bits and bytes

Business Data Structures in C/C++ Kirs and Pflughoeft

Bits and Bytes

Bits and BytesChapter 1


Bits & Bytes

OROff On

Bit = Binary Digit (or Binary Digit) = {0, 1}

BitsBits

Assume you wish to send a message using a Light SwitchA binary condition since the light switch can be either:

Any binary condition can be represented with a single light switch :

Good OR Bad Yes OR No

Male OR Female Dead OR Alive


But, what if there are more than two states? What if I want But, what if there are more than two states? What if I want to represent the conditions to represent the conditions GOOD, SO-SOGOOD, SO-SO, and B, and BADAD????????

Add more Light SwitchesSimple …..

If there are 2 light switches, the total combinations are:

#1 #2 #3 #4

Off0

Off0

Off0

On1

Off0

On1

On1

On1

Interpreted as:Bad

Interpreted as:So-So

Interpreted as:Good

Not Used


Bits and Bytes

If I can transmit 4 messages with two bits, how many could If I can transmit 4 messages with two bits, how many could I transmit if I had 3 bits? Or 4 bits?I transmit if I had 3 bits? Or 4 bits?

With 3-bits, there are 8 possible combinations:

000 100001 101010 110011 111

And, with 4-bits, there are 16 possible combinations:

0000 0100 1000 11000001 0101 1001 11010010 0110 1010 11100011 0111 1100 1111


Bits and Bytes

Is there any way to know how many messages we could Is there any way to know how many messages we could transmit for a given number of bits without having to test transmit for a given number of bits without having to test all possible combinations??all possible combinations??

As in Decimal (base 10), it is possible to determine how many messages can be transmitted for any number of decimal places. In Binary (base 2), the same calculations are made, but using bits (instead of decimals).

Decimal Number Number Number Places Messages Bits Messages

0 100 = 1 0 20 = 1 1 101 = 10 1 21 = 2 2 102 = 100 2 22 = 4 3 103 = 1,000 3 23 = 8 4 104 = 10,000 4 24 = 16 5 105 = 100,00 5 25 = 32 6 106 = 1,000,000 6 26 = 64 7 107 = 10,000,000 7 27 = 128 8 108 = 100,000,000 8 28 = 256 9 109 = 1,000,000,000 9 29 = 512 10 1010 = 10,000,000,000 10 210 = 1,024


Bits and Bytes

The General formula is:The General formula is:

I = Bn where: I = The amount of Information (messages) available B = The base we are working in (Decimal or Binary) n = The number of digits (decimals or bits) we have

Applying the formula to both decimal and binary values:Applying the formula to both decimal and binary values:

100 = 1 20 = 1101 = 10 21 = 2102 = 100 22 = 4103 = 1,000 23 = 8104 = 10,000 24 = 16105 = 100,000 25 = 32106 = 1,000,000 26 = 64107 = 10,000,000 27 = 128108 = 100,000,000 28 = 256109 = 1,000,000,000 29 = 5121010 = 10,000,000,000 210 = 1,024


Bits and Bytes

What if I Know how much information (I = Number of What if I Know how much information (I = Number of Messages) I want to transmit. How do I determine the Messages) I want to transmit. How do I determine the number of bits I need?number of bits I need?

Just reverse the process.

If I = 10n (decimal) OR I = 2n (binary) then log(I) = n log(10) log(I) = n log(2)

log(I) log(I) log(I) log(I)And n = = n = =

log(10) 1.000 log(2) 0.30103 = log(I) Since 100.30103 = 2

Information Decimals Needed Bits Needed

10 log(10) = 1.000 log(10)/log(2) = 1.000/.30103 = 3.32 50 log(50) = 1.699 log(50)/log(2) = 1.699/.30103 = 5.64 100 log(100) = 2.000 log(100)/log(2) = 2.000/.30103 = 6.64 500 log(500) = 2.699 log(500)/log(2) = 2.699/.30103 = 8.97 1,000 log(1000) = 3.000 log(1000)/log(2) = 3.000/.30103 = 9.9710,000 log(10000) = 4.000 log(10000)/log(2) = 4.000/.30103 = 13.29


Bits and Bytes

How can we have partial bits (or decimals)? For example, How can we have partial bits (or decimals)? For example, how can we have 5.64 bits to represent 50 messages?how can we have 5.64 bits to represent 50 messages?

