UCB CS61C : Machine Structures

Similar documents
CS61C : Machine Structures

Inequalities in MIPS (2/4) Inequalities in MIPS (1/4) Inequalities in MIPS (4/4) Inequalities in MIPS (3/4) UCB CS61C : Machine Structures

Lecture 7: Procedures

CS61C : Machine Structures

CS61C : Machine Structures

ECE 15B Computer Organization Spring 2010

CS 61C: Great Ideas in Computer Architecture More MIPS, MIPS Functions

Lecture 7: Procedures and Program Execution Preview

CS 110 Computer Architecture Lecture 6: More MIPS, MIPS Functions

CS 61C: Great Ideas in Computer Architecture More RISC-V Instructions and How to Implement Functions

CS 110 Computer Architecture MIPS Instruction Formats

EEM 486: Computer Architecture. Lecture 2. MIPS Instruction Set Architecture

CS 61C: Great Ideas in Computer Architecture Intro to Assembly Language, MIPS Intro

Reduced Instruction Set Computer (RISC)

MIPS Functions and Instruction Formats

Reduced Instruction Set Computer (RISC)

CS 61C: Great Ideas in Computer Architecture. Lecture 6: More MIPS, MIPS Func.ons. Instructor: Sagar Karandikar

CS 61C: Great Ideas in Computer Architecture. More MIPS, MIPS Functions

CS61CL : Machine Structures

Agenda. Strings: C vs. Java. Loading, Storing bytes. Support for Characters and Strings 6/29/2011

Final 9am instead of 9:30! Your grade will be ed by autograder! Jeremy Huddleston sb $t1 0($s2)!

I-Format Instructions (3/4) Define fields of the following number of bits each: = 32 bits

Chapter 2. Computer Abstractions and Technology. Lesson 4: MIPS (cont )

CS61C : Machine Structures

CS 61c: Great Ideas in Computer Architecture

UCB CS61C : Machine Structures

COMP 303 Computer Architecture Lecture 3. Comp 303 Computer Architecture

And in Review. Register Conventions (2/4) saved. Register Conventions (1/4) Register Conventions (4/4) Register Conventions (3/4) volatile

Do-While Example. In C++ In assembly language. do { z--; while (a == b); z = b; loop: addi $s2, $s2, -1 beq $s0, $s1, loop or $s2, $s1, $zero

Chapter 2: Instructions:

Branch Addressing. Jump Addressing. Target Addressing Example. The University of Adelaide, School of Computer Science 28 September 2015

Control Instructions. Computer Organization Architectures for Embedded Computing. Thursday, 26 September Summary

Control Instructions

ú There are CONVENTIONS when calling procedures! Stack main ()! { a(0);! }! void a (int m)! frame" Garcia, Spring 2013 UCB printf() (y==?

Review. Lecture #9 MIPS Logical & Shift Ops, and Instruction Representation I Logical Operators (1/3) Bitwise Operations

CS61C : Machine Structures

Lecture 5: Procedure Calls

CS61C : Machine Structures

CS61C Machine Structures. Lecture 12 - MIPS Procedures II & Logical Ops. 2/13/2006 John Wawrzynek. www-inst.eecs.berkeley.

Computer Architecture Instruction Set Architecture part 2. Mehran Rezaei

MIPS Datapath. MIPS Registers (and the conventions associated with them) MIPS Instruction Types

COE608: Computer Organization and Architecture

CISC 662 Graduate Computer Architecture. Lecture 4 - ISA MIPS ISA. In a CPU. (vonneumann) Processor Organization

ENCM 369 Winter 2013: Reference Material for Midterm #2 page 1 of 5

Thomas Polzer Institut für Technische Informatik

comp 180 Lecture 10 Outline of Lecture Procedure calls Saving and restoring registers Summary of MIPS instructions

The plot thickens. Some MIPS instructions you can write cannot be translated to a 32-bit number

101 Assembly. ENGR 3410 Computer Architecture Mark L. Chang Fall 2009

Course Administration

Lecture 2. Instructions: Language of the Computer (Chapter 2 of the textbook)

Instruction Set Architecture part 1 (Introduction) Mehran Rezaei

CSCI 402: Computer Architectures. Instructions: Language of the Computer (3) Fengguang Song Department of Computer & Information Science IUPUI.

