10-Sep-0194.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Sept. 20011...

10-Sep-01 94.201 - Fall 2001: copyright ©T. Pearce, D. Hutchinson, L. Marshall Sept. 2001 1

Representing Information in Computers: numbers: counting numbers, integers: signed

magnitude and 2’s complement (slides 2-29) characters: ASCII, etc. (slides 30-33) summary (slides 34)

Data Representation


base: set of n atomic symbols + interpretation of each symbol as a “value” (quantity)

number: a string of symbols (digits) + interpretation rule

digits are numbered right to left, start at 0• e.g. 4-digit number: d3d2d1d0

“value” of a digit depends on value of symbol and position of digit in string• value of digiti = (value of symboli) ni

“value” of number is sum of value of digits

Base-n Number Systems


(nothing new here!)

symbols = { 0, 1, 2, 3, , 9} – usual meaning 4-digit example: d3d2d1d0 = 1436

= d3 103 + d2 102 + d1 101 + d0 100

= 1 103 + 4 102 + 3 101 + 6 100

= 1 1000 + 4 100 + 3 10 + 6 1

= 1000 + 400 + 30 + 6

= 143610

Base-10 Example


set of symbols = { 0 , 1 } – usual meaning 4-digit number example: 10112

= 1 23 + 0 22 + 1 21 + 1 20

= 1 8 + 0 4 + 1 2 + 1 1

= 8 + 0 + 2 + 1

= 1110

Base-2 (binary) Number System


Binary Decimal Binary Decimal 0000 0 1000 8 0001 1 1001 9 0010 2 1010 10 0011 3 1011 11 0100 4 1100 12 0101 5 1101 13 0110 6 1110 14 0111 7 1111 15

Binary – Decimal Conversion Table

Memorize these!


What about the range of representation?• n-bit values can represent (at most) 2n different

counting numbers (why?)• if start representation at 0, then largest value

represented is 2n – 1 (why?)• recall 4-bit example: ( 24 = 16 )

– range = 0 . . 15

Range of Representation


Some Observations:• binary representation of counting numbers often

requires lots of bits

– example: 60110 = 10010110012 (10 bits!)

• converting binary decimal is not trivial• binary can be awkward and error prone for use by

people!

More about Binary


16 symbols = { 0 . . 9, A, B, C, D, E, F}• interpret: 0 . . 9 – usual meaning• A 10 B 11 C 12• D 13 E 14 F 15

Why use hexadecimal (hex) notation? shorthand for binary notation converting binary hexadecimal is easy

Hexadecimal (Base-16) Number System


Converting Binary Hex 4 binary digits make up the value of one hexadecimal

digit: • hex digits range 0 . . 15• 4-bit binary values range 0 . . 15

2 Steps• group bits in 4’s starting from least significant• replace each group by corresponding hex digit

Converting Hex Binary replace each hex digit by 4-digit binary rep

Converting Binary Hex


16-bits: 10001010011011102 (= 35,43810)1. group in 4’s 1000 1010 0110 1110

2. replace with hex 8 A 6 E 16

10-bits: 10010110012 (= 60110)1. group in 4’s 0010 0101 1001

2. replace with hex 2 5 9 16

Binary => Hex Examples


Binary Decimal Hex Binary Decimal Hex 0000 0 0 1000 8 8

0001 1 1 1001 9 9

0010 2 2 1010 10 A

0011 3 3 1011 11 B

0100 4 4 1100 12 C

0101 5 5 1101 13 D

0110 6 6 1110 14 E

0111 7 7 1111 15 F

Binary / Decimal / Hexadecimal

Memorize these!


Encoding of More Types of Information counting numbers, integers what are they? are these different from base-n systems?

Encoding of Counting Numbers (0, 1, 2, 3, …) simple mapping: binary number system! base-2 number system

Encoding of Integers positive and negative values must encode sign & magnitude not so easy!

