ec 413 computer organization - secure computing · department of electrical & computer engineering...

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Department of Electrical & Computer Engineering

EC 413 Computer Organization

Prof. Michel A. Kinsy

Lecture 2: Information Representation


Computing Systems §  A system is a set of related components that

works as a whole to achieve a goal §  A system contains:

•  Inputs •  Behavior •  Outputs

§  Behavior is a function that translates inputs to outputs

Behaviorinputs outputs: :


System Components §  Components are electronic blocks: analog,

digital, and mixed signal •  Analog system has values from a continuous set

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System Components §  Components are electronic blocks: analog,

digital, and mixed signal •  Digital system is a system in which signals have a

finite number of discrete values


Modern Computing Systems §  Modern computing systems are mainly digital

systems §  Digital representation means that everything is

represented by numbers only §  Digital representation is very suited for both

numerical and non-numerical information or data processing •  Numerical values: 0, 4, 5567, 24354543, -2 •  Non-numerical values: There will be light! (Each

letter in this sentence)


Modern Computing Systems §  Our physical world is analog and computing is

interacting with the physical world: •  So we need to convert physical signals to digital

then back to analog to communicate with the real world

-

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Bit & Byte §  A bit is the smallest unit of information

•  It represents one 2-way decision or a 2-way choice •  yes / no, true / false, on / off ... •  Abstraction of all of these is represented as 0 or 1

§  A single digit with one of two values §  Binary digit à bit

§  All information in a computer is stored and processed as bits •  High voltage/low voltage, current flowing/not

flowing §  A byte is 8 bits that are treated as a unit


Review of of decimal numbers §  "Decimal" numbers have 10 digits

•  0 1 2 3 4 5 6 7 8 9 §  One decimal digit represents one choice out of 10 §  2 decimal digits represents one choice out of 100

•  00 01 02 … 10 11 12 … 20 21 22 … 98 99 §  Decimal numbers are short-hands for sums of

powers of 10 §  1492 = 1 x 1000 + 4 x 100 + 9 x 10 + 2 x 1 §  = 1 x 103 + 4 x 102 + 9 x 101 + 2 x 100


Review of of decimal numbers §  Most us count in "base 10", using powers of 10 §  Number of items and number of digits are

tightly related: •  One dictates the other •  Maximum number value given number of digit is

§  basenumber of digits

§  For example BU ID number has 9 digits §  How many possible student numbers can we

form?

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Binary Number System §  It uses bits to represent numbers §  It follows the same rules as the decimal number

system except has only two digits: 0 and 1 §  Number representation are based on powers of 2

•  1, 2, 4, 8, 16, 32, … §  Instead of powers of 10 seen in the decimal system

•  1, 10, 100, 1000, … §  1 binary digit can represent one choice from 2,

count 2 things or specify 2 distinct values


Binary Number System §  1 binary digit (1-bit value)

•  0 and 1 §  2 binary digits can represent one choice from 4,

count 4 things or specify 4 distinct values •  00, 01, 10, 11

§  3 binary digits can represent one choice from 8, count 8 things or specify 8 distinct values •  000, 001, 010, 011, 100, 101, 110 , 111


Binary Number System §  Binary numbers are short-hands for sums of

powers of 2 •  11011 = 1 x 16 + 1 x 8 + 0 x 4 + 1 x 2 + 1 x 1 = 1 x 24 + 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20

§  Most of us count in "base 10", using powers of 10

§  Computer counts in "base 2", using power of 2

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Binary Number System §  8 bits (1 byte) can encode 28 = 256 different

values §  e.g., numbers 0 .. 255 §  2 bytes (16 bits) can encode 216 = 65536

values §  With 4 bytes (32 bits) can encode 232 =

4,294,967,296 values §  With 8 bytes (64 bits) can encode 264 =

18,446,744,073,709,551,616


Binary arithmetic §  It works just like decimal arithmetic

Addition

0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10

Subtraction 0 - 0 = 0 0 - 1 = -1 1 - 0 = 1 1 - 1 = 0

Small Problem We will see how to solve it later in the lecture


Converting binary to decimal §  Converting to decimal, so we can use

polynomial evaluation •  101101012 =1x27 + 0x26 + 1x25 + 1x24 + 0x23 + 1x22 + 0x21 + 1x20 =128 + 32 + 16 + 4 + 1 = 18110

