
Access the Instruction from Memory

CS 230: Computer Organization and Assembly Language
Single-Cycle Datapath and Control
Computer Organization and Design Instruction Sets - 2
COMP541 Datapaths I Montek Singh Mar 28, 2012.
IT 251 Computer Organization and Architecture
Introduction CPU performance factors
/ Computer Architecture and Design
ELEN 468 Advanced Logic Design
Computer Organization and Design Instruction Sets - 2
RISC Concepts, MIPS ISA Logic Design Tutorial 8.
Processor Architecture: Introduction to RISC Datapath (MIPS and Nios II) CSCE 230.
Computer Organization and Design Building a Computer!
Processor (I).
CS/COE0447 Computer Organization & Assembly Language
Computer Organization and Design Instruction Sets
CSCI206 - Computer Organization & Programming
CSCE 212 Chapter 5 The Processor: Datapath and Control
Pipelining Read Chapter
Building a Computer I wonder where this goes? MIPS Kit ALU 1
Yuanqing Cheng Great Ideas in Computer Architecture Lecture 11: RISC-V Processor Datapath Yuanqing Cheng
MIPS Processor.
Datapath & Control MIPS
ECE232: Hardware Organization and Design
Levels in Processor Design
Rocky K. C. Chang 6 November 2017
Guest Lecturer TA: Shreyas Chand
Lecture 9. MIPS Processor Design – Decoding and Execution
Systems Architecture I
COMS 361 Computer Organization
COSC 2021: Computer Organization Instructor: Dr. Amir Asif
Basic Building Blocks Multiplexer Demultiplexer Adder +
Computer Organization and Design Building a Computer!
COMP541 Datapaths I Montek Singh Mar 18, 2010.
Building a Computer! Don Porter Lecture 14.
Computer Architecture Processor: Datapath
Basic Building Blocks Multiplexer Demultiplexer Adder +
Computer Organization and Design Building a Computer!
Now that’s what I call dirty laundry
The Processor: Datapath & Control.
COMS 361 Computer Organization
What You Will Learn In Next Few Sets of Lectures
MIPS Processor.
CS/COE0447 Computer Organization & Assembly Language
Presentation transcript:

Building a Computer I wonder where this goes? MIPS Kit ALU 1 Instruction Memory A D 1

“Motivating Force” or “Inciting Incident” THIS IS IT! “Motivating Force” or “Inciting Incident” This is the point in the course where the PLOT actually begins. We are now ready to build a computer. The ingredients are all in place, now it is time to build a legitimate computer. One that executes instructions, much the way any other desktop, PDA, or other computer does.

Review: The MIPS ISA OP 6 5 16 26 The MIPS instruction set as seen from a Hardware Perspective 000000 rs rt rd func shamt R-type: ALU with Reguster operands Reg[rd]  Reg[rs] op Reg[rt] rs rt 001XXX immediate I-type: ALU with constant operand Reg[rt]  Reg[rs] op SEXT(immediate) Instruction classes distinguished by types: 3-operand ALU ALU w/immediate Loads/Stores Branches Jumps rs rt 10X011 immediate I-type: Load and Store Reg[rt]  Mem[Reg[rs] + SEXT(immediate)] Mem[Reg[rs] + SEXT(immediate)]  Reg[rt] rs rt 10X011 immediate I-type: Branch Instructions if (Reg[rs] == Reg[rt]) PC  PC + 4 + 4*SEXT(immediate) if (Reg[rs] != Reg[rt]) PC  PC + 4 + 4*SEXT(immediate) 00001X 26-bit constant J-type: jump PC  (PC & 0xf0000000) | 4*(immediate)

Design Approach Incremental Featurism + Each instruction class can be implemented using a simple component repertoire. We’ll try implementing data paths for each class individually, and merge them (using MUXes, etc). Steps: 1. 3-Operand ALU instructions 2. ALU w/immediate instructions 2. Load & Store Instructions 3. Jump & Branch instructions 4. Exceptions Our Bag of Components: Registers 1 Muxes ALU A B ALU & adders + Data Memory WD A RD R/W Register File (3-port) RA1 RA2 WA WE RD1 RD2 Instruction D Memories

A Few ALU Tweaks Let’s review the ALU that we built a few lectures ago. (With a few minor additions) A B 5-bit ALUFN Sub Bool Shft Math OP 0 XX 0 1 A+B 1 XX 0 1 A-B X X0 1 1 0 X X1 1 1 1 X 00 1 0 B<<A X 10 1 0 B>>A X 1 1 1 0 B>>>A X 00 0 0 A & B X 01 0 0 A | B X 10 0 0 A ^ B X 1 1 0 0 A | B Add/Sub Bidirectional Barrel Shifter Boolean Sub Bool 0 1 Shft 1 0 Math … 1 0 Flags V,C N Flag R Z Flag

Instruction Fetch/Decode Use a counter to FETCH the next instruction: PROGRAM COUNTER (PC) use PC as memory address add 4 to PC, load new value at end of cycle fetch instruction from memory º use some instruction fields directly (register numbers, 16-bit constant) º use bits <31:26> and <5:0> to generate controls PC 00 A Instruction 32 Memory + 1 30 +4 D 32 32 INSTRUCTION WORD FIELDS Can be built with half-adders OP[31:26], FUNC[5:0] Control Logic CONTROL SIGNALS

