notes2 memory_cpu

Imperial College
London
Department of Computing
Computer Systems 113
Architecture 110
Memory Organisation
CPU Organisation and Operation
Introduction to Assembly Programming
Dr. N. Dulay
November 2007
N. Dulay Main memory organisation (1)

Preliminaries
Welcome to the Computer Systems (113) / Architecture (110) course. Over the next two terms we’ll
study the basic operation and organisation of a computer. Hopefully you’ve all completed the self-
study on integer representations and arithmetic. Some preliminaries first:
How to contact me
There are 2 options: (i) see me at a tutorial, (2) email me using nd@doc.ic.ac.uk
Workload
The course consists of lectures (2 per week), tutorials (1 per week), a number of pieces of assessed
coursework and a lab exercise. The tutorials and lab exercise are not examined but are important as
they give you an opportunity to check on your progress and to discuss problems and ideas with the
tutorial helpers. The assessed coursework is marked and undertaken as homework. The lab exercise is
scheduled as part of the PPT programme.
Exams
There is a single combined Christmas test that normally has 1 architecture question. The main
examinations are held at the start of the summer term. Past exam papers can be accessed and viewed
via the departmental homepage.
Textbooks
Unfortunately there is no single ideal introductory textbook1
. The following textbooks are good
however:
− Structured Computer Organization (5th edition) by Andrew S. Tannenbaum, Prentice Hall
− Computer Organization and Architecture: Designing for Performance (7th Ed) by
William Stallings, Prentice-Hall.
The course also involves Intel Pentium assembly programming. For this you may wish to consider
acquiring a book about Pentium programming such as:
− Guide to Assembly Language Programming in Linux by S. Dandamudi, Springer.
An excellent web resource is Wikipedia at
http://en.wikipedia.org/
Website
The course notes, tutorials, slides etc will be available via the following URL:
https://www.doc.ic.ac.uk/~nd/architecture
Note: the URL starts https, not http. You need to print your own copies.
1 More ambitious students might also wish to consider Computer Organisation & Design by David. A. Patterson and
John. L. Hennessey, Morgan Kaufmann Publishing.

Main Memory (RAM) Organisation
Computers employ many different types of memory (semi-conductor, magnetic disks and tapes,
DVDs etc.) to hold data and programs. Each type has its own characteristics and uses. We will look
at the way that Main Memory (RAM) is organised and very briefly at the characteristics of Register
Memory and Disk Memory. Let’s locate these 3 types of memory in an abstract computer:
Disk DriveDisk Drive
Registers
Arithmetic & Logic
Unit
Control
Unit
RAM
Controller(s)
CPU Main Memory
I/O
RAM
Disk Drive
Register Memory
Registers are memories located within the Central Processing Unit (CPU). They are few in number
(there are rarely more than 64 registers) and also small in size, typically a register is less than 64 bits;
32-bit and more recently 64-bit are common in desktops.
The contents of a register can be “read” or “written” very quickly2
however, often an order of
magnitude faster than main memory and several orders of magnitude faster than disk memory.
Different kinds of register are found within the CPU. General Purpose Registers3
are available for
general4
use by the programmer. Unless the context implies otherwise we’ll use the term "register" to
refer to a General Purpose Register within the CPU. Most modern CPU’s have between 16 and 64
general purpose registers. Special Purpose Registers have specific uses and are either non-
programmable and internal to the CPU or accessed with special instructions by the programmer.
Examples of such registers that we will encounter later in the course include: the Program Counter
register (PC), the Instruction Register (IR), the ALU Input & Output registers, the Condition Code
(Status/Flags) register, the Stack Pointer register (SP). The size (the number of bits in the register) of
the these registers varies according to register type. The Word Size of an architecture is often (but not
always!) defined by the size of the general purpose registers.
2 e.g. less than a nanosecond (10
-9
sec)
3 Occasionally called Working Registers
4 Used for performing calculations, moving and manipulating data etc.

In contrast to main memory and disk memory, registers are referenced directly by specific instructions
or by encoding a register number within a computer instruction. At the programming (assembly)
language level of the CPU, registers are normally specified with special identifiers (e.g. R0, R1, R7,
SP, PC)
As a final point, the contents of a register are lost if power to the CPU is turned off, so registers are
unsuitable for holding long-term information or information that is needed for retention after a power-
shutdown or failure. Registers are however, the fastest memories, and if exploited can result in
programs that execute very quickly.
Main Memory (RAM)
If we were to sum all the bits of all registers within CPU, the total amount of memory probably would
not exceed 5,000 bits. Most computational tasks undertaken by a computer require a lot more
memory. Main memory is the next5
fastest memory within a computer and is much larger in size.
Typical main memory capacities for different kinds of computers are: PC 512MB6
, fileserver 2GB,
database server 8GB. Computer architectures also impose an architectural constraint on the
maximum allowable RAM. This constraint is normally equal to 2
WordSize
memory locations.
RAM7
(Random8
Access Memory) is the most common form of Main Memory. RAM is normally
located on the motherboard and so is typically less than 12 inches from the CPU. ROM (Read Only
Memory) is like RAM except that its contents cannot be overwritten and its contents are not lost if
power is turned off (ROM is non-volatile).
Although slower than register memory, the contents of any location9
in RAM can still be “read” or
“written” very quickly10
. The time to read or write is referred to as the access time and is constant for
all RAM locations.
In contrast to register memory, RAM is used to hold both program code (instructions) and data
(numbers, strings etc). Programs are “loaded” into RAM from a disk prior to execution by the CPU.
Locations in RAM are identified by an addressing scheme e.g. numbering the bytes in RAM from 0
onwards11
. Like registers, the contents of RAM are lost if the power is turned off.
5 Actually many computers systems also include Cache memory, which is faster than Main memory, but slower than
register memory. We will ignore Cache memories in this course.
6 1K = 2
10
= 1024, 1M = 2
20
, 1G = 2
30
, ‘B’ will be used for Bytes, and ‘b’ or ‘bit’ for bits, cf. 1MB and 1Mbit
7 There are many types of RAM technologies.
8 Random is a Misnomer. Direct Access Memory would have been a better term.
9 Typically a byte multiple.
10 e.g. less than 10 nanoseconds (10x10
-9
sec)
11 Some RAM locations (typically those with the lowest & highest addresses) may cause side-effects, e.g. cause data to
be transferred to/from external devices