We Can’t

The formula given should have been:

log(I) log(I) Where: n = = log(2) 0.30103 is the ceiling of the result (i.e., rounded up)

And the number of bits needed would be:

Messages Bits Needed

10 log(10)/log(2) = 1.000/.30103 = 3.32 = 4

50 log(50)/log(2) = 1.699/.30103 = 5.64 = 6

100 log(100)/log(2) = 2.000/.30103 = 6.64 = 7

500 log(500)/log(2) = 2.699/.30103 = 8.97 = 9

1,000 log(1000)/log(2) = 3.000/.30103 = 9.97 = 10

10,000 log(10000)/log(2) = 4.000/.30103 = 13.29 = 14


Bits and Bytes

Notice that we could have predicted that, for example, it would take 6 bits to represent 50 pieces of information since:

25 = 32 and 26 = 64

If we need 6 bits to represent 50 pieces of information, and we If we need 6 bits to represent 50 pieces of information, and we could represent 64 pieces of information, what happens to the could represent 64 pieces of information, what happens to the remaining 16 pieces of information??remaining 16 pieces of information??

They either remain unused, or are available for future use


Bits and Bytes

What does this have to do with Computers?What does this have to do with Computers?

If we were to look inside a computer (especially earlier ones) we might see a series of ‘doughnuts’:

Which were merely metal rings with wires running through them

Depending on whether there was voltage running through them or not (actually, high voltage or low voltage) the series represented a sequence of messages.

A BINARY SITUATION!


Bits and Bytes

Notice that since there are 5 ‘doughnuts’, there are 25 or 32 Combinations

Where and represent the different voltage states


Bits and Bytes

How Many bits (or ‘doughnuts’) do we really need?How Many bits (or ‘doughnuts’) do we really need?

Good question! What symbols/information do we wish to convey?

Pieces of InformationThe digits (0, …, 9) 10 The alphabet (a, …, z) 26The upper case alphabet (A, …, Z) 26Special characters (! + ( ) . ? / * - % & # =, etc.) 32 (?)

94

Since: n = log(I)/log(2) = log(94)/log(2) = 1.973/0.301 = 6.55we need 7 bits, which we could have predicted since:

26 = 64 and 27 = 128


Bits and Bytes

What about the remaining 34 (128 - 94) bits?What about the remaining 34 (128 - 94) bits?

There are a number of additional special characters and a number of ‘hidden’ characters which we didn’t account for:

Carriage Return (CR)Back Space (BS)End of File (EOF)etc.

So the additional bits will be used.

Are 7 bits normally used to represent a character set?Are 7 bits normally used to represent a character set?

Yes. The Standard coding scheme consists of 128 characters


Bits and Bytes

LIAR!! LIAR!! PANTS ON FIRE !!!LIAR!! LIAR!! PANTS ON FIRE !!!

Yes - sort of.

BUT, the standard character still contains only 128 characters, which requires 7-bits

Then Why does a byte contain 8-bits?Then Why does a byte contain 8-bits?

Doesn’t a byte represent a character? And isn’t a byte equal Doesn’t a byte represent a character? And isn’t a byte equal to 8 bits, not 7?to 8 bits, not 7?

• 1-Byte = 8-bits• A Byte is used to represent a character• A Byte is the basic addressable unit in RAM


Bits and Bytes

There are a few reasons. Primarily, however, it is because earlier machines suffered some reliability problems (remember what the term debugging really means)1:

1 In the days of vacuum, tubes, bugs were attracted to the heat given off by the tubes. Programmers frequently spent much of their time scrapping dead bugs off the circuitry, or ‘de-bugging’.

There were problems with storage and data transmission

One additional bit was added to help detect errors:

The Parity BitThe Parity Bit

How does adding one additional bit help detect errors?How does adding one additional bit help detect errors?