Other Encodings


use most significant bit (leftmost) to encode sign 0 = positive, 1 = negative use remaining n – 1 bits to encode magnitude using

binary number system (as for counting numbers) example: 8-bit signed-magnitude

100000012

magnitude = 1sign bit = 1 negative number

Thus this represents value – 1

Signed Magnitude Representation


Problems with Signed-Magnitude Encoding: two representations of 0!?

• positive 0 (sign bit = 0)• negative 0 (sign bit = 1)• is zero positive or negative?

performing arithmetic operations with signed-magnitude values can be awkward

signed-magnitude not often used in practice

Signed Magnitude (contd)


complement of one bit: b = 1 – b • complement of 0 = 1; complement of 1 = 0

complement of an n-bit value:• complement each bit

2’s complement of an n-bit value:• complement each bit, and then add 1• ignore any carry out of most significant bit (why?)

Binary Addition and Subtraction:

0 + 0 = 0; 1 + 0 = 1; 0 + 1 = 1; 1 + 1 = 0 (carry 1) 0 – 0 = 0; 1 – 0 = 1; 0 – 1 = 1 (borrow 1); 1 – 1 = 0

2’s Complement Encoding


1. original value = 100111010010 complement each bit: 011000101101 add 1 +1 2’s complement rep = 011000101110

2. original value = 0001 complement each bit: 1110 add 1 +1 2’s complement rep = 1111

2’s Complement Examples


Assume we have n bits available. use half of binary values for negative values (i.e. 2n–1

values), the rest for non-negative encode positive values (and 0) the same way as

counting numbers range 0 . . 2n – 1 – 1 uses all values that start with 0 use 2’s complement as negation operation! encoding of – “x” = 2’s complement of “x”

Encoding Integers in 2’s Complement


1. 8-bit 2’s complement encoding of –1

+110 0000 0001

complement: 1111 1110

add 1 + 1

so – 110 = 1111 1111

the largest binary representation is for the smallest negative number!

Integer Encoding Examples


2. 8-bit 2’s complement encoding of –128

+12810 1000 0000complement: 0111 1111add 1 + 1

–12810 1000 0000 the smallest binary representation starting with 1 (indicating

negative) is for the largest negative number that can be represented!

do we have two values with the same representation? –12810 = +12810 ? No, as +12810 cannot be represented in 8-bit 2’s complement (see next page!)

Integer Encoding Examples (contd)


range of numbers that can be represented is:

– 2n – 1, …, – 1, 0, …, 2n – 1 – 1 e.g. for 8-bit 2’s complement encodings:

n = 8: 28 = 256, 27 = 128 range: – 2n – 1, …, 2n – 1 – 1 = – 128, …, 127 remember: all negative value encodings start with

most significant bit = 1

We assume 8 bit representation in the following slides!

2’s Complement Range of Representation


Encoding of negative numbers seems strange!? Why do it like this?!

We want A. Negating a negative number = same positive

number we started with B. A single representation of zero C. Math (addition, subtraction, etc.) to work without

keeping track of representation being used (i.e. regular counting numbers vs. 2’s complement)

More on 2’s Complement


A. Negating a negated number should give the original number:

with 2’s complement does – (– x) = x?

Example 1: (8 bits)

– 110 1111 1111complement 0000 0000add 1 + 1

+110 0000 0001 – ( – 1 ) = 1 it works!

2’s Complement (contd)


A. (continued) Negating a negated number should give the original number:

Special Case!Example 2: (8 bits)

– 12810 1000 0000complement 0111 1111add 1 + 1

– 12810 1000 0000 (we saw this already!)

Should be +12810 (which cannot be represented in 8 bits 2’s complement)

What happened? Overflow (more later!)