Disk Memory
Disk memory12
is used to hold programs and data over the longer term. The contents of a disk are
NOT lost if the power is turned off. Typical hard disk capacities range from 40GB to over 500 GB
(5x10
29
). Disks are much slower than register and main memory, the access-time (known as the seek-
time) to data on disk is typically between 2 and 4 milli-seconds, although disk drives can transfer
thousands of bytes in one go achieving transfer rates from 25MB/s to 500MB/s.
Disks can be housed internally within a computer “box” or externally in an enclosure connected by a
fast USB or firewire cable13
. Disk locations are identified by special disk addressing schemes (e.g.
track and sector numbers).
Summary of Characteristics
12 Some authors refer to disk memory as disk storage.
13 For details about how disks and other storage devices work, check out Tanenbaum or Stallings.

SRAM, DRAM, SDRAM, DDR SDRAM
There are many kinds of RAM and new ones are invented all the time. One of aims is to make RAM
access as fast as possible in order to keep up with the increasing speed of CPUs.
SRAM (Static RAM) is the fastest form of RAM but also the most expensive. Due to its cost it is not
used as main memory but rather for cache memory. Each bit requires a 6-transistor circuit.
DRAM (Dynamic RAM) is not as fast as SRAM but is cheaper and is used for main memory. Each
bit uses a single capacitor and single transistor circuit. Since capacitors lose their charge, DRAM
needs to be refreshed every few milliseconds. The memory system does this transparently. There are
many implementations of DRAM, two well-known ones are SDRAM and DDR SDRAM.
SDRAM (Synchronous DRAM) is a form of DRAM that is synchronised with the clock of the CPU’s
system bus, sometimes called the front-side bus (FSB). As an example, if the system bus operates at
167Mhz over an 8-byte (64-bit) data bus , then an SDRAM module could transfer 167 x 8 ~
1.3GB/sec.
DDR SDRAM (Double-Data Rate DRAM) is an optimisation of SDRAM that allows data to be
transferred on both the rising edge and falling edge of a clock signal. Effectively doubling the amount
of data that can be transferred in a period of time. For example a PC-3200 DDR-SDRAM module
operating at 200Mhz can transfer 200 x 8 x 2 ~ 3.2GB/sec over an 8-byte (64-bit) data bus.
ROM, PROM, EPROM, EEPROM, Flash
In addition to RAM, they are also a range of other semi-conductor memories that retain their contents
when the power supply is switched off.
ROM (Read Only Memory) is a form of semi-conductor that can be written to once, typically in bulk
at a factory. ROM was used to store the “boot” or start-up program (so called firmware) that a
computer executes when powered on, although it has now fallen out-of-favour to more flexible
memories that support occasional writes. ROM is still used in systems with fixed functionalities, e.g.
controllers in cars, household appliances etc.
PROM (Programmable ROM) is like ROM but allows end-users to write their own programs and
data. It requires a special PROM writing equipment. Note: users can only write-once to PROM.
EPROM (Erasable PROM). With EPROM we can erase (using strong ultra-violet light) the contents
of the chip and rewrite it with new contents, typically several thousand times. It is commonly used to
store the “boot” program of a computer, known as the firmware. PCs call this firmware, the BIOS
(Basic I/O System). Other systems use Open Firmware. Intel-based Macs use EFI (Extensible
Firmware Interface).
EEPROM (Electrically Erasable PROM). As the name implies the contents of EEPROMs are erased
electrically. EEPROMSs are also limited to the number of erase-writes that can be performed (e.g,
100,000) but support updates (erase-writes) to individual bytes whereas EPROM updates the whole
memory and only supports around 10,000 erase-write cycles.
FLASH memory is a cheaper form of EEPROM where updates (erase-writes) can only be performed
on blocks of memory, not on individual bytes. Flash memories are found in USB sticks, flash cards
and typically range in size from 32M to 2GB. The number of erase/write cycles to a block is typically
several hundred thousand before the block can no longer be written.