CENG3420 Lecture 03 Review

The plot thickens. Some MIPS instructions you can write cannot be translated to a 32-bit number

ECE232: Hardware Organization and Design. Computer Organization - Previously covered

Machine Language Instructions Introduction. Instructions Words of a language understood by machine. Instruction set Vocabulary of the machine

Numbers: positional notation. CS61C Machine Structures. Faux Midterm Review Jaein Jeong Cheng Tien Ee. www-inst.eecs.berkeley.

MIPS R-format Instructions. Representing Instructions. Hexadecimal. R-format Example. MIPS I-format Example. MIPS I-format Instructions

CS61C : Machine Structures

CS 61C: Great Ideas in Computer Architecture. MIPS Instruction Formats

CISC 662 Graduate Computer Architecture. Lecture 4 - ISA

UCB CS61C : Machine Structures

Architecture II. Computer Systems Laboratory Sungkyunkwan University

CS61C : Machine Structures

EE 361 University of Hawaii Fall

CS/COE1541: Introduction to Computer Architecture

Bonus slides. Garcia, Spring 2014 UCB. CS61C L11 Introduction to MIPS : Procedures II & Logical Ops (18)

CS 110 Computer Architecture Review for Midterm I

Subroutines. int main() { int i, j; i = 5; j = celtokel(i); i = j; return 0;}

Introduction to the MIPS. Lecture for CPSC 5155 Edward Bosworth, Ph.D. Computer Science Department Columbus State University

MIPS%Assembly% E155%

UCB CS61C : Machine Structures

MIPS Instruction Set Architecture (2)

Instructions: Language of the Computer

Clever Signed Adder/Subtractor. Five Components of a Computer. The CPU. Stages of the Datapath (1/5) Stages of the Datapath : Overview

Computer Architecture. The Language of the Machine

ECE 331 Hardware Organization and Design. Professor Jay Taneja UMass ECE - Discussion 3 2/8/2018

Lecture 4: MIPS Instruction Set

Instructions: Assembly Language

COMPUTER ORGANIZATION AND DESIGN

Chapter 2. Instructions: Language of the Computer. Adapted by Paulo Lopes

Chapter 2A Instructions: Language of the Computer

More C functions and Big Picture [MIPSc Notes]

CSE Lecture In Class Example Handout

Computer Organization and Structure. Bing-Yu Chen National Taiwan University

ECE232: Hardware Organization and Design

Instructions: MIPS ISA. Chapter 2 Instructions: Language of the Computer 1

CS61C - Machine Structures. Lecture 6 - Instruction Representation. September 15, 2000 David Patterson.

Instruction Set Design and Architecture

ECE260: Fundamentals of Computer Engineering

CS 351 Exam 2 Mon. 11/2/2015

MIPS Instruction Set

Review (1/2) IEEE 754 Floating Point Standard: Kahan pack as much in as could get away with. CS61C - Machine Structures

Computer Organization and Structure. Bing-Yu Chen National Taiwan University

Instruction Set Architecture

CS222: MIPS Instruction Set

Overview. Introduction to the MIPS ISA. MIPS ISA Overview. Overview (2)

UC Berkeley CS61C : Machine Structures

Where Are We Now? Linker (1/3) Linker (2/3) Four Types of Addresses we ll discuss. Linker (3/3)

Decoding example Binary=>Decimal=>Assembly=>C? Start at program at address 4,194,304 (2 22 )

Transcription:

inst.eecs.berkeley.edu/~cs61c UCB CS61C : Machine Structures Lecture 10 Introduction to MIPS Procedures I Sr Lecturer SOE Dan Garcia 2014-02-14 If cars broadcast their speeds to other vehicles (and the speeds of cars were automatically controlled you could still steer) a simple in-car algorithm could help dissolve traffic jams as soon as they occur!. Key idea be optimistic leaving the jam and defensive leading into it. www.technologyreview.com/blog/arxiv/27166/!

