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CSC  8400:  Computer  Systems    

Represen3ng  and  Manipula3ng  Informa3on  

Background:    

Number  Systems  

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Analog  vs.  Digital  System  q  Analog  Signals  - Value  varies  con1nuously  

q  Digital  Signals  - Value  limited  to  a  finite  set  - Digital  systems  more  robust  

q  Binary  Signals  - Has  at  most  2  values  - Used  to  represent  bit  values  - Bit  1me  T  needed  to  send  1  bit  

Why  Bits  (Binary  Digits)?  

q  Computers  are  built  using  digital  circuits  -  Inputs  and  outputs  can  have  only  two  values  -  True  (high  voltage)  or  false  (low  voltage)  -  Represented  as  1  and  0  

q  Can  represent  many  kinds  of  informa1on  -  Boolean  (true  or  false)  - Numbers  (23,  79,  …)  -  Characters  (‘a’,  ‘z’,  …)  -  Pixels,  sounds  -  Internet  addresses  

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Coding  q  A  single  binary  input  can  have  two  values:  1  or  0  

q  More  bits  =  more  combina1ons    

 

 

  0 0

0 1

1 0

1 1

Coding  q  How  many  values  can  you  represent  on  3  bits?    

q  What  about  n  bits?  

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Binary  Numbers  

q  How  do  you  figure  out  what  the  value  of  1110  is?  - Same  way  you  do  for  4173,  for  instance  

q  Decimal  (base  10)  - Each  digit  represents  a  power  of  10  - 417310  =  4  x  103  +  1  x  102  +  7  x  101  +  3  x  100  

q  Binary  (base  2)  - Each  bit  represents  a  power  of  2  - 11102  =  1  x  23  +  1  x  22  +  1  x  21  +  0  x  20    =    1410  

 

0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 10010

Coun3ng  in  Binary   0 = 1 = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 9 = 10 = 11 = 12 = 13 = 14 = 15 = 16 = 17 = 18 =

0000 0000000 000 00 00 00 00 0 0 0 0 0 0 0 0

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Number  Systems  

Base Character Set 2 {0,1} 3 {0,1,2} 4 {0,1,2,3} 5 {0,1,2,3,4} 6 {0,1,2,3,4,5} 7 {0,1,2,3,4,5,6} 8 {0,1,2,3,4,5,6,7} 9 {0,1,2,3,4,5,6,7,8} 10 {0,1,2,3,4,5,6,7,8,9} 11 {0,1,2,3,4,5,6,7,8,9,A} 12 {0,1,2,3,4,5,6,7,8,9,A,B} 13 {0,1,2,3,4,5,6,7,8,9,A,B,C} 14 {0,1,2,3,4,5,6,7,8,9,A,B,C,D} 15 {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E} 16 {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}

Binary  (Base  2)  to  Decimal  

q  Sum  up  (bit*weight):  

28 27 26 25 24 23 22 21 20

256 128 64 32 16 8 4 2 11 0 0 0 0 0 1 1

1 0 1 0 0 1 1 0 0

100000112 = _________________________________10 ? 1010011002 = _________________________________10 ?

Binary Weights

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Powers  of  2  

q  Memorize!  

210 29 28 27 26 25 24 23 22 21 20

1024 512 256 128 64 32 16 8 4 2 1

280 270 260 250 240 230 220 210

Yotta Zetta Exa Peta Tera Giga Mega KiloY Z E P T G M K

Octal  (Base  8)  to  Decimal  

q  Sum  up  (digit*weight):  

84 83 82 81 80

4096 512 64 8 12 5 7

1 0 2 2 6

2578 = _________________________________10 ? 102268 = _________________________________10 ?

Octal weights

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Hexadecimal  (Base  16)  to  Decimal  

q  Sum  up  (hex  digit*weight):  

163 162 161 160

4096 256 16 1A 2 B

1 4 A 6

A2B16 = _________________________________10 ? 14A616 = _________________________________10 ?