Main Memory Organisation
Main memory can be considered to be organised as a matrix of bits. Each row represents a memory
location, typically this is equal to the word size of the architecture, although it can be a word multiple
(e.g. 2xWordsize) or a partial word (e.g. half the wordsize). For simplicity we will assume that
data within main memory can only be read or written a single row (memory location) at a time.
For a 96-bit memory we could organise the memory as 12x8 bits, or 8x12 bits or, 6x16 bits, or even
as 96x1 bits or 1x96 bits. Each row also has a natural number called its address14
which is used for
selecting the row:
Address <–––––––– 8 bit –––––––>
0
1
2
3
4
5
6
7
8
9
10
11
Address <––––––––––––– 12 bit –––––––––––––>
0
1
2
3
4
5
6
7
Address <––––––––––––––––––– 16 bit –––––––––––––––––––>
0
1
2
3
4
5
14 The concept of an address is very important to properly understanding how CPUs work.

Byte Addressing
Main-memories generally store and recall rows, which are multi-byte in length (e.g. 16-bit word = 2
bytes, 32-bit word = 4 bytes). Many architectures, however, make main memory byte-addressable
rather than word addressable. In such architectures the CPU and/or the main memory hardware is
capable of reading/writing any individual byte. Here is an example of a main memory with 16-bit
memory locations15
. Note how the memory locations (rows) have even addresses.
Word Address 16 bit = 2 bytes
0
2
4
6
8
10
12
14
16
18
20
Byte Ordering
The ordering of bytes within a multi-byte data item defines the endian-ness of the architecture.
In BIG-ENDIAN systems the most significant byte of a multi-byte data item always has the lowest
address, while the least significant byte has the highest address.
In LITTLE-ENDIAN systems, the least significant byte of a multi-byte data item always has the
lowest address, while the most significant byte has the highest address.
In the following example, table cells represent bytes, and the cell numbers indicate the address of that
byte in main memory. Note: by convention we draw the bytes within a memory word left-to-right for
big-endian systems, and right-to-left for little-endian systems.
Word
Address
Big-Endian Word
Address
Little-Endian
0 0 1 2 3 0 3 2 1 0
4 4 5 6 7 4 7 6 5 4
8 8 9 10 11 8 11 10 9 8
12 12 13 14 15 12 15 14 13 12
MSB –––––––––-> LSB MSB –––––––––––> LSB
15 To avoid confusion we will use the term memory word for a word-sized memory location.

Note: an N-character ASCII string value is not treated as one large multi-byte value, but rather as N
byte values, i.e. the first character of the string always has the lowest address, the last character has
the highest address. This is true for both big-endian and little-endian. An N-character Unicode string
would be treated as N two-byte value and each two-byte value would require suitable byte-ordering.
Example: Show the contents of memory at word address 24 if that word holds the number given by
122E 5F01H in both the big-endian and the little-endian schemes?
Big Endian Little Endian
MSB –––––––––> LSB MSB –––––––––> LSB
24 25 26 27 27 26 25 24
Word 24 12 2E 5F 01 Word 24 12 2E 5F 01
Example: Show the contents of main memory from word address 24 if those words hold the text
JIM SMITH.
Big Endian Little Endian
+0 +1 +2 +3 +3 +2 +1 +0
Word 24 J I M Word 24 M I J
Word 28 S M I T Word 28 T I M S
Word 32 H ? ? ? Word 32 ? ? ? H
The bytes labelled with ? are unknown. They could hold important data, or they could be don’t care
bytes – the interpretation is left up to the programmer.
Unfortunately computer systems16
, in use today are split between those that are big-endian, and those
that are little-endian17
. This leads to problems when a big-endian computer wants to transfer data to a
little-endian computer. Some architectures, for example the PowerPC and ARM, allow the endian-
ness of the architecture to be changed programmatically.
Word Alignment
Although main-memories are generally organised as byte-addressed rows of words and accessed a
row at a time, some architectures, allow the CPU to access any word-sized bit-group regardless of its
byte address. We say that accesses that begin on a memory word boundary are aligned accesses
while accesses that do not begin on word boundaries are unaligned accesses.
Address Memory (16-bit) word
0 MSB LSB Word starting at Address 0 is Aligned
2
4 MSB Word starting at Address 5 is Unaligned
6 LSB
16 The interested student might want to read the paper, “On Holy Wars and a Plea for Peace”, D. Cohen, IEEE
Computer, Vol 14, Pages 48-54, October 1981.
17 The Motorola 68000 architecture is big-endian, while the Intel Pentium architecture is little-endian.