B. How many representations are there for 0 in 2’s complement?

010 0000 0000

complement 1111 1111

add 1 + 1 (ignore carry out of “1”)

010 0000 0000



C. (Example) Add: 1810 + 7510 (should be 9310 )Convert to Hex and Binary:

1810 = 1*16 + 2 = 12h = 0001 00102

7510 = 4*16 + 11 = 4Bh = 0100 10112

Add in Hex and Convert back to Decimal:

12h + 4Bh = 5Dh = 5*16 + 13 = 9310

Add in Binary and Convert back to Decimal:

0001 00102 + 0100 10112 = 0101 11012 =

1*64 + 1*16 + 1*8 + 1*4 + 1*1 = 9310

BinaryAddition worked!

Binary Math Example


C. (continued) What if the binary numbers on the previous page were in 2’s complement representation?

0001 00102

0100 10112

0101 11012

No difference as all these numbers are positive (most significant bit is “0”)

2’s Complement Math (contd)


Add: 1010 + 23710 (should be 24710 )

Convert to Hex and Binary:

1010 = 0*16 + 10 = 0Ah = 0000 10102

23710 = 14*16 + 13 = EDh = 1110 11012

Add in Hex and Convert back to Decimal:

0Ah + EDh = F7h = 15*16 + 7 = 24710

Add in Binary and Convert back to Decimal:

0000 10102 + 1110 11012 = 1111 01112 =

1*128 + 1*64 + 1*32 + 1*16 + 1*4 + 1*2 + 1*1 = 24710

As expected, BUT...

More Binary Arithmetic


C. (continued) What if these binary numbers were in 2’s complement representation?

0000 10102 positive number so no change = 1010

1110 11012 negative number so complement and add 1

= – (0001 00102 + 1) = – 0001 00112 = – (1*16 + 1*2 + 1*1)10

= – 1910

Thus we would want the answer to be:

1010 + – 1910 = – 910

converting – 910 to 2’s complement gives:

0000 10012 + 1 = 1111 01102 + 1 = 1111 01112

Which is exactly what we had on the previous page!

More 2’s Complement Arithmetic


C. (continued)

This “proves”: addition works if the (Hex or) binary numbers

represent counting numbers or 2’s complement number!

left as an exercise: subtraction works, too!

Math works regardless of representation - this is why 2’s complement is used!

Why 2’s Complement


Encoding of Characters: (slides 30-33) want to represent “displayable” characters:

characters: { a, A, b, B , . . . , z, Z } decimal (10) digits: {0, 1, . . . , 8, 9 } punctuation: ! , . – ? / : ; math symbols: + * = ( – and / above) brackets: ( ) [ ] { } < > other symbols: @ # $ % ^ & \ | ~ “ ” etc. about 90 characters need to be represented

various encoding schemes have been used ASCII encoding – international standard American Standard Code for Information Interchange

Encoding of Characters


7-Bit ASCII Encoding• 7 bits to encode each character (128 codes)• often extended to 8-bit (byte) values by making most significant bit

(msb) = 0• Note: all codes are given as Hex values

• 00 – 1F non-displayable control char’s 00 NULL 07 BELL 08 backspace 09 tab others – often serve special purposes in communication

applications

Encoding of Characters

0A line feed 0C form feed 0D carriage return


7-Bit ASCII Encoding (continued) 20 = blank space “ ”; 21 = “!” ; 2E = “.” 30 – 39 decimal digit characters

30 = “0”, …, 39 = “9” 41 – 5A upper case letter characters

41 = “A”, …, 5A = “Z” 61 – 7A lower case letter characters

61 = “a”, …, 7A = “z”

Encoding of Characters (contd)


ASCII Example: “94.201 is FUN!”• hex encoding: 39 34 2E 32 30 31 20 69 73 20 46 55 4E 21• binary encoding: … 00110010 …. 01110011 … 00100001

hex is often used a shorthand for binary ASCII encoding is summarized in ASCII Table (on

web site)

Other Character Encoding Schemes: IBM standardized an 8-bit scheme (256 characters) as

defacto standard (PCs) Java: unicode – 16-bit scheme (65,536 characters)

• multi-lingual character sets

Encoding of Characters (contd)


fixed-width binary values are used to represent information in computers

the computer works with the binary representations (NOT the information!!)

the same binary value can be used to represent different information: Programmer must be aware!