3-Operand ALU Data Path rs rt rd WERF! 000000 100XXX 00000 R-type: ALU with Reguster operands Reg[rd]  Reg[rs] op Reg[rt] PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> RA1 Register RA2 Rd: <15:11> WA File WD RD1 RD2 WE WERF 32 32 Control Logic A B ALU ALUFN WERF! ALUFN WERF 32

Shift Instructions rs rt rd 000000 000XXX shamt R-type: ALU with Register operands sll: Reg[rd]  Reg[rt] (shift) shamt sllv: Reg[rd]  Reg[rt] (shift) Reg[rs] PC 00 Instruction Memory A D +4 Rt: <20:16> Rs: <25:21> RA1 Register RA2 Rd: <15:11> WA File WD RD1 RD2 WE WERF shamt:<10:6> Control Logic 1 ASEL A B ALU ALUFN ASEL! ALUFN WERF ASEL 32

ALU with Immediate rs rt 001XXX I-type: ALU with constant operand Reg[rt]  Reg[rs] op SEXT(immediate) PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> BSEL Rd:<15:11> Rt:<20:16> 1 RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 shamt:<10:6> imm: <15:0> SEXT SEXT 1 BSEL Control Logic SEXT A B ALU BSEL! BSEL ALUFN SEXT 15 14 13 12 … 3 2 1 0 ALUFN ... WERF ASEL 31 30 … 17 16 15 14 13 12 … 3 2 1 0

Load Instruction rs rt 100011 I-type: Load immediate I-type: Load Reg[rt]  Mem[Reg[rs] + SEXT(immediate)] PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> BSEL Rd:<15:11> Rt:<20:16> 1 RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 shamt:<10:6> Imm: <15:0> SEXT SEXT Control Logic 1 BSEL SEXT A B BSEL ALU WD R/W Wr ALUFN WDSEL ALUFN Data Memory Wr 32 Adr RD WERF ASEL 32 0 1 2 WDSEL

Store Instruction rs rt 10X011 I-type: Store immediate I-type: Store Mem[Reg[rs] + SEXT(immediate)]  Reg[rt] PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> BSEL Rd:<15:11> Rt:<20:16> 1 RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 SEXT BSEL shamt:<10:6> Imm: <15:0> Control Logic 32 No WERF! SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr WERF ASEL WDSEL 0 1 2

PC<31:29>:J<25:0>:00 JMP Instructions PC<31:29>:J<25:0>:00 00001X 26-bit constant J-type: j: PC  (PC & 0xf0000000) | 4*(immediate) jal: PC  (PC & 0xf0000000) | 4*(immediate); Reg[31]  PC + 4 PCSEL 6 5 4 3 2 1 PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> WASEL J:<25:0> Rd:<15:11> 1 2 3 Rt:<20:16> RA1 Register RA2 “31” WA WA File WD “27” RD1 RD2 WE WERF ASEL 1 SEXT BSEL shamt:<10:6> Imm: <15:0> Control Logic PCSEL WASEL SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr WERF ASEL PC+4 32 WDSEL 0 1 2

PC<31:29>:J<25:0>:00 BEQ/BNE Instructions PC<31:29>:J<25:0>:00 rs rt BT 10X011 immediate PCSEL 1 2 3 4 5 6 R-type: Branch Instructions if (Reg[rs] == Reg[rt]) PC  PC + 4 + 4*SEXT(immediate) if (Reg[rs] != Reg[rt]) PC  PC + 4 + 4*SEXT(immediate) PC 00 Instruction Memory A D That “x4” unit is trivial. I’ll just wire the input shifted over 2–bit positions. +4 Rs: <25:21> Rt: <20:16> WASEL Rd:<15:11> Rt:<20:16> 1 2 3 “31” “27” J:<25:0> RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 SEXT BSEL shamt:<10:6> Imm: <15:0> Z x4 Why add, another adder? Couldn’t we reuse the one in the ALU? Nope, it needs to do a subtraction. Control Logic + PCSEL WASEL BT SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr Z WERF ASEL PC+4 32 WDSEL 0 1 2

Jump Indirect Instructions PC<31:29>:J<25:0>:00 000000 rs rt rd 00100X 00000 JT BT PCSEL 1 2 3 4 5 6 R-type: Jump Indirect, Jump and Link Indirect jr: PC  Reg[rs] jalr: PC  Reg[rs], Reg[rd]  PC + 4 PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> WASEL Rd:<15:11> Rt:<20:16> 1 2 3 “31” “27” J:<25:0> RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 SEXT BSEL shamt:<10:6> Imm: <15:0> JT Z x4 Control Logic + PCSEL WASEL BT SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr Z WERF ASEL PC+4 32 WDSEL 0 1 2