Reading an unaligned word from RAM requires (i) reading of adjacent words, (ii) selecting the
required bytes from each word and (iii) concatenating those bytes together => SLOW. Writing an
unaligned word is more complex and slower18
. For this reason some architectures prohibit unaligned
word accesses. e.g. on the 68000 architecture, words must not be accessed starting from an odd-
address (e.g. 1, 3, 5, 7 etc), on the SPARC architecture, 64-bit data items must have a byte address
that is a multiple of 8.
Memory Modules, Memory Chips
So far, we have looked at the logical organisation of main memory. Physically RAM comes on small
memory modules (little green printed circuit-boards about the size of a finger). A typical memory
module holds 512MB to 2GB. The computer’s motherboard will have slots to hold 2, 4 maybe 8
memory modules. Each memory module is itself comprised of several memory chips. For example
here are 3 ways of forming a 256x8 bit memory module.
1
1
1
1
0
0
1
1
Eight
256 x 1bit RAMs
1
1
1
1
0
0
1 1
256 x
4bit
RAM
256 x
4bit
RAM256 x 8bit RAM
1
1
1
0
1
1
1
0
In the first case, main memory is built with a single memory chip. In the second, we use two memory
chips, one gives us the most significant 4 bits, the other, the least significant 4 bits. In the third we
use 8 memory chips, each chip gives us 1 bit - to read an 8 bit memory word, we would have to
access all 8 memory chips simultaneously and concatenate the bits.
On PCs, memory modules are known as DIMMs (dual inline memory modules) and support 64-bit
transfers. The previously generation of modules were called SIMMs (single inline memory modules)
and supported 32-bit data transfers.
Example: Given Main Memory = 1M x 16 bit (word addressable),
RAM chips = 256K x 4 bit
Module 0 Module 1 Module 2 Module 3
∧

2
18

C
H
I
P
0
C
H
I
P
1
C
H
I
P
2
C
H
I
P
3
C
H
I
P
4
C
H
I
P
5
C
H
I
P
6
C
H
I
P
7
C
H
I
P
8
C
H
I
P
9
C
H
I
P
10
C
H
I
P
11
C
H
I
P
12
C
H
I
P
13
C
H
I
P
14
C
H
I
P
15
18 Describe a method for doing an unaligned word write operation.

∨
4x4 bits 4x4 bits 4x4 bits 4x4 bits
RAM chips per memory module = Width of Memory Word = 16/4 = 4
Width of RAM Chip
18 bits are required to address a RAM chip (since 256K = 2
18
= Length of RAM Chip )
A 1Mx16 bit word-addressed memory requires 20 address bits (since 1M =2
20
)
Therefore 2 bits (=20–18) are needed to select a module.
The total number of RAM Chips = (1M x 16) / (256K x 4) = 16.
Total number of Modules = Total number of RAM chips / RamChipsPerModule = 16/4 = 4
Interleaved Memory
When memory consists of several memory modules, some address bits will select the module, and the
remaining bits will select a row within the selected module.
When the module selection bits are the least significant bits of the memory address we call the
resulting memory a low-order interleaved memory.
When the module selection bits are the most significant bits of the memory address we call the
resulting memory a high-order interleaved memory.
Interleaved memory can yield performance advantages if more than one memory module can be
read/written at a time:-
(I) for low-order interleave if we can read the same row in each module. This is good for a single
multi-word access of sequential data such as program instructions, or elements in a vector,
(ii) for high-order interleave, if different modules can be independently accessed by different units.
This is good if the CPU can access rows in one module, while at the same time, the hard disk
(or a second CPU) can access different rows in another module.
Example: Given that Main Memory = 1Mx8bits, RAM chips = 256K x 4bit. For this memory we
would require 4x2=8 RAM chips. Each chip would require 18 address bits (ie. 2
18
= 256K) and the
full 1Mx16 bit memory would requires 20 address bits (ie. 2
20
= 1M )

CPU Organisation & Operation
The Fetch-Execute Cycle
The operation of the CPU19
is usually described in terms of the Fetch-Execute cycle20
.
Fetch-Execute Cycle The cycle raises many interesting questions, e.g.
Fetch the Instruction What is an Instruction? Where is the Instruction? Why does it
need to be fetched? Isn't it okay where it is? How does the
computer keep track of instructions? Where does it put the
instruction it has just fetched?
Increment the Program
Counter
What is the Program Counter? What does the Program Counter
count? Increment by how much? Where does the Program
Counter point to after it is incremented?
Decode the Instruction Why does the instruction need to be decoded? How does it get
decoded?
Fetch the Operands What are operands? What does it mean to fetch? Is this fetching
distinct from the fetching in Step 1 above? Where are the
operands? How many are there? Where do we put the operands
after we fetch them?
Perform the Operation Is this the main step? Couldn't the computer simply have done this
part? What part of the CPU performs this operation?
Store the results What results? Where from? Where to?
Repeat forever Repeat what? Repeat from where? Is it really an infinite loop?
Why? How do these steps execute any instructions at all?
In order to appreciate the operation of a computer we need to answer such questions and to consider
in more detail the organisation of the CPU.
Representing Programs
Each complex task carried out by a computer needs to be broken down into a sequence of simpler
tasks and a binary machine instruction is needed for the most primitive tasks. Consider a task that
adds two numbers21
, held in memory locations designated by B and C22
and stores the result in memory
location designated by A.
A = B + C
This assignment can be broken down (compiled) into a sequence of simpler tasks or assembly
instructions, e.g:
19 Central Processing Unit.
20 Sometimes called the Fetch-Decode-Execute Cycle.
21 Let’s assume they are held in two’s complement form.
22 A, B and C are actually main memory addresses, i.e. natural binary numbers.
N. Dulay CPU organisation and operation (12)

