Transcript Week 1
CS35101 Computer Architecture Spring 2006 Week 1 Slides adapted from: Mary Jane Irwin (www.cse.psu.edu/~mji) Course url: www.cs.psu.edu/~cg331 [Adapted from Dave Patterson’s UCB CS152 slides] Course Administration Instructor: Paul J Durand [email protected] http://www.cs.kent.edu/~durand Labs: Accounts on loki or neptune Texts: Computer Organization and Design: The Hardware/Software Interface, Third Edition, Patterson and Hennessy Course Goals and Structure Introduction to the major components of a computer system, how they function together in executing a program, how they are designed. MIPS assembler programming using the spim system spim Assembler and Simulator spim is a self-contained assembler and simulator for the MIPS R2000/R3000 It provides a simple assembler, debugger and a simple set of operating system services It implements both a simple, terminal-style interface and a visual windowing interface Available as xspim on unix - installed on the CS unix machines loki, hermes, neptune, poseidon PCSpim on Windows - can be downloaded and installed on your own PC from www.cs.wisc.edu/~larus/SPIM/pcspim.exe Sorry, there is no Macintosh version of spim Head’s Up This week’s material Course introduction - Reading assignment – PH 1.1 through 1.3 and A.9 through A.10 (on cd) Reminders Make sure your unix account is operational; change your password to something you can remember and that is secure (must be six to eight alphanumeric characters) Check out the course homepage (www.cs.kent.edu/~durand) Next week’s material Introduction to MIPS assembler - Reading assignment - PH 2.1 through 2.7, omit 2.6 What You Should Already Know How to write, compile and run programs in a higher level language (C, C++, Java, …) How to create, organize, and edit files and run programs on Unix How to represent and operate on positive and negative numbers in binary form (two’s complement, sign magnitude, etc.) Logic design How to design combinational components (Boolean algebra, logic minimization, decoders and multiplexors) Below the Program High-level language program (in C) swap (int v[], int k) (int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; ) Assembly language program (for MIPS) swap: sll add lw lw sw sw jr C compiler $2, $5, 2 $2, $4,$2 $15, 0($2) $16, 4($2) $16, 0($2) $15, 4($2) $31 Machine (object) code (for MIPS) 000000 00000 00101 0001000010000000 000000 00100 00010 0001000000100000 . . . assembler Advantages of Higher-Level Languages Higher-level languages Allow the programmer to think in a more natural language and for their intended use (Fortran for scientific computation, Cobol for business programming, Lisp for symbol manipulation, …) Improve programmer productivity – more understandable code that is easier to debug and validate Improve program maintainability Allow programmers to be independent of the computer on which they are developed (compilers and assemblers can translate high-level language programs to the binary instructions of any machine) Emergence of optimizing compilers that produce very efficient assembly code optimized for the target machine As a result, very little programming is done today at the assembler level Machine Organization Capabilities and performance characteristics of the principal Functional Units (FUs) e.g., register file, ALU, multiplexors, memories, ... The ways those FUs are interconnected e.g., buses Logic and means by which information flow between FUs is controlled The machine’s Instruction Set Architecture (ISA) Register Transfer Level (RTL) machine description Major Components of a Computer Processor Control Datapath Devices Memory Input Output Impacts of Advancing Technology Processor increases about 30% per year 2x every 1.5 years Memory logic capacity: performance: DRAM capacity: 4x every 3 years memory speed: 1.5x every 10 years cost per bit: decreases about 25% per year Disk capacity: increases about 60% per year Example: Growth in DRAM Chip Capacity 1000000 256,000 Kbit capacity 100000 64,000 16,000 10000 4,000 1000 1,000 256 100 64 10 1980 1982 1984 1986 1988 1990 1992 Year of introduction 1994 1996 1998 2000 Below the Program High-level language program (in C) swap (int v[], int k) . . . Assembly language program swap: sll $2, $5, 2 add $2, $4, $2 lw $15, 0($2) lw $16, 4($2) sw $16, 0($2) sw $15, 4($2) jr $31 one-to-many (for MIPS) Machine (object) code (for MIPS) 000000 000000 100011 100011 101011 101011 000000 00000 00100 00010 00010 00010 00010 11111 00101 00010 01111 10000 10000 01111 00000 0001000010000000 0001000000100000 0000000000000000 0000000000000100 0000000000000000 0000000000000100 0000000000001000 C compiler one-to-one assembler Input Device Inputs Object Code 000000 000000 100011 100011 101011 101011 000000 Processor Control Datapath Devices Memory Input Output 00000 00100 00010 00010 00010 00010 11111 00101 00010 01111 10000 10000 01111 00000 0001000010000000 0001000000100000 0000000000000000 0000000000000100 0000000000000000 0000000000000100 0000000000001000 Object Code Stored in Memory Memory Processor Control Datapath 000000 000000 100011 100011 101011 101011 000000 00000 00100 00010 00010 00010 00010 11111 00101 00010 01111 10000 10000 01111 00000 0001000010000000 0001000000100000 0000000000000000 0000000000000100 0000000000000000 0000000000000100 0000000000001000 Devices Input Output Processor Fetches an Instruction Processor fetches an instruction from memory Memory Processor Control Datapath 000000 000000 100011 100011 101011 101011 000000 00000 00100 00010 00010 00010 00010 11111 00101 00010 01111 10000 10000 01111 00000 0001000010000000 0001000000100000 0000000000000000 0000000000000100 0000000000000000 0000000000000100 0000000000001000 Where does it fetch from? Devices Input Output Control Decodes the Instruction Control decodes the instruction to determine what to execute Processor Devices Control 000000 00100 00010 0001000000100000 Memory Input Datapath Output Datapath Executes the Instruction Datapath executes the instruction as directed by control Processor Devices Control 000000 00100 00010 0001000000100000 Memory Input Datapath contents Reg #4 ADD contents Reg #2 results put in Reg #2 Output Processor Organization Control needs to have the Ability to input instructions from memory Logic and means to control instruction sequencing Logic and means to issue signals that control the way information flows between datapath components Logic and means to control what operations the datapath’s functional units perform Datapath needs to have the Components - functional units (e.g., adder) and storage locations (e.g., register file) - needed to execute instructions Components interconnected so that the instructions can be accomplished Ability to load data from and store data to memory Where does it load and store from and to? What Happens Next? Memory Processor Control Datapath 000000 000000 100011 100011 101011 101011 000000 00000 00100 00010 00010 00010 00010 11111 00101 00010 01111 10000 10000 01111 00000 0001000010000000 0001000000100000 0000000000000000 0000000000000100 0000000000000000 0000000000000100 0000000000001000 Fetch Exec Decode Devices Input Output Output Data Stored in Memory At program completion the data to be output resides in memory Processor Memory Control Datapath Devices Input 00000100010100000000000000000000 00000000010011110000000000000100 00000011111000000000000000001000 Output Output Device Outputs Data Processor Control Datapath Devices Memory Input Output 00000100010100000000000000000000 00000000010011110000000000000100 00000011111000000000000000001000 The Instruction Set Architecture software instruction set architecture hardware Stopped here 1/19 The interface description separating the software and hardware. MIPS R3000 Instruction Set Architecture Instruction Categories Registers Load/Store Computational Jump and Branch Floating Point R0 - R31 - coprocessor PC HI Memory Management Special LO 3 Instruction Formats: all 32 bits wide OP rs rt OP rs rt OP rd sa immediate jump target Q: How many already familiar with MIPS ISA? funct