Hexadecimal Weight

0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22

Coun3ng  in  Octal   0 = 1 = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 9 = 10 = 11 = 12 = 13 = 14 = 15 = 16 = 17 = 18 =

00 00 00 00 00 00 00 00 0 0 0 0 0 0 0 0 0 0 0

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0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12

Coun3ng  in  Hexadecimal   0 = 1 = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 9 = 10 = 11 = 12 = 13 = 14 = 15 = 16 = 17 = 18 =

00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0 0 0

Decimal  to  Binary  

q  Use  the  Placement  Method  

1024 512 256 128 64 32 16 8 4 2 1Powers of 2

15510 = _________________________________ 2 ?

•  128 goes into 155 once leaving 27 to be placed

1 ? ? ? ? ? ? ?

•  64 and 32 are too big (make them 0)•  16 goes in once leaving 11

1 0 0 1 ? ? ? ?

•  and so on

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You  Try  It  …  

1024 512 256 128 64 32 16 8 4 2 1Powers of 2

58310 = _________________________________ 2 ?

Hexadecimal Benefits q  It is often convenient to write binary (base-2) numbers

as hexadecimal (base-16) numbers instead. -  fewer digits -- four bits per hex digit -  less error prone -- easy to corrupt long string of 1’s and 0’s

Binary Hex Decimal 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7

Binary Hex Decimal 1000 8 8 1001 9 9 1010 A 10 1011 B 11 1100 C 12 1101 D 13 1110 E 14 1111 F 15

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Converting from Binary to Hexadecimal q  Every four bits is a hex digit. -  start grouping from right-hand side

011101010001111010011010111

7D4F8A3

This is not a new machine representation, just a convenient way to write the number.

1.  Convert  the  hex  value  0x7A8BF7D6  into  its  binary  equivalent:    

       

     

2.  Convert  the  binary     10 0110 1110 1001 0100 1100 0101 11112      

                 to  hex:  

 

Exercises  

0111 1010 1000 1011 1111 0111 1101 0110

0010 0110 1110 1001 0100 1100 0101 1111

6 E 9 4 C 5 F 2

7 A 8 B F 7 D 6

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3.  Convert  from  binary  to  octal:    

10 111 010 100 110 001 011 1112        

Exercises  (contd.)  

010 111 010 100 110 001 011 111

7 2 4 6 1 3 7 2

Number  Systems  q  The  binary,  hexadecimal  (hex)  and  octal  system  share  one  common  feature  –  they  are  all  based  on  powers  of  2.    

q  Each  digit  in  the  hex  system  is  equivalent  to  a  four-­‐digit  binary  number  and  each  digit  in  the  octal  system  is  equivalent  to  a  3-­‐digit  binary  number.    

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Prac3ce  

Hex Decimal Binary

10

240

11111111

Binary  Addi3on  q  Let’s  review  decimal  addi1on  -  From  right  to  leg,  we  add  each  pair  of  digits  - We  write  the  sum,  and  add  the  carry  to  the  next  column  

1 9 8

+ 2 6 4

Sum

Carry

2

1

6

1

4

0

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Binary  Addi3on  q  From  right  to  leg,  we  add  each  pair  of  bits  

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

q  We  write  the  sum,  and  add  the  carry  to  the  next  column  

1 9 8

+ 2 6 4

Sum

Carry

0 1 1

+ 0 0 1

Sum

Carry 2

1

6

1

4

0

Base 10 Base 2

Data  Representa3on  

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What kinds of data do we need to represent?

- Numbers – signed, unsigned, integers, floating point, complex, rational, irrational, …

- Text – characters, strings, … -  Images – pixels, colors, shapes, … - Sound - Logical – true, false -  Instructions - …

q  Data type: -  representation and operations within the computer

Word-Oriented Memory Organization

q  Addresses Specify Byte Locations -  Address of first byte in word -  Addresses of successive

words differ by 4 (32-bit) or 8 (64-bit)

0000 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011

32-bitWords Bytes Addr.

0012 0013 0014 0015

64-bitWords

Addr =??

Addr =??

Addr =??

Addr =??

Addr =??

Addr =??