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Converting decimal to binary §  Divide by target base (2 in this case) §  Remainders become digits in the new

representation §  Digits produced in right to left order §  Quotient is used as next dividend §  Stop when the quotient becomes zero, but use

the corresponding remainder


Converting decimal to binary 97 ÷ 2 à quotient = 48, remainder = 1 (LSB) 48 ÷ 2 à quotient = 24, remainder = 0 24 ÷ 2 à quotient = 12, remainder = 0 12 ÷ 2 à quotient = 6, remainder = 0 6 ÷ 2 à quotient = 3, remainder = 0 3 ÷ 2 à quotient = 1, remainder = 1 1 ÷ 2 à quotient = 0 (Stop) remainder = 1 (MSB)

Answer=11000012


Hexadecimal notation §  Binary number can be long and hard to read, so

hexadecimal numbers were introduced §  It combines 4 bits into a single digit, written in

base 16 §  A more compact and more readable

representation of the same information §  Hexadecimal number system uses the symbols

A, B, C, D, E, F for the digits 10, 11, 12, 13, 14, and 15, respectively

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Numbers and Bases Decimal Base-10 Binary Base-2 Hexadecimal Base-16

0 00000 01 00001 12 00010 23 00011 34 00100 45 00101 56 00110 67 00111 78 01000 89 01001 910 01010 A11 01011 B12 01100 C13 01101 D14 01110 E15 01111 F16 10000 1017 10001 11


Text Representation §  ASCII: American

Standard Code for Information Interchange •  It is an arbitrary

but agreed-upon representation in the US

•  It is now widely used around the world


Information Representation §  Unsigned and signed Integer, floating point

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Unsigned Integer §  An unsigned integer containing n bits can have

a value between 0 and 2n - 1 §  Memory addresses are always represented by

unsigned integers §  Given the binary number 11001:

•  = 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20 •  = 16 + 8 + 0 + 0 + 1 •  = 25


Signed Integer Representation §  Positive value representation alone are not the

problem §  Negative value representation is the challenge

•  To represent negative values, computer systems allocate the high-order bit to indicate the sign of a value

•  The high-order bit is the leftmost bit in a binary number

•  It is also called the most significant bit •  The remaining bits contain the value of the number


Signed Integer Representation §  There are three ways in which signed binary

numbers may be expressed: •  Signed magnitude •  One’s complement •  Two’s complement

§  In an 8-bit binary number, signed magnitude representation places the absolute value of the number in the 7 bits to the right of the sign bit

•  4 = 0 0000100 •  -4 = 1 0000100

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Signed magnitude Representation §  Arithmetic operations on signed magnitude

numbers are performed inside the computer in much the same way as humans carry out pencil and paper arithmetic

§  Sum of 74 and 46 •  1- Convert 74 and 46 to binary •  2- Arrange as a sum but separate the sign bits from

the magnitude bits 0 1 0 0 1 0 1 0 0 + 0 1 0 1 1 1 0 = 0 1 1 1 1 0 0 0


Signed magnitude Representation §  In this example, two values were picked such as

the sum would fit into seven bits §  If that is not the case, there is a problem §  Sum of 102 and 46

•  The carry from the seventh bit overflows and it is discarded

•  Erroneous result: 102 + 46 = 36 0 1 1 0 0 1 1 0 0 + 0 1 0 1 1 1 0 = 0 0 1 0 0 1 0 0


Signed magnitude Representation §  Signed magnitude representation is easy for

people to understand •  But it requires complicated computer hardware •  Another disadvantage of signed magnitude is that

it allows two different representations for zero: positive zero and negative zero

§  So computers systems uses complement systems for numeric value representation

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One’s complement §  With one’s complement

addition, the carry bit is carried around and added to the sum

§  Sum of 48 and – 19 •  19 in binary is 00010011 •  -19 in one’s complement is

11101100 •  Answer is 29

0 0 1 1 0 0 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 1 0 0 1 0 0 0 1 1 1 0 1

+ = +


One’s complement §  It amounts to little more than flipping the bits of a

binary number §  In an 8-bit binary number using One’s

complement •  4 = 0 0000100 •  -4 = 1 1111011

§  In one’s complement, as with signed magnitude, negative values are indicated by a 1 in the high order bit

•  Complement systems are useful because they eliminate the need for subtraction


One’s complement §  Although the end carry around adds some

complexity, one’s complement is simpler to implement in hardware than signed magnitude.