Bits and Bytes

Assume that we wished to send the series of bits:

1001100

But, because of transmission errors, actually sent the message:

1001101

How can we tell that an error was made? How do we know How can we tell that an error was made? How do we know that the sequence that the sequence 10011011001101 was not the true message? was not the true message?

As it stands now, we can’t.


Bits and Bytes

If we were to send a transmission using an extra bit:

1001100 1 Parity-Bit

We could determine if the message was correctly transmitted by counting the total number of on bits

E.G. If the total number of on bits is an EVEN number, the message was correctly transmitted.

Since the message sent contains 4 bits (an even number) the message sent was correct.


Bits and Bytes

IF, however, we received the message:

1001101 1 Parity-Bit

We know it is incorrect because the message contains 5 (an odd number) bits

Other examples using EVEN Parity:

Message Sent:

1101101 1

Mess. Received:

1101101 1

No. Bits:

6 (Even) Correct

0001100 0 0101100 0 3 (Odd) Incorrect

1101011 1 1001011 1 5 (Odd) Incorrect

0101110 0 1010110 0 4 (Even) Correct


Bits and Bytes

What gives? The last message:What gives? The last message:

Message Sent:

0101110 0

Mess. Received:

1010110 0

No. Bits:

4 (Even) Correct

Was Was NOTNOT correct, even though the total number of on bits correct, even though the total number of on bits received was even???received was even???

Yes - The system is NOT perfect, but if there are thousands or millions of messages sent, it is highly unlikely that mistakes will not be caught.

All it takes is one incorrect message.


Bits and Bytes

Must Parity always be equal??Must Parity always be equal??

No, it can be ODD (or there can be NO parity). That decision is made by the software designer.

If we look at our previous examples using ODD parity:

Message Sent:

1101101 0

Mess. Received:

1101101 0

No. Bits:

5 (Odd) Correct

0001100 1 0101100 1 4 (Even) Incorrect

1101011 0 1001011 0 4 (Even) Incorrect

0101110 1 1010110 1 5 (Odd) Correct

Notice that errors can still go undetected.


Bits and Bytes

There was one other problem with bytes:There was one other problem with bytes:• CompatibilityCompatibilityGiven the

binary sequences:

0000000000000100000100000011

1111110111110111111101111111

Manufacturers Interpreted them differentlyManufacturers Interpreted them differently

Manufact.#1:

ABCD

6789

Manufact.#2:

0123

vxyz

Manufact.#3:

+-*?

TABCRLFFF


Bits and Bytes

Which is the Correct Interpretation???Which is the Correct Interpretation???

Each is equally CorrectEach is equally Correct

• 0000010 CouldCould be either a ‘C’ OR a ‘2’• The letter ‘C’ CouldCould be pronounced either ‘cee’ OR ‘ess’

What’s the Solution ???What’s the Solution ???

ASCIIASCII

The AAmerican SStandard CCode for

IInformation IInterchange


Bits and Bytes

Sample ASCII Codes:Sample ASCII Codes:Binary Sequence0000000

Value0

CharacterNULL

Description . NULL/Tape feed

0000111 7 BEL Rings Bell0001000 8 BS Back Space

0001101 13 CR Carriage Return

0011011 27 ESC Escape

0100000 32 SP Space

0110000 48 0 Zero0110001 49 1 One

1000001 65 A Capital ‘A’1000010 66 B Capital ‘B’

1100001 97 a Lower Case ‘a’1100010 98 b Lower Case ‘b’


Bits and Bytes

A Preview of Things to Come:A Preview of Things to Come:

For the first Exam MemorizeMemorize the Numeric Values for:

• NULL Value: 0• BEL (Ring The Bell) Value: 7 • BS (Backspace) Value: 8• CR (Carriage Return) Value: 13• ESC (Escape) Value: 27• SP (Space) Value: 32• The digits (0, 1, …, 9) NOTE: The Digit 0 (zero) has the value: 48• The Uppercase Alphabet NOTE: The Character ‘A’ has the value: 65• The Lowercase Alphabet NOTE: The Character ‘a’ has the value: 97


Bits and Bytes

Are We limited to only 128 (= 2Are We limited to only 128 (= 277) characters ??) characters ??

