2e3 numbersystems week1 lecture2

8/21/2019 2E3 NumberSystems Week1 Lecture2

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2E3: Computer Engineering IINumber Systems

Dr. Ivana Dusparichttp://www.scss.tcd.ie/Ivana.Dusparic/2E3/

Introduction to Number Systems

Why different number systems?

Why do we need to know them?

Binary, Octal, Hexadecimal – signed vs. unsigned

Conversion between number systems – Binary to/from decimal

– Binary to/from hexadecimal – Hexadecimal to/from decimal

C++ notation

Binary Operations – AND, OR, XOR, NOT

– Bitwise shift left/right


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Introduction to Number Systems

Humans use decimal number system

– 10 fingers

Computers – binary – 2 physical states– 0V and 5V– charge / NO charge on transistor gate

How to represent decimal digits using only 2states?– 3 or 4 bits depending on the digit – 4th bit often wasted– Use octal (3 bits) or hexadecimal (4 bits)

Binary

Unsigned vs. signed binary integers

– Unsigned – positive integers only

Base 2 notation:

– E.g. 3710 = 1001012

We normally deal in decimal = base 10

– 3710 = 3×101 + 7×100

– 40310 = 4×102 + 0×101 + 3×100


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Unsigned Binary Integers

100101

1’s2’s4’s8’s16’s32’s

Can do the same for any base, say base 2

– 1001012 = 1×25 + 0×24 + 0×23 + 1×22 + 0×21 + 1×20

– 1001012 = 3210 + 010 + 010 + 410 + 010 + 110

– 1001012 = 3710


Always remember what base you are operating in

– “Value” of a number is meaningless otherwise

Convert decimal to binary:

– Keep dividing by the base and read the remainders

9

4

2

1

0

182

2

2

2

2

372 1

0

1

0

0

1

Read frombottom 1001012


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pictorial view of a 4 bitunsigned binary integer

4 bit unsigned binaryinteger range:

00002 .. 111120 .. 15

n bit unsigned binaryinteger range:

0.. 2n-1

Unsigned Binary Integers - Exercise

Practise converting unsigned binary integers todecimal: – 01012 – 111001102

Practise converting from decimal to unsigned binaryintegers:

– 3410 – 12310 – 1010

Do not use calculator for conversion, but use it todouble-check the correctness of your result!

= 510

= 23010

= 1000102

= 11110112

= 10102


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Signed Binary Integers

Positive and negative integers

2's complement notationThe 2's complement of the

number behaves like the negativeof the original number

Convert -5 to a 2's complement

binary number by inverting bitsand adding 1

5 01012

invert bits 10102

add 1 00012

-5 10112

invert bits 01002

add 1 01012

5 01012


Same effect achieved by subtracting

from 0

4 bit signed binary integer positive range: 00002 .. 01112 0 .. 7 negative range: 11112 .. 10002 -1 .. -8

n bit signed binary integer range: -2n-1 .. 2n-1 – 1 Most significant bit indicates integer negative - sign

bit Note asymmetrical range – only one zero

zero 00002

subtract 5 01012

-5 10112


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pictorial view of a 4 bitsigned binary integer

if unsigned, inner ring

would have values 0 ..15

value depends on

how binary bits areinterpreted

Signed Binary Integers - Exercise

Practise converting 8-bit signed binary integers todecimal: – 000001012 – 111111012

Practise converting from decimal to signed 8-bitbinary integers: – -2810 – -12310 – -1010

= 510

= -310

= 111001002= 100001012= 111101102

To help you double-check your result

http://www.rsu.edu/faculty/PMacpherson/Programs/twos.html


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Hexadecimal

base 16

Same principles apply: – 15616 = 1×16

2 + 5×161 + 6×160

– 15616 = 25610 + 8010 + 610 – 15616 = 34210

But: – each binary digit has 2 possible values: {0, 1}

– each decimal digit has 10 possible values: {0, 1, … 8, 9}

– each hexadecimal digit must have 16 possible values:{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }

Hexadecimal

012345678

9ABCDEF

012345678

9101112131415

Base 16 Base 10

=

BINARY HEX

0000 0

0001 1

0010 2

0011 3

0100 4

0101 5

0110 6

0111 7

1000 8

1001 9

1010 A

1011 B

1100 C

1101 D

1110 E

1111 F


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Hexadecimal

Hexadecimal to binary

Easier way to handle large binary numbers bygrouping 4 binary bits into a hexadecimal digit

Consider the following 16 bit unsigned binary integer

– 1111 1010 1100 11102 = FACE16 = 0xFACE

Hexadecimal

Hexadecimal to decimal

Now convert 0xFACE to decimal– F×163 + A×162 + C×162 + E×160

– 15×163 + 10×162 + 12×161 + 14×160

– 15×4096 + 10×256 + 12×16 + 14×1– 64,206

What decimal value is 0xFACE if interpreted as a 16bit signed binary integer??– 1111 1010 1100 1110 – leading bit =1, negative number

- flip bits and add 1 to get positive equivalent, and add minussign


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Hexadecimal

Convert decimal to hexadecimal

Convert -2008510 to hexadecimal (assume 16 bitsigned number

2008516

125516

7816

416

0

5

7

E

4

read from

bottom4E7516

20085 4E7516

invert bits B18A16

add 1 000116

-20085 B18B16

Hexadecimal Exercises

Convert to hexadecimal:

– 1010 – 3410 – 15010

Covert to decimal:

– 4516 – F16 – 1EA16

= A16

= 2216

= 9616

= 6910

= 1510

= 49016


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1

Hexadecimal Exercises

Convert from hex to binary:

– 12316 – FF16

Convert from binary to hex:

– 11012 – 01001102 – 1100000011111111111011102

= 1001000112

= 111111112

= D16

= 2616

= C0FFEE16

Hexadecimal

Hex is a convenient way to express pixelvalues:

Can ignore

11111111 11111111 00111111 00000111 Bit 0Bit 31

Red Green BlueAlpha

1111 1111 1111 1111 0011 1111 0000 0111F F F F 3 F 0 7

color = 0xFF3F07;


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1

Hexadecimal

Each colour channel (R, G, B) has 8 bits – Therefore 256 levels = {0, 255}

50% grey = all colours at 50%: – {50%, 50%, 50%}

– {12810, 12810, 12810} ← note rounding

– {100000002, 100000002, 100000002}

– {8016, 8016, 8016}

– int color = 0x808080;

Blue:– int color = 0x0000FF;

Purple:– int color = 0xFF00FF;

Number Systems in C++

Naming conventions

C/C++ syntax

4 bits nybble

8 bits byte

16 bits word

32 bits double word

64 bits quad word

C/C++ base decimal equivalent

1234 decimal 1234

0x1234 hexadecimal 4660

01234 octal (base 8) 2322

leading 0 indicates octal integer (beware!)


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VC++ can handle 8, 16, 32 and 64 bit signed and unsigned arithmetic

int is CPU(and OS) dependent and maps to native word size of CPU

can determine a type's size using the sizeof() operator e.g. sizeof(int)

Type name # bytes range

char 1 –128 to 127

unsigned char 1 0 to 255

short 2 –32,768 to 32,767

unsigned short 2 0 to 65,535

int 4 –2,147,483,648 to 2,147,483,647

unsigned int 4 0 to 4,294,967,295

long long 8 –9,223,372,036,854,775,808 to 9,223,372,036,854,775,807

unsigned long long 8 0 to 18,446,744,073,709,551,615


To output in hex:std::cout


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Bit Manipulation

Required in any software talking directly to hardware

– E.g. graphics / audio etc.

Illustrate concepts in binary but generally work in hex

Binary operators:

– Take 2 parameters

Unary operators:

– Takes 1 parameter

– NOT operator is unary – Others = ++, --

a=~b;!~NOT

a=b>>c;n/a>>Shift Right

a=b


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Bitwise AND: &

a = b AND c

– a is TRUE if b and c areTRUE

– a is FALSE otherwise

E.g.

Useful for clearing bits:

0 0

0 1

0 1

0

1

AND

Truth table01101101

& 00111100

00101100

01101101& 11110001

01100001

Bit mask

if(a && b)

{

…

}

if(a & b)

{

…

}

≠≠≠≠

if a==true and b==true if (a&b) ≠ 0

Bitwise OR: |

a = b OR c

– a is TRUE if b or c areTRUE


E.g.

Useful for setting bits:

0 1

1 1

0 1

0

1

OR

Truth table

01101101

| 00111100

01111101

01101101

| 00001110

01101111

Bit mask

if(a || b)

{

…

}

if(a | b)

{

…

}

≠≠≠≠

if a==true or b==true if (a|b) ≠ 0


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Bitwise XOR/Exclusive OR: ^

a = b XOR c

– a is TRUE if b or c areTRUE but not both


E.g.

Useful for flipping bits:

0 1

1 0

0 1

0

1

XOR

Truth table01101101

^ 0011110001010001

01101101

^ 00001110

01100011

Bit mask

Bitwise NOT/ 1’s Complement: ~

a = NOT b

– a is TRUE if b is FALSE


E.g.

Useful for flipping all bits

of a value

1 0

0 1NOT

Truth table

~01101101 = 10010010

if(!a)

{

…

}

if(~a)

{

…

}

≠≠≠≠

if a≠true if (~a) ≠ 0


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Exercise

Try these (give answers in base of operands):

– 1111 00112 & 0011 11002 – 1010 01012 | 0101 01012 – ~(11001)2 assuming 8-bit

– 110012 ^ 1111112 8-bit also

– 810 | 1610 – 516 | 1516

– 23416 & F0F16 – ~(1FE)16 12-bit = 3 hex digits

Use calculator to double-check your results

= 0011 00002

= 1111 01012

= 1110 01102

= 0010 01102

= 2410

= 1516

= 20416= E0116

Size of Operands

Must be careful with bit operations:

– Always know the size of the operands

– E.g. 11112 & 1112 = ?

– What is ~100102 ?

– So must be clear what size of operand you’re workingon – is it 5 bits, a byte, or a word, or something else ?

11112& 1112

?1112

11112& 01112

01112

100102 = 0001 00102 = 0000 0000 0001 00102

011012 ≠ 1110 11012 ≠ 1111 1111 1110 11012


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Bit-wise Shift Left

Shift left operator: > 1 = 4310 = 430 / 101

B=2: 10112 >> 2 = 102(1110) >> 2 = (210) = 11 / 2

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2e3 numbersystems week1 lecture2

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