§  But it still has the disadvantage of having two different representations for zero: positive zero and negative zero

§  Two’s complement solves this problem

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Two’s complement §  To express a value in two’s complement: §  If the number is positive, convert it to binary and

it is done §  If the number is negative, find the one’s

complement of the number and then add 1 •  4 = 0 0000100 •  -4 = 1 1111100


Two’s complement §  With Two’s complement

arithmetic §  Just add the two binary numbers §  Discard any carries emitting from

the high order bit §  Sum of 48 and – 19

•  19 in binary is 00010011 •  -19 in one’s complement is

11101100 •  -19 in two’s complement is

11101101 •  Answer is 29

0 0 1 1 0 0 0 0 1 1 1 0 1 1 0 1 = 0 0 0 1 1 1 0 1


Overflow §  In a computing system, resources are finite §  There is always the risk that the result of a

calculation becomes too large to be stored in the computer

§  Overflow can not always be prevented §  It can always be detected §  In complement arithmetic, an overflow condition

is easy to detect.

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Overflow §  Using two’s complement binary arithmetic, the

sum of 104 and 46 is

§  The nonzero carry from the seventh bit overflows into the sign bit, resulting in an erroneous value: 104 + 46 = -106

0 1 1 0 1 0 0 0 0 0 1 0 1 1 1 0 = 1 0 0 1 0 1 1 0


Overflow §  Good programmers stay alert for it §  Rule for detecting signed two’s complement

overflow •  Carry in and carry out of the sign bit are different

§  Rule for detecting unsigned number overflow •  There is carry out of the leftmost bit •  1111 + 1 = 0000


Representation Range §  3 bits

•  Signed: -3 , 3 •  1’s: -3 , 3 •  2’s: -4 , 3

§  6 bits •  Signed: -31, 31 •  1’s: -31, 31 •  2’s: -32, 31

§  5 bits: •  Signed: -15, 15 •  1’s: -15, 15 •  2’s: -16, 15

§  8 bits •  Signed: -127, 127 •  1’s: -127, 127 •  2’s: -128, 127

§  Formula for calculating the range •  Signed & 1’s: -(2n-1 – 1) , (2n-1 – 1) •  2’s: -2n-1 , (2n-1 – 1)

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Floating-Point Representation §  Signed magnitude, one’s complement, and

two’s complement representations alone are not useful in scientific or business applications

§  Real number values over a wide range §  Numerals to the right of a base point represent

negative powers of the base

0.112 = 1 x 2-1 + 1 x 2-2 = 0.5 + 0.25

= 0.75


Floating-Point Representation §  Converting 0.8125 to binary §  This method also works with any base. Just use

the target base as the multiplier •  0.8125 * 2 = 1.6250 •  0.6250 * 2 = 1.2500 •  0.2500 * 2 = 0. 5000 •  0.5000 * 2 = 1. 0000

§  Conversion is done when the product is zero or the desired number of binary places is reached

Reading from top to bottom is: 0.812510 = 0.11012


Floating-Point Representation §  Computer systems use a form of scientific notation

for floating-point representation §  Numbers written in scientific notation have three

components: •  Sign Mantissa Exponent

+ 1.5 x 10-1

§  Computer representation of a floating-point number consists of three fixed-size fields:

Sign Exponent Significand

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Floating-Point Representation

§  The one-bit sign field is the sign of the stored

value §  The size of the exponent field, determines the

range of values that can be represented §  The size of the significand determines the

precision of the representation




§  The IEEE-754 single precision floating point

standard uses an 8-bit exponent and a 23-bit significand

§  The IEEE-754 double precision standard uses an 11-bit exponent and a 52-bit significand




§  The significand of a floating-point number is always preceded by an implied binary point

§  The significand always contains a fractional binary value

§  The exponent indicates the power of 2 to which the significand is raised


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Floating-Point Representation §  Express 3210 in the simplified 14-bit floating-

point representation §  32 = 25 = 1.0 x 25 = + 0.1 x 26 §  Exponent field = 1102 = 610 §  Significand field 1

§  How do we express fractional numbers 0.5 =2-1?