• The STANDARD ASCII Character Set Consists of 128 Characters (as given in Addendum 1.1)

Yes and no:

There is an EXTENDED ASCII Character set which uses ALL 8-bits (1-byte) available (parity is NOT an issue)• The extended ASCII Character set consists of 256

(= 28) characters (See Addendum 1.2)

• The Majority of the characters included in the extended ASCII character set are extensions of the Greco-Roman Alphabet (e.g., ß, Ü, å) or ‘graphics’ characters (e.g., )


Bits and Bytes

What does the term ‘ASCII file’ Mean ??What does the term ‘ASCII file’ Mean ??An ASCII File assumes that every 8-bits (1-byte) in the file are grouped together according to the ASCII tables

Aren’t ALL Files ASCII Files ??Aren’t ALL Files ASCII Files ??

NONO - As we will see later, not all data is stored according to ASCII formats

That Helps (sort-of) to explain why when we display non-ASCII files we sometimes get characters such as , , , , , and


Bits and Bytes

Do ALL computers use ASCII to Represent Do ALL computers use ASCII to Represent Symbols???Symbols???

NONO - Although most do.

IBM had the first Coding Scheme (dating back to 1880)

EBCDICEBCDIC

EExtended BBinary CCoded DDecimal IInterchange CCode

EBCDIC is still used (?) in IBM Mainframes and to store data on large reel-to-reel Tape Drives


Bits and Bytes

And so that’s it ??And so that’s it ??There is only ASCII and EBCDIC ??There is only ASCII and EBCDIC ??

Well, … NoWell, … NoIt became obvious that Even the Extended ASCII and Character Sets were insufficient

How So – Kimo Sabi ??How So – Kimo Sabi ??

Suppose you wanted to represent ALL the characters used by ALL the languages in the World ---

How Many Are there ????

I Don’t know, How Many ??I Don’t know, How Many ?? I Don’t know, Either --I Don’t know, Either --

But it’s a lot !!!


Bits and Bytes

Enter Unicode (1990):Enter Unicode (1990):If we were to use 16-bits, instead of 8, to represent characters we could represent:

216 = 65,536 Characters

AHA!! So Everyone is using Unicode now -- Right ??AHA!! So Everyone is using Unicode now -- Right ??

Well, … NoWell, … No

Well, why not ?? Well, why not ??

Life is not so simple …Life is not so simple …


Bits and Bytes

There are a lot of problems still be worked out:There are a lot of problems still be worked out:

• There is a lot of disagreement about what should be included

(Even though there are 65,536 combinations, you would be surprised at how quickly those combinations can be used up)

• The large number of characters in this set poses a severe problem for a font vendor

(No fonts – No Characters)

• By doubling the number of bits (or bytes), we are doubling the storage and processing requirements

• Result: It will take years to get this straightened out


Bits and Bytes

SO – What have we learned ????SO – What have we learned ????• What a bit is• How a bit corresponds to computer architecture• How combinations of bits can be used to store information• How to calculate how much information a given number of bits yields• How to calculate how many bits we need to store information• What a byte is and why it is 8-bits• What parity is and why it is/was necessary• What ASCII is and why it was developed• What EBCDIC is• What Unicode is and why it was developed• … And many other things in between …

Do I have to know this stuff ?? Do I have to know this stuff ??

Of Course not !! – I just like to waste my time Of Course not !! – I just like to waste my time andand yours !! yours !!


Bits and Bytes

So what do we need to do ??So what do we need to do ??• Make sure you THOUROGHLY understand ALL of the

concepts covered in these slides

• Answer ALL of the relevant questions on the Review Page

• Memorize the assigned ASCII codes

• Submit your References

• Submit your Question(s)

• Look at the Bits/Bytes/ASCII C/C++ Programming Assignment (it’s not due yet, but it can’t hurt to look at it)

??? Any Questions ??? (Please!!)??? Any Questions ??? (Please!!)


Bits and Bytes

bits and bytes

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