PC<31:29>:J<25:0>:00 Loose Ends PC<31:29>:J<25:0>:00 rs rt JT BT 001XXX immediate PCSEL 1 2 3 4 5 6 I-type: set on less than & set on less than unsigned immediate slti: if (Reg[rs] < SEXT(imm)) Reg[rt]  1; else Reg[rt]  0 sltiu: if (Reg[rs] < SEXT(imm)) Reg[rt]  1; else Reg[rt]  0 PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> Reminder: To evaluate (A < B) we first compute A-B and look at the flags. LT = N  V LTU = C WASEL Rd:<15:11> Rt:<20:16> 1 2 3 “31” “27” J:<25:0> RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 SEXT BSEL shamt:<10:6> Imm: <15:0> Z N JT V C x4 Control Logic + PCSEL WASEL BT SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr N V C Z WERF ASEL PC+4 32 WDSEL 0 1 2

PC<31:29>:J<25:0>:00 More Loose Ends PC<31:29>:J<25:0>:00 000000 rs rt rd 10101X 00000 JT BT PCSEL 1 2 3 4 5 6 R-type: set on less than & set on less than unsigned slt: if (Reg[rs] < Reg[rt]) Reg[rd]  1; else Reg[rd]  0 sltu: if (Reg[rs] < Reg[rt]) Reg[rd]  1; else Reg[rd]  0 PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> WASEL Rd:<15:11> Rt:<20:16> 1 2 3 “31” “27” J:<25:0> RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF ASEL 1 SEXT BSEL shamt:<10:6> Imm: <15:0> N JT Z V C x4 Control Logic + PCSEL WASEL BT SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr N V C Z WERF ASEL PC+4 32 WDSEL 0 1 2

PC<31:29>:J<25:0>:00 LUI Ends PC<31:29>:J<25:0>:00 rt JT BT 001XXX 00000 immediate PCSEL 1 2 3 4 5 6 I-type: Load upper immediate lui: Reg[rt]  Immediate << 16 PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> WASEL Rd:<15:11> Rt:<20:16> 1 2 3 “31” “27” J:<25:0> RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF Imm: <15:0> SEXT JT SEXT Z N V C x4 shamt:<10:6> Control Logic + “16” 1 2 ASEL 1 BSEL PCSEL WASEL BT SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr N V C Z WERF ASEL PC+4 32 WDSEL 0 1 2

Reset, Interrupts, and Exceptions FIRST, we need some way to get our machine into a known initial state. This doesn’t mean that all registers will be initialized, just that we’ll know where to fetch the first instruction. We’ll call this control input, RESET We’d also like RECOVERABLE INTERRUPTS for • FAULTS (eg, Illegal Instruction) - CPU or SYSTEM generated [synchronous] • TRAPS & system calls (eg, read-a-character) - CPU generated [synchronous] • I/O events (eg, key struck) - externally generated [asynchronous] EXCEPTION GOAL: Interrupt running program, invoke exception handler, return to continue execution. These are “Software” notions of synchrony and asynchrony. (Implemented as an “agreed upon” Illegal instruction)

PC<31:29>:J<25:0>:00 Exceptions 0x80000000 0x80000040 PC<31:29>:J<25:0>:00 0x80000080 JT BT Reset: PC  0x80000000 PCSEL 1 2 3 4 5 6 Bad Opcode: Reg[27]  PC+4; PC  0x80000040 IRQ: Reg[27]  PC+4; PC  0x80000080 PC 00 Instruction Memory A D +4 Rs: <25:21> Rt: <20:16> WASEL Rd:<15:11> Rt:<20:16> 1 2 3 “31” “27” J:<25:0> RA1 Register RA2 WA WA File WD RD1 RD2 WE WERF Imm: <15:0> RESET SEXT N JT SEXT IRQ Z V C x4 ASEL 2 shamt:<10:6> “16” 1 Control Logic + 1 BSEL PCSEL WASEL BT SEXT A B BSEL ALU Data Memory RD WD R/W Adr Wr ALUFN WDSEL ALUFN Wr N V C Z WERF ASEL LSEL PC+4 32 WDSEL 0 1 2

MIPS: Our Final Version WA PC +4 Instruction Memory A D Register File RA1 RA2 RD1 RD2 ALU B WD WE ALUFN Control Logic Data Memory RD R/W Adr Wr WDSEL 0 1 2 BSEL J:<25:0> PCSEL WERF 00 PC+4 Rt: <20:16> Imm: <15:0> ASEL SEXT + x4 BT Z WASEL Rd:<15:11> Rt:<20:16> 1 2 3 PC<31:29>:J<25:0>:00 JT N V C Rs: <25:21> shamt:<10:6> 4 5 6 “16” IRQ 0x80000080 0x80000040 0x80000000 RESET “31” “27” This is a complete 32-bit processor. Although designed in one class lecture, it executes the majority of the MIPS R2000 instruction set. Executes one instruction per clock All that’s left is the control logic design

MIPS Control The control unit can be built as a large ROM X 1 4 00 6 3 Instruction R E S T I Q Z N V C P L S E X T WA S E L W D S E L ALUFN Sub Bool Shft Math W R W E R F A BSEL X 1 4 00 6 3 add sll andi lw sw beq