Instructor: Randy H. Katz hap://inst.eecs.berkeley.edu/~cs61c/fa13. Fall Lecture #7. Warehouse Scale Computer

Transcription

1 CS 61C: Great Ideas in Computer Architecture Everything is a Number Instructor: Randy H. Katz hap://inst.eecs.berkeley.edu/~cs61c/fa13 9/19/13 Fall Lecture #7 1 New- School Machine Structures (It s a bit more complicated!) So4ware Parallel Requests Assigned to computer e.g., Search Katz Parallel Threads Assigned to core e.g., Lookup, Ads Parallel Instruc]ons >1 one ]me e.g., 5 pipelined instruc]ons Parallel Data >1 data one ]me e.g., Add of 4 pairs of words Hardware descrip]ons All one ]me Programming Languages Harness Parallelism & Achieve High Performance Hardware Today s Lecture Warehouse Scale Computer Core Memory Input/Output Instruc]on Unit(s) Cache Memory Core 9/19/13 Fall Lecture #7 2 Computer (Cache) Core Func]onal Unit(s) A 0 +B 0 A 1 +B 1 A 2 +B 2 A 3 +B 3 Smart Phone Logic Gates 1

2 Big Idea #1: Levels of Representa]on/ Interpreta]on Machine Interpreta4on High Level Language Program (e.g., C) Compiler Assembly Language Program (e.g., MIPS) Assembler Machine Language Program (MIPS) Hardware Architecture DescripCon (e.g., block diagrams) Architecture Implementa4on Logic Circuit DescripCon (Circuit SchemaCc Diagrams) temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; lw $t0, 0($2) lw $t1, 4($2) sw $t1, 0($2) sw $t0, 4($2) We are here! Anything can be represented as a number, i.e., data or instruc]ons ! 9/19/13 Fall Lecture #7 3 Review C is func]on oriented; code reuse via func]ons Jump and link (jal) invokes, jump register (jr $ra) returns Registers $a0-$a3 for arguments, $v0-$v1 for return values Stack for spilling registers, nested func]on calls, C local (automa]c) variables 9/19/13 Fall Lecture #7 4 2

3 Agenda Memory Heap Everything is a Number Administrivia Overflow and Real Numbers Technology Break Instruc]ons as Numbers Assembly Language to Machine Language And in Conclusion, 9/19/13 Fall Lecture #7 5 Agenda Memory Heap Everything is a Number Administrivia Overflow and Real Numbers Technology Break Instruc]ons as Numbers Assembly Language to Machine Language And in Conclusion, 9/19/13 Fall Lecture #7 6 3

4 What If a Func]on Calls a Func]on? Recursive Func]on Calls? Would clobber values in $a0 to $a1 and $ra What is the solu]on? 9/19/13 Fall Lecture #7 7 Alloca]ng Space on Stack C has two storage classes: automa]c and sta]c Automa@c variables are local to func]on and discarded when func]on exits. Sta@c variables exist across exits from and entries to procedures Use stack for automa]c (local) variables that don t fit in registers Procedure frame or ac@va@on record: segment of stack with saved registers and local variables Some MIPS compilers use a frame pointer ($fp) to point to first word of frame (29 of 32, 3 ler!) 9/19/13 Fall Lecture #7 8 4

5 Stack Before, During, Arer Call 9/19/13 Fall Lecture #7 9 Recursive Func]on Factorial int fact (int n) { if (n < 1) return (1); else return (n * fact(n-1)); } 9/19/13 Fall Lecture #7 10 5

6 Recursive Func]on Factorial Fact: L1: # adjust stack for 2 items # Else part (n >= 1) addi $sp,$sp,-8 # arg. gets (n 1) # save return address addi $a0,$a0,-1 sw $ra, 4($sp) # call fact with (n 1) # save argument n jal fact sw $a0, 0($sp) # return from jal: restore n # test for n < 1 lw $a0, 0($sp) slti $t0,$a0,1 # restore return address # if n >= 1, go to L1 lw $ra, 4($sp) beq $t0,$zero,l1 # adjust sp to pop 2 items # Then part (n==1) return 1 addi $sp, $sp,8 addi $v0,$zero,1 # return n * fact (n 1) # pop 2 items off stack mul $v0,$a0,$v0 addi $sp,$sp,8 # return to the caller # return to caller jr $ra jr $ra mul is a pseudo instruc@on 9/19/13 Fall Lecture #7 11 Op]mized Func]on Conven]on To reduce expensive loads and stores from spilling and restoring registers, MIPS divides registers into two categories: 1. Preserved across func]on call Caller can rely on values being unchanged $ra, $sp, $gp, $fp, saved registers $s0- $s7 2. Not preserved across func]on call Caller cannot rely on values being unchanged Return value registers $v0,$v1, Argument registers $a0- $a3, temporary registers $t0- $t9 9/19/13 Fall Lecture #7 12 6