8-bit example: 1111 00002 represents unsigned integer 24010

2s complement signed integer –1610

ASCII “ ” and others...

Data Representation Summary(Important Concepts)


Manipulating Information in Computers: always operate on fixed-width binary representation arithmetic: add, subtract (slides 36-40) logic: and, or, xor (slides 41-42) shift and rotate (slides 43-45)

Data Manipulation(slides 35-45)


1-bit operations – previous slides bitwise add/subtract values of width n to give result of width

n 8-bit Unsigned Integer Examples:

11710 = 0111 01012 13310 = 1000 01012

+ 9910 = 0110 00112 – 5110 = 0011 00112

21610 = 1101 10002 8210 = 0101 00102

do exactly the same thing for 2’s complement signed integers!

for 8-bit result: ignore any carry/borrow out/in @ msb (why?)

Addition and Subtraction


8-bit Signed Integer (2’s Complement) Examples: –110 = 1111 11112 3210 = 0010 00002

+ 110 = 0000 00012 – 6510 = 0100 00012

010 = 1 0000 00002 – 3310 = 1 1101 11112

= 0000 00002 = 1101 11112

in addition, ignore carry at msb for 8-bit result in subtraction, ignore borrow at msb for 8-bit result computers often implement subtraction using “negate and add”, i.e.

X – Y = X + (– Y) 3210 = 0010 00002

+ – 6510 = 1011 11112

– 3310 = 1101 11112

Addition and Subtraction


Problem: limited representation range due to fixed width

overflow occurs when the result of an operation is outside the range that can be represented

8-bit Unsigned Integer Example:

25510 = 1111 11112

+ 110 = 0000 00012

010 = 0000 00002 ?? 25610 ?? carry @ msb is important in this case! need 9 bits to represent 25610

used 8-bits: Overflow Occurred!

Overflow


But there is another interpretation, in 2’s complement:

–110 = 1111 11112

+ 110 = 0000 00012

010 = 1 0000 00002

= 0000 00002

The result is correct if the values are interpreted as signed integers! ( – 1 + 1 = 0 )

Note: Overflow depends on the interpretation of the values!

Overflow (contd)


Another example: unsigned signed

0111 11112 127 127

+ 0000 00012 + 1 + 1

1000 00002 128 – 128

OK Overflow

in this case there is overflow only in the signed representation

Overflow (contd)


AND, OR, XOR (exclusive or):

1-bit truth tables:

a b a AND b a OR b a XOR b

0 0 0 0 0

0 1 0 1 1

1 0 0 1 1

1 1 1 1 0

Logical Operations(slides 41-42)


8-bit examples:

1011 0110 1011 0110

AND 1100 0011 OR 1100 0011

1000 0010 1111 0111

1011 0110

XOR 1100 0011

0111 0101

Logical Operations (contd)


values can be manipulated by shifting left – towards msb right – towards lsb (least significant bit)

shifting example: value: 1001 1010 shifted left 4 bits: 1010 0000

question: should injected values be 0 or 1? shift left: always inject 0’s shift right:

logical: inject 0’s arithmetic: inject msb value

Shifting and Rotating(Slides 43-45)


Shift left 2 bits• value: 1001 1010

• shifted value: 0110 1000

Logical shift right 2 bits• value: 1001 1010


Arithmetic shift right 2 bits• value: 1001 1010


More Shifting


rotate: like shift, except bit that is shifted “out” of the value gets injected as the new bit

all bits in original value are saved left – towards msb right – towards lsb

rotating example: rotate right 3 bitsvalue: 1001 1110 rotated 1st bit: 0100 1111rotated 2nd bit: 1010 0111rotated 3rd bit: 1101 0011

Shifting and Rotating (contd)

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