R! Review MIPS Machine Language Instruction: 32 bits representing a single instruction opcode! rs! rt! rd! shamt! funct! I! opcode! rs! rt! immediate! J! opcode! target address! Branches use PC-relative addressing, Jumps use absolute addressing. Disassembly is simple and starts by decoding opcode field. (more next lecture) CS61C L10 Introduction to MIPS : Procedures I (2)

C functions main() { int i,j,k,m;... i = mult(j,k);... m = mult(i,i);...! }! What information must compiler/programmer keep track of? /* really dumb mult function */! int mult (int mcand, int mlier){ int product = 0; while (mlier > 0) { product = product + mcand; mlier = mlier -1; } return product; }! What instructions can accomplish this? CS61C L10 Introduction to MIPS : Procedures I (3)

Function Call Bookkeeping Registers play a major role in keeping track of information for function calls. Register conventions: ú Return address $ra! ú Arguments $a0, $a1, $a2, $a3! ú Return value $v0, $v1! ú Local variables $s0, $s1,, $s7! The stack is also used; more later. CS10 : The Beauty and Joy of Computing http://inst.eecs.berkeley.edu/~cs39n/fa10/ 2010-02-01 @ Faculty Lunch

Instruction Support for Functions (1/6) C M I P S... sum(a,b);... /* a,b:$s0,$s1 */ } int sum(int x, int y) {!return x+y; }! address (shown in decimal) 1000 1004 1008 1012 1016 2000 2004! In MIPS, all instructions are 4 bytes, and stored in memory just like data. So here we show the addresses of where the programs are stored. CS61C L10 Introduction to MIPS : Procedures I (5)

Instruction Support for Functions (2/6) C M I P S... sum(a,b);... /* a,b:$s0,$s1 */ } int sum(int x, int y) {!return x+y; }! address (shown in decimal) 1000 add $a0,$s0,$zero # x = a 1004 add $a1,$s1,$zero # y = b 1008 addi $ra,$zero,1016 #$ra=1016 1012 j sum!! #jump to sum 1016 2000 sum: add $v0,$a0,$a1 2004 jr $ra! # new instruction! CS61C L10 Introduction to MIPS : Procedures I (6)

Instruction Support for Functions (3/6) C M I P S... sum(a,b);... /* a,b:$s0,$s1 */ } int sum(int x, int y) {!return x+y; }! Question: Why use jr here? Why not use j? Answer: sum might be called by many places, so we can t return to a fixed place. The calling proc to sum must be able to say return here somehow.!2000 sum: add $v0,$a0,$a1 2004 jr $ra! # new instruction! CS61C L10 Introduction to MIPS : Procedures I (7)

Instruction Support for Functions (4/6) Single instruction to jump and save return address: jump and link (jal) Before: 1008 addi $ra,$zero,1016 #$ra=1016 1012 j sum!!! #goto sum! After: 1008 jal sum # $ra=1012,goto sum! Why have a jal? ú Make the common case fast: function calls very common. ú Don t have to know where code is in memory with jal! CS61C L10 Introduction to MIPS : Procedures I (8)

Instruction Support for Functions (5/6) Syntax for jal (jump and link) is same as for j (jump): jal!label! jal should really be called laj for link and jump : ú Step 1 (link): Save address of next instruction into $ra! Why next instruction? Why not current one? ú Step 2 (jump): Jump to the given label CS61C L10 Introduction to MIPS : Procedures I (9)

Instruction Support for Functions (6/6) Syntax for jr (jump register): jr register! Instead of providing a label to jump to, the jr instruction provides a register which contains an address to jump to. Very useful for function calls: ú jal stores return address in register ($ra) ú jr $ra jumps back to that address CS61C L10 Introduction to MIPS : Procedures I (10)

Nested Procedures (1/2) int sumsquare(int x, int y) {!return mult(x,x)+ y; }! Something called sumsquare, now sumsquare is calling mult. So there s a value in $ra that sumsquare wants to jump back to, but this will be overwritten by the call to mult. Need to save sumsquare return address before call to mult. CS61C L10 Introduction to MIPS : Procedures I (11)

Nested Procedures (2/2) In general, may need to save some other info in addition to $ra. When a C program is run, there are 3 important memory areas allocated: ú Static: Variables declared once per program, cease to exist only after execution completes. E.g., C globals ú Heap: Variables declared dynamically via malloc ú Stack: Space to be used by procedure during execution; this is where we can save register values CS61C L10 Introduction to MIPS : Procedures I (12)