+ 00110 10000000


General Form §  Representation

•  Bits to right of “binary point” represent fractional powers of 2 Represents rational number:

bi bi–1 b2 b1 b0 b–1 b–2 b–3 b–j• • •• • • .124

2i–12i

• • •

• • •

1/21/41/8

2–j


Floating-Point Representation §  To provide for negative exponents, a biased

exponent is used §  The IEEE-754 single precision floating point

standard uses bias of 127 over its 8-bit exponent

•  An exponent of 255 indicates •  Infinity if significand is zero •  NaN, “not a number,” often used to flag an error

condition if the significand is nonzero §  The double precision standard has a bias of

1023 over its 11-bit exponent

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Floating-Point Representation §  Sum of 1210 and 1.2510 using the 14-bit floating-

point representation •  1210 = 0.1100 x 24 •  1.2510 = 0.101 x 21 = 0.000101 x 24 •  Thus, the sum is 0.110101 x 24


Floating-Point Representation §  Using the same number of bits, unsigned integers

can express twice as many values as signed numbers

§  Using two’s complement allows for one type of hardware/process to add both signed and unsigned numbers

§  IEEE-754 floating point standards allow two representations for zero

•  Programmers should avoid testing a floating-point value for equality to zero

•  Negative zero does not equal positive zero


IEEE-754 Single Precision §  Single precision (32-bit) format

§  Normalized rule: number represented is (-1)S×1.F×2E-127

•  An exponent of 255 indicates §  Infinity if significand is zero §  NaN, “not a number,” often used to flag an error condition if

the significand is nonzero

§  Example: +101100.111è+1.01100111×25

S E F

1 8 23

0 10000100 01100111000000000000000

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IEEE-754 Single Precision all zeros

all ones

S E M Represents 0 all zeros all zeros 0 1 all zeros all zeros -0 0 all ones all zeros 1 all ones all zeros

0 or 1 all ones non-zero NaN

∞∞−

E = 0

E = 255


IEEE-754 Single Precision §  Examples

•  0.0 §  0 00000000 00000000000000000000000

•  1.0 (1.0 x 20) §  0 01111111 00000000000000000000000

•  0.5 (0.1 binary = 1.0 x 2-1) §  0 01111110 00000000000000000000000

•  0.75 (0.11 binary = 1.1 x 2-1) §  0 01111110 10000000000000000000000

•  3.0 (11 binary = 1.1*21) §  0 10000000 10000000000000000000000

•  -0.375 (-0.011 binary = -1.1*2-2) §  1 01111101 10000000000000000000000

•  1 10000011 01000000000000000000000 = - 1.01 * 24 = -20.0


Other Representation ( ) ( ) ( )

( ) ( )22

52210

1011011.121011011.111.11011075.54

×≅

×==

0 0 1 0 1 1 1 0 1

Sign of the number

mantissa Sign of the exponent

exponent

( ) ( )( ) ( )201102

5210

21100.1

21100.102832.0−

−

×=

×≅

10 bit word (sign, sign of exponent, 4 for exponent, 4 for mantissa)

0 1 0 1 1 0 1 1 0 0

Sign of the number

mantissa Sign of the exponent

exponent

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Floating Point Operation •  Now consider adding 0.5 and -0.4375

1.  Conversions §  0.5 = 0.12 = 1.0002 × 2–1 §  - 0.4375 = - 0.011102 = - 1.1102 × 2–2

2.  New Representation §  0.5 + –0.4375 => 1.0002 × 2–1 + –1.1102 × 2–2

3.  Align binary points §  Shift number with smaller exponent §  1.0002 × 2–1 + –0.1112 × 2–1

4.  Add significands §  1.0002 × 2–1 + –0.1112 × 2–1 = 0.0012 × 2–1

5.  Normalize result & check for over/underflow §  1.0002 × 2–4, with no over/underflow

6.  Round and renormalize if necessary §  1.0002 × 2–4 (no change) = 0.0625


Floating Point Hardware Unit

Small ALU

Exponentdifference

Control

ExponentSign Fraction

Big ALU


0 1 0 1 0 1

Shift right

0 1 0 1

Increment ordecrement

Shift left or right

Rounding hardware


S t i l l n o r m a l i z e d ?

4 . R o u n d t h e s i g n i f i c a n d t o t h e a p p r o p r i a t en u m b e r o f b i t s

Y e sO v e r f l o w o ru n d e r f l o w ?

S t a r t

N o

Y e s

D o n e

1 . C o m p a r e t h e e x p o n e n t s o f t h e t w o n u m b e r s .S h i f t t h e s m a l l e r n u m b e r t o t h e r i g h t u n t i l i t se x p o n e n t w o u l d m a t c h t h e l a r g e r e x p o n e n t

2 . A d d t h e s i g n i f i c a n d s

3 . N o r m a l i z e t h e s u m , e i t h e r s h i f t i n g r i g h t a n di n c r e m e n t i n g t h e e x p o n e n t o r s h i f t i n g l e f t

a n d d e c r e m e n t i n g t h e e x p o n e n t

N o E x c e p t i o n


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