7 Where is the Stack in Memory? MIPS conven]on Stack starts in high memory and grows down Hexadecimal (base 16) : 7fff fffc hex MIPS programs (text segment) in low end hex Sta@c data segment (constants and other sta]c variables) above text for sta]c variables MIPS conven]on global pointer ($gp) points to sta]c (30 of 32, 2 ler! will see when talk about OS) Heap above sta]c for data structures that grow and shrink ; grows up to high addresses 9/19/13 Fall Lecture #7 13 MIPS Memory Alloca]on 9/19/13 Fall Lecture #7 14 7

8 Register Alloca]on and Numbering 9/19/13 Fall Lecture #7 15 Which statement is FALSE? MIPS uses jal to invoke a func]on and jr to return from a func]on jal saves PC+1 in $ra! The callee can use temporary registers ($ti) without saving and restoring them! 16 8

9 Which statement is FALSE? MIPS uses jal to invoke a func]on and jr to return from a func]on jal saves PC+1 in $ra! The callee can use temporary registers ($ti) without saving and restoring them! 17 Agenda Memory Heap Everything is a Number Administrivia Overflow and Real Numbers Technology Break Instruc]ons as Numbers Assembly Language to Machine Language Summary 9/19/13 Fall Lecture #7 18 9

10 Key Concepts Inside computers, everything is a number But everything is of a fixed size 8- bit bytes, 16- bit half words, 32- bit words, 64- bit double words, Integer and floa]ng point opera]ons can lead to results too big to store within their representa]ons: overflow/underflow 9/19/13 Fall Lecture #7 19 Number Representa]on Value of i- th digit is d Base i where i starts at 0 and increases from right to ler: = 1 10 x x x = 1x x x1 10 = = Binary (Base 2), Hexadecimal (Base 16), Decimal (Base 10) different ways to represent an integer We use 1 two, 5 ten, 10 hex to be clearer (vs. 1 2, 4 8, 5 10, ) 9/19/13 Fall Lecture #

11 Number Representa]on Hexadecimal digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F FFF hex = 15 ten x 16 ten ten x 16 ten ten x 16 ten 0 = 3840 ten ten + 15 ten = 4095 ten two = FFF hex = 4095 ten May put blanks every group of binary, octal, or hexadecimal digits to make it easier to parse, like commas in decimal 9/19/13 Fall Lecture #7 21 Signed and Unsigned Integers C, C++, and Java have signed integers, e.g., 7, - 255: int x, y, z; C, C++ also have unsigned integers, which are used for addresses 32- bit word can represent 2 32 binary numbers Unsigned integers in 32 bit word represent 0 to (4,294,967,295) 9/19/13 Fall Lecture #

12 Unsigned Integers two = 0 ten two = 1 ten two = 2 ten two = 2,147,483,645 ten two = 2,147,483,646 ten two = 2,147,483,647 ten two = 2,147,483,648 ten two = 2,147,483,649 ten two = 2,147,483,650 ten two = 4,294,967,293 ten two = 4,294,967,294 ten two = 4,294,967,295 ten 9/19/13 Fall Lecture #7 23 Signed Integers and Two s Complement Representa]on Signed integers in C; want ½ numbers <0, want ½ numbers >0, and want one 0 Two s complement treats 0 as posi]ve, so 32- bit word represents 2 32 integers from ( 2,147,483,648) to (2,147,483,647) Note: one nega]ve number with no posi]ve version Book lists some other op]ons, all of which are worse Every computers uses two s complement today Most significant bit (lermost) is the sign bit, since 0 means posi]ve (including 0), 1 means nega]ve Bit 31 is most significant, bit 0 is least significant 9/19/13 Fall Lecture #

13 Sign Bit Two s Complement Integers two = 0 ten two = 1 ten two = 2 ten two = 2,147,483,645 ten two = 2,147,483,646 ten two = 2,147,483,647 ten two = 2,147,483,648 ten two = 2,147,483,647 ten two = 2,147,483,646 ten two = 3 ten two = 2 ten two = 1 ten 9/19/13 Fall Lecture #7 25 Twos Complement Examples Assume for simplicity 4 bit width, - 8 to +7 represented (- 2) (- 2) Overflow! (- 1) Underflow! Carry into MSB = Carry Out MSB Carry into MSB = Carry Out MSB 9/19/13 Fall Lecture #

14 Suppose we had a 5 bit word. What integers can be represented in two s complement? - 32 to to to +15! 27 Suppose we had a 5 bit word. What integers can be represented in two s complement? - 32 to to to +15! 28 14

15 MIPS Logical Instruc]ons Useful to operate on fields of bits within a word e.g., characters within a word (8 bits) Opera]ons to pack /unpack bits into words Called logical opera@ons Logical operations C operators Java operators MIPS instructions Bit-by-bit AND & & and Bit-by-bit OR or Bit-by-bit NOT ~ ~ nor Shift left << << sll Shift right >> >>> srl 9/19/13 Fall Lecture #7 29 Bit- by- bit Defini]on OperaCon Input Input Output AND AND AND AND OR OR OR OR NOR NOR NOR /19/13 NOR 1 Fall Lecture #