C Memory Allocation Address $sp stack pointer 0 Stack Heap Static Code CS61C L10 Introduction to MIPS : Procedures I (13) Space for local vars, saved procedure information Explicitly created space, i.e., malloc() Variables declared once per program; e.g., globals (doesn t change size) Program (doesn t change size)

Using the Stack (1/2) So we have a register $sp which always points to the last used space in the stack. To use stack, we decrement this pointer by the amount of space we need and then fill it with info. So, how do we compile this? int sumsquare(int x, int y) {! return mult(x,x)+ y; }! CS61C L10 Introduction to MIPS : Procedures I (14)

Using the Stack (2/2) Hand-compile sumsquare: push pop mult:... addi $sp,$sp,-8 # space on stack sw $ra, 4($sp)! # save ret addr sw $a1, 0($sp)! # save y add $a1,$a0,$zero # mult(x,x) jal mult!! # call mult lw $a1, 0($sp)! # restore y add $v0,$v0,$a1 # mult()+y lw $ra, 4($sp)! # get ret addr addi $sp,$sp,8 # restore stack jr $ra CS61C L10 Introduction to MIPS : Procedures I (15) int sumsquare(int x, int y) {!return mult(x,x)+ y; }!

Steps for Making a Procedure Call 1. Save necessary values onto stack. 2. Assign argument(s), if any. 3. jal call 4. Restore values from stack. CS61C L10 Introduction to MIPS : Procedures I (16)

Rules for Procedures Called with a jal instruction, returns with a jr $ra! Accepts up to 4 arguments in $a0, $a1, $a2 and $a3! Return value is always in $v0 (and if necessary in $v1) Must follow register conventions So what are they? CS61C L10 Introduction to MIPS : Procedures I (17)

Basic Structure of a Function Prologue entry_label: addi $sp,$sp, -framesize sw $ra, framesize-4($sp) # save $ra save other regs if need be " "! ra Body Epilogue!!...! (call other functions ) restore other regs if need be lw $ra, framesize-4($sp) # restore $ra addi $sp,$sp, framesize jr $ra! memory CS61C L10 Introduction to MIPS : Procedures I (18)

MIPS Registers The constant 0 $0 $zero Reserved for Assembler $1 $at Return Values $2-$3 $v0-$v1 Arguments $4-$7 $a0-$a3 Temporary $8-$15 $t0-$t7 Saved $16-$23 $s0-$s7 More Temporary $24-$25 $t8-$t9 Used by Kernel $26-27 $k0-$k1 Global Pointer $28 $gp Stack Pointer $29 $sp Frame Pointer $30 $fp Return Address $31 $ra (From COD green insert) Use names for registers -- code is clearer! CS61C L10 Introduction to MIPS : Procedures I (19)

Other Registers $at: may be used by the assembler at any time; unsafe to use $k0-$k1: may be used by the OS at any time; unsafe to use $gp, $fp: don t worry about them Note: Feel free to read up on $gp and $fp in Appendix A, but you can write perfectly good MIPS code without them. CS61C L10 Introduction to MIPS : Procedures I (20)

Peer Instruction int fact(int n){! if(n == 0) return 1; else return(n*fact(n-1));}! When translating this to MIPS 1) We COULD copy $a0 to $a1 (& then not store $a0 or $a1 on the stack) to store n across recursive calls. 2) We MUST save $a0 on the stack since it gets changed. 3) We MUST save $ra on the stack since we need to know where to return to 123! a) FFF! b) FFT! c) FTF! c) FTT! d) TFF! d) TFT! e) TTF! e) TTT! CS61C L10 Introduction to MIPS : Procedures I (21)

And in Conclusion Functions called with jal, return with jr $ra. The stack is your friend: Use it to save anything you need. Just leave it the way you found it! Instructions we know so far Arithmetic: add, addi, sub, addu, addiu, subu! Memory: lw, sw, lb, sb! Decision: beq, bne, slt, slti, sltu, sltiu! Unconditional Branches (Jumps): j, jal, jr! Registers we know so far ú All of them! CS61C L10 Introduction to MIPS : Procedures I (22)