CS 232: Computer Architecture II

Prof. Laxmikant (Sanjay) Kale

CS 232 Objectives:• Learn about how a computer system, and specifically its

processor, works– Assembly language programming

– Instruction Set Architecture

– Basic Processor design: Data-path and control

– An overview of advanced topics

• with specific topics covered in depth

• Example: Pipelining, Caches, I/O, Multiprocessors

Some of the Dos and Don’t• Must use the class web page regularly

– Important announcements, syllabus, lecture notes (slides), assignments

• Also, must read the class newsgroup regularly– Frequently asked questions

– Timely response

– Make sure you don’t post solutions

– No trash tolerated!

Lets begin with numbers• Say you wanted to build a machine that calculates the

exact square of any given number quickly.– 100 years ago

– May want to use the same idea for other things

• calculate cubes?

• Other number calculations?

• Need to decide:– How are you going to represent the numbers?

– What are you going to build the machine out of?

• Need a basic “smart” component

Representing numbers• Focus on integers• Examine our decimal number system:

– Much nicer compared to roman numbers • How do you add two large numbers in roman representation?

• With weighted positions (ones, tens, hundreds) addition is easy.– In fact, we can describe a simple procedure (“algorithm”) for doing it.

– But choice of “10” as the base is somewhat arbitrary

• What base is better for our machine?– Base 10: need 10 symbols

– Base 2: need 2 symbols (0,1)

• We can probably make machines that can represent 0 and 1– so that they don’t mix with each other

Binary number representation• You have done this in CS231

– 11011 is 27

– 35 is 100011

– Addition and subtraction rules are the same as decimal numbers

• Let us agree to use this in our machine– Still have the problem of having to translate between decimal

system (that we understand) and binary system, that our machine understands

– Assume we will do it manually, for now

Smart component: Switch• A switch is either on or off

– To really compose switches to build a machine, we need to be able to control a switch

• (I.e. if switch A is ON, it will also change switch B to the ON position).

– Let us assume we have electricity

• Basics: Voltage, Current, Flows if switch is on

• Make a controllable switch – such that if input voltage is “High”, the switch is “ON” (closed)

– Let us assume “High Voltage” represents 1 and “Low” 0.

Gates• We can now connect switches together:

– Two switches (A and B) connected in series:

• If both are “ON”, the output is HIGH

• So, if the input to both switch A and switch B is High, the output of the composite circuit is High

• Let us call this the “AND” circuit (or AND gate)

– You can also connect switches such that:

• If either of the switch is ON, the output is HIGH

• Connect the switches in parallel

• So, if input to A is HIGH or input to B is HIGH, output is High: OR gate

– NOT gate:

• Input High, Output Low, and vice versa

Gates to Adders• We can now connect gates together to make an adder:

– Half adder (ignores carry):

• Given two inputs, if exactly one of them 1, then output 1

• Also output carry if both are 1

• How can we connect gates together to make this circuit?

– Full adder:

• uses carry

• 3 inputs lines, two output lines:

– Multi-bit adder:

• Feed the carry of least significant bit to the next higher one

So, we have an adder• But we needed to square a number….• The “adder” is just a simple calculating device

– We could connect a lot more of those to make a multiplier, or a squarer.

– But, we can reuse the same hardware.

– Challenge: just use one adder (say 32 bit adder).

• We need one more type of device:– To store intermediate results

– “Memory”: or registers

– We tell the adder to repeatedly add M to a running total.

– So, we need to store the running total and M somewhere

– Also need to to remember how many times to add (Count)

How to make registers• Switches can be connected to make a memory device

– Latches and Flip-Flops

NOR

NOR

Reset

Set

A register is a set latches grouped together

What does it look like now?• We have:

– 3 32 bit “registers” : M, result, count

– 1 32 bit adder

– How are they connected?

• We need to bring back M and result to the inputs of the adder

• Store the sum back in result

• bring “count” and “1” to the adder

• store sum in count

• We can use gates to select which register is connected to the input of the adder: multiplexers

– Also, must decide when the count has reached M

– How to tell when to connect which input to the adder??

Multiplexers• Multiplexers connect one of their inputs to the output,

– and allow one to choose which ones to select

– can be implemented with gates as wellInput

Control

Instructions and “Stored Program”• Let us have a “control unit”

– Tells which inputs are to be connected,

– Where to store the output

• How does the control unit know what to do?– Store instructions for it in another set of registers?

– What is an instruction?

• Identifies Input registers, and output register,

RegistersMultiplexor

Adder

Control Unit

Instructions

The SIS CPU

Registers

Multiplexor Adder

Control Unit

Instructionsmemory

with multiplexor

Multiplexor

PC: Program Counter

Let us add memory• When the amount of data is large:

– Can’t keep on adding registers

– Memory: a linear, random-access storage

– Cheaper, slower than registers

– Now we need to bring the data from memory into registers and vice versa

The SIS CPU

Registers

Multiplexors Adder

Control Unit

Instructionsmemory

with multiplexor

Multiplexor

PC

DataMemory

Load and StoreUnit

The instruction set so far• Add R1, R2, R3

– Add contents of registers R1 and R2 and store result in R3

• Load R1, R2– Load contents of R2 with memory location indicated by R1

• Store R1,R2– Store contents of R2 in memory location indicated by R1

• Need to decide when to stop adding (for squaring)– It suffices to compare two registers, and change PC based on the

result

– BEQ R1,R2, R3

• If contents of R1 and R2 are identical, change PC to contents of R3

A few more instructions• Loading fixed numbers into registers:

– have to be careful: instructions have only 16 bits, of which 4 are opcode

– LoadThis R1, Data

• It is called : LoadImmediate (LDI R1, Data)

• Data can be 8 bits only

• Where does it go? In the lower order 8 bits of the register?

• A couple more branching instructions– BNE R1, R2, R3 // branch to R3 if not equal

– BR R1 // unconditional branch

The machine language• Each instruction is:

– 16 bits

– 1 3-4 bit field to specify operation (need only 3 for now)

– 3 4-bit fields to identify operand registers

– Assign meanings to opcode bit combinations:

0001: Load

0010: Store

0011: Add

0100: BEQ

0101: BNE

0110: BR

1000: Input

1001: Output

1010: LDI

1011: SetZero

The squaring program• Read N

– and output square of N

Input R1

setzero R2 // count = 0;

setzero R3 // result = 0

LDI R5, 00001010 // ten

LDI R4, 00000001 // R4 has one

add R1, R3, R3

ADD R4,R2, R2

BNE R1,R2, R5

Output R3

Stop

0

2

4

6

8

10

12

14

16

18

1000 0001 xxxx xxxx

Complete the program

Question: is this program correct? Check it and report: correct if needed

The point of this exercise:• Get the “big picture”

– Bottom up:

• we know exactly how the whole machine works, assuming only that a “controllable switch” is available

– The entire machine architecture, with all the fundamental components is introduced in a simple setting

– A simpler machine language to get you started

• Later (soon!)– We switch to a real machine language (MIPS, in Chapter 3)

– Chapter 3 is about instruction set architecture,

• So we will then ignore how it can be implemented

• return to implementation in Chapter 5

–

Another program:• Suppose we want to print squares of M numbers, starting with N

– input sequence: 5, 3– output sequence: 25, 36, 49

• Write in Higher language• Then:

– convert to symbolic lang.– then to machine language

• We will program in – The “symbolic language”– Called the assembly language– Assemblers translate this

Higher level language:

read n, m

x = n;

while (x<=m) {

r=0; c=0;

while (c!=x)

{ r = r+x; c= c+1;}

output r;

x = x+1;}