1 cs232: computer architecture ii fall 2011 intel i7 quad-core

13
1 CS232: Computer Architecture II Fall 2011 Intel i7 Quad-core

Upload: mitchell-knight

Post on 03-Jan-2016

223 views

Category:

Documents


0 download

TRANSCRIPT

1

CS232: Computer Architecture II

Fall 2011

Intel i7 Quad-core

2

Who we are

Lecturer:Prof. Craig Zilles [email protected]

Teaching Assistants/Section Instructor: Room 0212 SC

Ryan Cunningham noon sectionDwight Guth 3pm sectionCharles Tucker 1pm section

http://www.cs.illinois.edu/class/cs232— MP 1 released on Thursday, due next Thursday.

• Make sure you have an EWS account!

3

Computer architecture is about building and analyzing computer systems

Instruction Set Architecture is bridge between hardware and software— Study the MIPS ISA in detail— Learn what compilers do when they translate high-level code into

assembly (we won’t learn how they do it)— Learn how HLL program constructs are represented to the

machine Key techniques: Pipelining, Caching, Virtual Memory Tuning complex code for performance Hardware support for parallelism

What is computer architecture about?

Memory

Processor

Input/Output

CompilerHLL ASM

4

The ISA is an interface between software and hardware— the hardware “promises” to implement all ISA instructions— the software uses ISA primitives to build complex programs

Later in the semester, we’ll see how the ISA affects the hardware design

Instruction set architectures

Software

Hardware

ISA

5

We study the MIPS instruction set architecture to illustrate concepts in assembly language and machine organization—concepts are not MIPS-specific—MIPS is just convenient because it is real, yet simple (unlike

x86)

MIPS ISA is used in many places, primarily in embedded systems—routers from Cisco—game machines like the Nintendo 64 and Sony Playstation 2 &

PSP

Why MIPS?

6

What you will need to learn for Exam 1

You must become “fluent” in MIPS assembly:— Translate from C++ to MIPS and MIPS to C++

Example: Translate the following recursive C++ function into MIPS

int pow(int n, int m) {

if (m == 1)

return n;

return n * pow(n, m-1);

}

How are arguments passed?

How are values returned?

How are complex expressionsbroken into simple instructions?

How is recursion done?

7

MIPS: register-to-register, three address

MIPS is a register-to-register, or load/store, architecture—destination and sources of instructions must all be registers—special instructions to access main memory (later)

MIPS uses three-address instructions for data manipulation—each ALU instruction contains a destination and two sources

For example, an addition instruction (a = b + c) has the form:

add a, b, c

operation

destination sources

operands

8

MIPS register file

MIPS processors have 32 registers, each of which holds a 32-bit value—register addresses are 5 bits long

More registers might seem better, but there is a limit to the goodness:—more expensive: because of registers themselves, plus extra

hardware like muxes to select individual registers—instruction lengths may be affected D data

Write

D address

A address B address

A data B data

32 32 Register File

55

5

32

32 32

9

MIPS register names

MIPS register names begin with a $. There are two naming conventions:—by number:

$0 $1 $2 … $31

—by (mostly) two-character names, such as:

$a0-$a3 $s0-$s7 $t0-$t9 $sp $ra

Not all of the registers are equivalent:—e.g., register $0 or $zero always contains the value 0—some have special uses, by convention ($sp holds “stack

pointer”)

You have to be a little careful in picking registers for your programs—for now, stick to the registers $t0-$t9

10

Basic arithmetic and logic operations

The basic integer arithmetic operations include the following:

add sub mul div

And here are a few bitwise operations:

and or xor nor

Remember that these all require three register operands; for example:

add $t0, $t1, $t2 # $t0 = $t1 + $t2

mul $s1, $s1, $a0 # $s1 = $s1 x $a0

Note: a full MIPS ISA reference can be found in Appendix A (linked from website)

11

Complex arithmetic expressions may require multiple MIPS operations

Example: t0 (t1 t2) (t3 t4)

add $t0, $t1, $t2 # $t0 contains $t1 + $t2sub $t6, $t3, $t4 # temp value $t6 = $t3 - $t4mul $t0, $t0, $t6 # $t0 contains the final product

Temporary registers may be necessary, since each MIPS instructions can access only two source registers and one destination—in this example, we could re-use $t3 instead of introducing $t6

—must be careful not to modify registers that are needed again later

Larger expressions

12

How are registers initialized?

Special MIPS instructions allow you to specify a signed constant, or “immediate” value, for the second source instead of a register—e.g., here is the immediate add instruction, addi:

addi $t0, $t1, 4 # $t0 = $t1 + 4

Immediate operands can be used in conjunction with the $zero register to write constants into registers:

addi $t0, $0, 4 # $t0 = 4

Shorthand: li $t0, 4 # $t0 = 4(pseudo-instruction)

13

Our first MIPS program

Let’s translate the following C++ program into MIPS:

void main() {

int i = 516;

int j = i*(i+1)/2;

i = i + j;

}main: # start of main li $t0, 516 # i = 516 addi $t1, $t0, 1 # i + 1 mul $t1, $t0, $t1 # i * (i + 1) li $t2, 2 div $t1, $t1, $t2 # j = i*(i+1)/2 add $t0, $t0, $t1 # i = i + j

jr $ra # return