16 Examples If register $t2 contains two And register $t1 contains two What is value of $t0 arer: and $t0,$t1,$t2 # reg $t0 = reg $t1 & reg $t2 9/19/13 Fall Lecture #7 31 Examples If register $t2 contains two And register $t1 contains two What is value of $t0 arer: and $t0,$t1,$t2 # reg $t0 = reg $t1 & reg $t two 9/19/13 Fall Lecture #

17 Examples If register $t2 contains two And register $t1 contains two What is value of $t0 arer: or $t0,$t1,$t2 # reg $t0 = reg $t1 reg $t2 9/19/13 Fall Lecture #7 33 Examples If register $t2 contains two And register $t1 contains two What is value of $t0 arer: or $t0,$t1,$t2 # reg $t0 = reg $t1 reg $t two 9/19/13 Fall Lecture #

18 Examples If register $t2 contains two And register $t1 contains two What is value of $t0 arer: nor $t0,$t1,$zero # reg $t0 = ~ (reg $t1 0) 9/19/13 Fall Lecture #7 35 Examples If register $t2 contains two And register $t1 contains two What is value of $t0 arer: nor $t0,$t1,$zero # reg $t0 = ~ (reg $t1 0) two 9/19/13 Fall Lecture #

19 Shiring Shir ler logical moves n bits to the ler (insert 0s into empty bits) Same as mul]plying by 2 n for two s complement number For example, if register $s0 contained two = 9 ten If executed sll $s0, $s0, 4, result is: two = 144 ten And 9 ten 2 ten 4 = 9 ten 16 ten = 144 ten Shir right logical moves n bits to the right (insert 0s into empty bits) NOT same as dividing by 2n (nega]ve numbers fail) 9/19/13 Fall Lecture #7 37 Shiring Shir right arithme]c moves n bits to the right (insert high order sign bit into empty bits) For example, if register $s0 contained two = 25 ten If executed sra $s0, $s0, 4, result is: 9/19/13 Fall Lecture #

20 Shiring Shir right arithme]c moves n bits to the right (insert high order sign bit into empty bits) For example, if register $s0 contained two = 25 ten If executed sra $s0, $s0, 4, result is: two = 1 ten 9/19/13 Fall Lecture #7 39 Shiring Shir right arithme]c moves n bits to the right (insert high order sign bit into empty bits) For example, if register $s0 contained two = - 25 ten If executed sra $s0, $s0, 4, result is: 9/19/13 Fall Lecture #

21 Shiring Shir right arithme]c moves n bits to the right (insert high order sign bit into empty bits) For example, if register $s0 contained two = - 25 ten If executed sra $s0, $s0, 4, result is: two = - 2 ten Unfortunately, this is NOT same as dividing by 2n Fails for odd nega]ve numbers C arithme]c seman]cs is that division should round towards 0 9/19/13 Fall Lecture #7 41 Impact of Signed and Unsigned Integers on Instruc]on Sets What (if any) instruc]ons affected? Load word, store word? branch equal, branch not equal? and, or, sll, srl? add, sub, mult, div? sl] (set less than immediate)? 9/19/13 Fall Lecture #

22 C provides two sets of operators for AND (& and &&) and two sets of operators for OR ( and ) while MIPS doesn t. Why? Logical opera]ons AND and OR do & and while condi]onal branches do && and The previous statement has it backwards: && and logical ops, & and are branches They are redundant and mean the same thing: && and are simply inherited from the programming language B, the predecessor of C! 43 C provides two sets of operators for AND (& and &&) and two sets of operators for OR ( and ) while MIPS doesn t. Why? Logical opera]ons AND and OR do & and while condi]onal branches do && and The previous statement has it backwards: && and logical ops, & and are branches They are redundant and mean the same thing: && and are simply inherited from the programming language B, the predecessor of C! 44 22

23 Peer Instruc]on Answer C provides two sets of operators for AND (& and &&) and two sets of operators for OR ( and ) while MIPS doesn t. Why? Logical opera]ons AND and OR implement & and while condi]onal branches implement && and Reason: e.g., && is logical and: true && true is true and everything else is false & is bitwise and: e.g., (1010 two & 1000 two ) = 1000 two 9/19/13 Fall Lecture #7 45 Agenda Memory Heap Everything is a Number Administrivia Overflow and Real Numbers Instruc]ons as Numbers Technology Break Assembly Language to Machine Language And in Conclusion, 9/19/13 Fall Lecture #

24 Administrivia HW #3 Due 11:59:59 Project #1 Part 1 Due 11:59:59 Midterm on the horizon: 10/17 (4 weeks), 6-9 PM It s going to be completely mul]ple choice! 9/19/13 Fall Lecture #7 47 And in Conclusion, Program can interpret binary number as unsigned integer, two s complement signed integer, floa]ng point number, ASCII characters, Unicode characters, Integers have largest posi]ve and largest nega]ve numbers, but represent all in between Two s comp. weirdness is one extra nega]ve numinteger and floa]ng point opera]ons can lead to results too big to store within their representa]ons: overflow/underflow Floa]ng point is an approxima]on of reals Everything is a (binary) number in a computer Instruc]ons and data; stored program concept 9/19/13 Fall Lecture #