Assembly
Instruction
Effect
LOAD R2, B Copy the contents of memory location designated by B into Register 2
ADD R2, C Add the contents of the memory location designated by C to the contents of
Register 2 and put the result back into Register 2
STORE R2, A Copy the contents of Register 2 into the memory location designated by A.
Each of these assembly instructions needs to be encoded into binary for execution by the Central
Processing Unit (CPU). Let’s try this encoding for a simple architecture called TOY1.
TOY1 Architecture
TOY1 is a fictitious architecture with the following characteristics:
1024 x 16-bit words of RAM maximum. RAM is word-addressable.
4 general purpose registers R0, R1, R2 and R3. Each general purpose register is 16-bits (the
same size as a memory location).
16 different instructions that the CPU can decode and execute, e.g. LOAD, STORE, ADD,
SUB and so on. These different instructions constitute the Instruction Set of the
Architecture.
The representation for integers will be two’s complement.
For this architecture, the architect (us) needs to define a coding scheme23
for instructions. This is
termed the Instruction Format. Lets look at an example before we consider how we arrived at it.
Here’s our instruction format for TOY1:
TOY1 Instruction Format
TOY1 instructions are 16-bits (so they will fit into a main-memory word). Each instruction is divided
into a number of instruction fields that encode a different piece of information for the CPU.
Field Name
Field Width
OPCODE
4-bits
REG
2-bits
ADDRESS
10-bits
The OPCODE24
field identifies the CPU operation required. Since TOY1 only supports 16
instructions, these can be encoded as a 4-bit natural number. For TOY1, opcodes 1 to 4 will be25
:
0001 = LOAD 0010 = STORE 0011 = ADD 0100 = SUB
The REG field defines a General CPU Register. Arithmetic operations will use 1 register operand
and 1 main memory operand, results will be written back to the register. Since TOY1 has 4 registers;
these can be encoded as a 2-bit natural number:
23 Most architectures actually have different instruction formats for different categories of instruction.
24 Operation Code
25 The meaning of CPU operations is defined in the Architecture’s Instruction Set Manual.

00 = Register 0 01 = Register 1 10 = Register 2 11 = Register 3
The ADDRESS field defines the address of a word in RAM. Since TOY1 can have upto 1024
memory locations; a memory address can be encoded as a 10-bit natural number.
If we define addresses 200H, 201H and 202H for A, B and C, we can encode the example above as:
Assembly Instruction Machine Instruction
LOAD R2, [201H] 0001 10 10 0000 0001
ADD R2, [202H] 0011 10 10 0000 0010
STORE R2, [200H] 0010 10 10 0000 0000
Memory Placement of Program and Data
In order to execute a TOY1 program, its instructions and data needs to placed within main memory26
.
We’ll place our 3-instruction program in memory starting at address 080H and we’ll place the
variables A, B and C at memory words 200H, 201H, and 202H respectively. Such placement results
in the following memory layout prior to program execution. For convenience, memory addresses and
memory contents are also given in hex.
Memory Address
in binary & hex
Machine Instruction
OP Reg Address
Assembly
Instruction
0000 1000 0000
0 8 0
10 0000 0001
1 A 0 1
LOAD R2, [201H]
0000 1000 0001
0 8 1
10 0000 0010
3 A 0 2
ADD R2, [202H]
0000 1000 0010
0 8 2
10 0000 0000
2 A 0 0
STORE R2, [200H]
Etc Etc Etc
0010 0000 0000
2 0 0
0000 0000 0000 0000
0 0 0 0
A = 0
0010 0000 0001
2 0 1
0000 0000 0000 1001
0 0 0 9
B = 9
0010 0000 0010
2 0 2
0000 0000 0000 0110
0 0 0 6
C = 6
Of course, the big question is “How is such a program executed by the TOY1 CPU?”
26 The Operating System software is normally responsible for undertaking this task.

CPU Organisation
ALU
General
Registers
Instruction Decoder
Program Counter
Instruction Register
Control Unit
Input Register 1
InternalBus
Central Processing Unit (CPU) Memory
Address Bus
Data Bus
Input Register 2
Output
Register
000
001
002
3FD
3FE
3FF
R0
R1
R2
R3
$
$
Addr.
Control Bus
Read/Write
The Program Counter (PC) is a special register that holds the address of the next instruction to be
fetched from Memory (for TOY1, the PC is 10-bits wide). The PC is incremented27
to "point to" the
next instruction while an instruction is being fetched from main memory.
The Instruction Register (IR) is a special register that holds each instruction after it is fetched from
main memory. For TOY1, the IR is 16-bits since instructions are 16-bit wide.
The Instruction Decoder is a CPU component that decodes and interprets the contents of the
Instruction Register, i.e. its splits whole instruction into fields for the Control Unit to interpret. The
Instruction decoder is often considered to be a part of the Control Unit.
The Control Unit is the CPU component that co-ordinates all activity within the CPU. It has
connections to all parts of the CPU, and includes a sophisticated timing circuit.
The Arithmetic & Logic Unit (ALU) is the CPU component that carries out arithmetic and logical
operations e.g. addition, comparison, boolean AND/OR/NOT.
The ALU Input Registers 1 & 2 are special registers that hold the input operands for the ALU.
The ALU Output Register is a special register that holds the result of an ALU operation. On
completion of an ALU operation, the result is copied from the ALU Output register to its final
destination, e.g. to a CPU register, or main-memory, or to an I/O device.
The General Registers R0, R1, R2, R3 are available for the programmer to use in his/her programs.
Typically the programmer tries to maximise the use of these registers in order to speed program
execution. For TOY1, the general registers are the same size as memory locations, i.e. 16-bits.
27 By the appropriate number of memory words.

The Buses serve as communication highways for passing information within the CPU (CPU internal
bus) and between the CPU and the main memory (the address bus, the data bus, and the control
bus). The address bus is used to send addresses from the CPU to the main memory; these addresses
indicate the memory location the CPU wishes to read or write. Unlike the address bus, the data bus is
bi-directional; for writing, the data bus is used to send a word from the CPU to main-memory; for
reading, the data bus is used to send a word from main-memory to the CPU. For TOY1, the Control
bus28
is used to indicate whether the CPU wishes to read from a memory location or write to a
memory location. For simplicity we’ve omitted two special registers, the Memory Address Register
(MAR) and the Memory Data Register (MDR). These registers lie at the boundary of the CPU and
Address bus and Data bus respectively and serve to buffer data to/from the buses.
Buses can normally transfer more than 1-bit at a time. For the TOY1, the address bus is 10-bits (the
size of an address), the data bus is 16-bits (size of a memory location), and the control bus is 1-bit (to
indicate a memory read operation or a memory write operation).
Interlude: the Von Neumann Machine Model
Most computers conform to the von Neumann’s machine model, named after the Hungarian-
American mathematician John von Neumann (1903-57).
In von Neumann’s model, a computer has 3 subsystems (i) a CPU, (ii) a main memory, and (iii) an
I/O system. The main memory holds the program as well as data and the computer is allowed to
manipulate its own program29
. In the von-Neumann model, instructions are executed sequentially
(one at a time).
In the von-Neumann model a single path exists between the control until and main-memory, this
leads to the so-called "von Neumann bottleneck" since memory fetches are the slowest part of an
instruction they become the bottleneck in any computation.
Instruction Execution (Fetch-Execute-Cycle Micro-steps)
In order to execute our 3-instruction program, the control unit has to issue and coordinate a series of
micro-instructions. These micro-instructions form the fetch-execute cycle. For our example we will
assume that the Program Counter register (PC) already holds the address of the first instruction,
namely 080H.
28 Most control-buses are wider than a single bit, these extras bits are used to provide more sophisticated memory
operations and I/O operations.
29 This type of manipulation is not regarded as a good technique for general assembly programming.

LOAD R2, [201H]
0000 1000 0000
0 8 0
10 0000 0001
1 A 0 1
Copy the value in memory word
201H into Register 2
Control Unit Action
FETCH INSTRUCTION30
Data flows
PC to Address Bus31
080H 080H Address Bus
0 to Control Bus32
0 0 Control Bus
Address Bus to Memory 080H 080H Memory
Control Bus to Memory 0 READ 0 Memory
Increment PC33
080 INC 081H PC becomes PC+134
Memory [080H] to Data Bus 1A01H 1A01H Data Bus
Data Bus to Instruction Register 1A01H 1A01H Instruction Register
DECODE INSTRUCTION
IR to Instruction Decoder 1A01H 1A01H Instruction Decoder
Instruction Decoder to Control
Unit35
1, 2,
201H
1, 2, 201H Control Unit
EXECUTE INSTRUCTION36
Control Unit to Address Bus 201H 201H Address Bus
0 to Control Bus 0 0 Control Bus
Memory [201H] to Data bus 0009H 0009H Data Bus
Data Bus to Register 2 0009H 0009H Register 2
30 The micro-steps in the Fetch and Decode phases are common for all instructions.
31 This and the next 4 micro-steps initiate a fetch of the next instruction to be executed, which is to found at memory
address 80H. In practice a Memory Address Register (MAR) acts as an intermediate buffer for the Address,
similarly a Memory Data Register (MDR) buffers data to/from the data bus.
32 We will use 0 for a memory READ request, and 1 for a memory WRITE request.
33 For simplicity, we will assume that the PC is capable of performing the increment internally. If not, the Control
Unit would have to transfer the contents of the PC to the ALU, get the ALU to perform the increment and send the
results back to the PC. All this while we are waiting for the main-memory to return the word at address 80H.
34 Since TOY1’s main-memory is word-addressed, and all instructions are 1 word. If main-memory was byte-
addressed we would need to add 2.
35 The Instruction decoder splits the instruction into the individual instruction fields OPCODE, REG and ADDRESS
for interpretation by the Control Unit.
36 The micro-steps for the execute phase actually perform the operation.

ADD R2, [202H]
0000 1000 0001
0 8 1
10 0000 0002
3 A 0 2
Add37
the value in memory word
202H to Register 2
Control Unit Action
FETCH INSTRUCTION
Data flows
PC to Address Bus 081H 081H Address Bus
Increment PC 081H INC 082H PC becomes PC+1
Memory [081H] to Data Bus 3A02H 3A02H Data Bus
DECODE INSTRUCTION
IR to Instruction Decoder 3A02H 3A02H Instruction Decoder
Unit
3, 2,
202H
EXECUTE INSTRUCTION
Register 2 to ALU Input Reg 1 0009 0009 ALU Input Reg 1
Memory [202H] to Data bus 0006H 0006H Data Bus
Data Bus to ALU Input Reg 2 0006H 0006H ALU Input Reg 2
Control Unit to ALU ADD 000FH Output Register
ALU Output Reg to Register 2 000F 000FH Register 2
37 Using two’s complement arithmetic.

STORE R2, [200H]
0000 1000 0001
0 8 2
10 0000 0000
2 A 0 0
Copy the value in Register 2 into
memory word 202H
Control Unit Action
FETCH INSTRUCTION
Data flows
PC to Address Bus 082H 082H Address Bus
Increment PC 082H INC 083H PC becomes PC+1
Memory [082] to Data Bus 2A00H 2A00H Data Bus
DECODE INSTRUCTION
IR to Instruction Decoder 2A00 2A00 Instruction Decoder
Unit
2, 2,
200H
EXECUTE INSTRUCTION
Register 2 to Data Bus 000FH 000FH Data Bus
Data Bus to Memory 000FH 000FH Memory
Control Bus to Memory 1 WRITE 1 Memory

TOY1 Programming
How is computer such as TOY1 programmed? We’ll consider this question with some examples.
Let’s first define a basic Instruction Set for the TOY1 architecture38
:
OP Code Assembler Format Action
0000 STOP Stop Program Execution
0001 LOAD Rn, [addr] Rn = Memory [addr]
0010 STORE Rn, [addr] Memory [addr ] = Rn
0011 ADD Rn, [addr] Rn = Rn + Memory [addr]
0100 SUB Rn, [addr] Rn = Rn – Memory [addr]
0101 GOTO addr PC = addr
0110 IFZER Rn, addr IF Rn = 0 THEN PC = addr
0111 IFNEG Rn, addr IF Rn < 0 THEN PC = addr
Example 1: Multiplication
Given these instructions lets write a TOY1 assembly program, which will perform the following
assignment:
A = B * C
where A, B and C denote integers placed at memory words 100H, 101H and 102H respectively. The
first point to observe with this example is that a multiply operation is not available in the TOY1
instruction set! Therefore we need to consider if we can use other instructions to carry out the
multiplication. The obvious solution is to use repeated addition:
B*C = B
N=1
C
∑
Example: 12 * 3 = 12 + 12 + 12
12 * 1 = 12
12 * 0 = 0
38 Note: Only half of the possible sixteen instructions are defined. The remaining 8 will be defined later. The STOP
instruction does not make use of the Register or Address fields, while the GOTO instruction does not make use of
the Register field.

Let’s first write the multiplication algorithm in Pseudo Code
;Given:A, B, C
;Pre: C >= 0 Why do we have thispre-condition?
;Post:A = B * C
sum = 0 ;sum willaccumulate the answer
n = C ;n willindicatehow many additionsremain
loop
exit when n <= 0
sum = sum + B
n = n -1
end loop
A = sum
Let’s try translating (compiling) this Pseudo Code to TOY1 instructions. Since we have 4 general
registers, it is worthwhile allocating frequently used variables to them as this will lead to faster
execution. Let’s allocate Register 1 to hold 'sum', and Register 2 to hold 'n'.
sum = 0
The first assignment sum=0 yields our first problem. How do we get zero (or any constant) into a
Register?
The only instruction that we can use to set a register is LOAD Rn, addr. Therefore we must reserve
a memory word and pre-set it to zero before program execution begins. Lets place zero in memory
word 200H. Now to perform sum = 0 we have:
LOAD R1, [200H] ; sum = 0
Let’s place instructions starting at memory word 80H:
Address Assembler Instruction Comment39
80H LOAD R1, [200H] ; sum = 0
200H 0 ; holds zero
n = C
The next statement is n = C. This is easy to translate:
81H LOAD R2, [102H] ; n = C
39 Try to comment each line of an assembly-language program.

exit when n <= 0
What does the loop exit when n <= 0 statement mean in TOY1 terms? Lets consider a simpler
example first: loop exit when n = 0. On the TOY1 this statement has a simpler translation, namely:
loop
exit when n = 0
instructions
end loop
Address Instruction
L0 IFZER R2, Ly
L1 instructions
... ...
Lx GOTO L0
Ly
Note: GOTO alters the Program Counter register thereby causing an unconditional branch in the
order of program execution. IFZER alters the Program Counter only if the contents of the specified
Register are zero. To handle exit when n <= 0 we need to skip to the end of the loop if R2 is zero
or if R2 is negative:
loop
exit when n <= 0
instructions
end loop
Address Instruction
L0 IFZER R2, Ly
L1 IFNEG R2, Ly
L2 instructions
... ...
Lx GOTO L0
Ly
For our example we now have the following assembly program:
Address Assembler Instruction Comment
80H LOAD R1, [200H] ; sum = 0
81H LOAD R2, [102H] ; n = C
82H IFZER R2, Ly ; exit when n<=0
83H IFNEG R2, Ly ; we will define Ly when we can
..... .................. ...............
Lx GOTO 82H ; end loop
Ly ....
.....
100H A ; holds A
101H B ; holds B
102H C ; holds C
.....
200H 0 ; holds 0

sum = sum + B
n = n – 1
Let’s continue with: sum = sum + B. This is easy, namely
84H ADD R1, [101H] ; sum = sum + B
For n = n – 1 we will assume that location 201H is pre-set to the constant 1.
85H SUB R2, [201H] ; n = n–1
201H 1 ; Holds the value 1
end loop
A = sum
Adding STORE R1, 100H for A = sum and a STOP instruction we arrive at the final program:
Addr Assembler Instruct. Comment Machine Instruction
80H LOAD R1, [200H] ; sum = 0 0001 0110 0000 0000
81H LOAD R2, [102H] ; n = C 0001 1001 0000 0010
82H IFZER R2, 87H ; exit when n = 0 0110 1000 1000 0111
83H IFNEG R2, 87H ; exit when n < 0 0111 1000 1000 0111
84H ADD R1, [101H] ; sum = sum + B 0011 0101 0000 0001
85H SUB R2, [201H] ; n = n - 1 0100 1010 0000 0001
86H GOTO 82H ; end loop 0101 0000 1000 0010
87H STORE R1, [100H] ; A = sum 0010 0101 0000 0000
88H STOP ; End of program 0000 0000 0000 0000
... ...
100H A ; Holds A initial value of A
101H B ; Holds B initial value of B
102H C ; Holds C initial value of C
... ...
200H 0 ; Holds Zero 0000 0000 0000 0000
201H 1 ; Holds One 0000 0000 0000 0001

Multiplication (An improvement)
The multiply program will work correctly but can be improved. Consider 3 * 1000 if C is greater than
B then it will be faster to compute 1000 * 3. How can we adapt our program to handle this case?
Consider and work through the following solution:
sum = 0
if B <= C then
big=C, n=B
else C < B
big=B, n=C
end if
loop exit when n <= 0
sum = sum + big
n = n – 1
end loop
A = sum
Addr Instruction
80H LOAD R1, [200H] ; sum=0
81H LOAD R0, [102H ] ; if C<B
82H SUB R0, [101H] ; then ELSE
83H IFNEG R0, 88H
84H LOAD R0, [102H] ; then
85H STORE R0, [202H ]; big = C
86H LOAD R2, [101H] ; n = B
87H GOTO 8BH
88H LOAD R0, [101H] ; else
89H STORE R0, [202H ]; big=B
8AH LOAD R2, [102H ]; n=C
8BH etc ; loop....
...
202H ... ; Holds big
Example 2: Vector Sum
Write a sequence of TOY1 instructions (and constants) to sum 100 integers stored consecutively
starting at memory word 200H. The sum is to be left in Register 0.
Again, lets first write the Pseudo Code for the problem:
sum = 0
n = 100
addr = 200H
loop
exit when n <= 0
sum = sum + RAM [addr]
addr = addr + 1
n = n -1
end loop
Looking at this code, we find that the main "difficulty" is how to perform
There doesn't appear to be any way of accessing memory words based on a "Variable". We need
therefore to extend TOY1 to include an indirect addressing capability40
.
40 In fact there is a way of writing this program without resorting to indirect memory access instructions. Can you
think what the way might be?

Indirect Addressing Instructions for TOY1
OP Code Assembler Format Action
1001 LOAD Rn, [Rm] Rn = Memory [Rm]41
1010 STORE Rn, [Rm] Memory [Rm] = Rn
1011 ADD Rn, [Rm] Rn = Rn + Memory [Rm]
1100 SUB Rn, [Rm] Rn = Rn – Memory [Rm]
A second Instruction Format is also needed for these instructions42
. We will use the following:
OPCODE REGn REGm Unused43
4-bits 2-bits 2-bits 8-bits
Example: Given this format the TOY1 instruction ADD R1, [R2] would be coded as
1011 01 10 0000 0000 in binary or B600H in hexadecimal.
Vector Sum Example Contd.
The vector sum example is now straightforward. This program will be placed at 0FH onwards, and
the registers allocated as follows: R0 for 'sum', R1 for 'n', R2 for 'addr'
sum = 0
n = 100
addr = 200H
loop
exit when n <= 0
addr = addr + 1
n = n - 1
end loop
; Result in Register R0
0 0 ; Holds 0
1 1 ; Holds 1
2 100 ; Holds 100
3 200H ; Holds 200H
...
0FH LOAD R0, [0] ; sum = 0
10H LOAD R1, [2] ; n = 100
11H LOAD R2, [3] ; addr = 200H
12H IFZER R1, 18H ; exit when n<=0
13H IFNEG R1, 18H
14H ADD R0, [R2] ; sum = sum+...
15H ADD R2, [1] ; addr = addr + 1
16H SUB R1, [1] ; n = n – 1
17H GOTO 12H ; end loop
18H STOP
41 Memory [Rm] denotes the contents of the memory word whose address is given by Register m.
42 Is having more than one instruction format a good idea?
43 For these instructions 8-bits are unused. A more advanced CPU could allow two address-indirect instructions to be
encoded into one word, thus skipping one instruction fetch.

notes2 memory_cpu

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