0000

0004

0008

0012

0000

0008

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Data Representations q  Sizes of C Objects (in Bytes)

C Data Type Sparc/Unix Typical 32-bit Intel IA32 int 4 4 4 long int 8 4 4 char 1 1 1 short 2 2 2 float 4 4 4 double 8 8 8 long double 8 8 10/12 char * 8 4 4 (or any other pointer)

Address vs. Value q Sometimes we want to deal with the address

of a memory location, rather than the value it contains.

q Adding a column of numbers. - R2 contains address of first location. - Read value, add to sum, and

increment R2 until all numbers have been processed.

q R2 is a pointer -- it contains the address of data we’re interested in.

x3107 x2819 x0110 x0310 x0100 x1110 x11B1 x0019

x3100

x3101

x3102

x3103

x3104

x3105

x3106

x3107

x3100

R2

address

value

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Byte Ordering q  How should bytes within multi-byte word be ordered in

memory?

q  Conventions - Sun, Mac are “Big Endian” machines

o  Least significant byte has highest address - Alpha, PC are “Little Endian” machines

o  Least significant byte has lowest address

Byte Ordering Example q  Big Endian - Least significant byte has highest address

q  Little Endian - Least significant byte has lowest address

q  Example - Variable x has 4-byte representation 0x01234567 - Address given by &x is 0x100

0x100 0x101 0x102 0x103

01 23 45 67

0x100 0x101 0x102 0x103

67 45 23 01

Big Endian

Little Endian

01 23 45 67

67 45 23 01

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Integers  

Unsigned Integers q  An n-bit unsigned integer represents 2n values:

from 0 to 2n-1. 22 21 20 0 0 0 0 0 0 1 1 0 1 0 2 0 1 1 3 1 0 0 4 1 0 1 5 1 1 0 6 1 1 1 7

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Signed  Integers  q  How  do  computers  differen1ate  between  posi1ve  and  nega1ve  integers?    - Posi1ve  integers  have  most  significant  bit  0  - Nega1ve  integers  have  most  significant  bit  1  

q  Nega1ve  integer  representa1ons:  1.  Sign  Magnitude  2.  One’s  Complement  3.  Two’s  Complement  

1.  Sign  Magnitude  q  Use  the  legmost  bit  to  store  the  sign  - Zero  for  posi1ve  number  - One  for  nega1ve  number  

q  Examples  

0 0 1 0 1 1 0 0 è 44 1 0 1 0 1 1 0 0 è -44

q  Hard  to  do  arithme1c  this  way,  so  it  is  rarely  used  -  What  is  the  result  of  44  –  44?  

Sign Magnitude

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1.  Sign  Magnitude  (contd.)  

0 0 1 0 1 1 0 0 è 44 1 0 1 0 1 1 0 0 è -44

q  For  numbers  represented  on  n  bits:  - Range  of  posi1ve  integers:  - Range  of  nega1ve  integers:  

Sign Magnitude

from 0 to (2n-1 – 1)

from –(2n-1 – 1) to –1

2.  One’s  Complement  q  Legmost  bit  is  0  for  posi1ve  numbers  

0 0 1 0 1 1 0 0 è 44

q  To  obtain  the  corresponding  nega1ve  number  (-­‐44),  flip  every  bit:  

1 1 0 1 0 0 1 1 è -44

q  So  -­‐44  is  the  one’s  complement  of  44.  

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2.  One’s  Complement  (contd.)  q  What  is  the  result  of  44  –  44?  

0 0 1 0 1 1 0 0 ( 44) 1 1 0 1 0 0 1 1 (-44)

_______________

q  Issue:  two  different  representa1ons  for  zero    

3.  Two’s  Complement  q  Legmost  bit  is  0  for  posi1ve  numbers  

  0 0 1 0 1 1 0 0 è 44

q  To  obtain  the  corresponding  nega1ve  number  -­‐44,  add  1  to  the  one’s  complement  of  44:  

1 1 0 1 0 0 1 1 è one’s complement + 0 0 0 0 0 0 0 1 _______________ 1 1 0 1 0 1 0 0 è two’s complement

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3.  Two’s  Complement  (contd.)  q  What  is  the  result  of  44  –  44?  

0 0 1 0 1 1 0 0 ( 44) 1 1 0 1 0 1 0 0 (-44)

_______________

q  Used  by  most  computer  systems  q  For  numbers  represented  on  n  bits:  

- Range  of  posi1ve  integers:  - Range  of  nega1ve  integers:  

from 0 to (2n-1 – 1)

from –(2n-1 – 1) to –1

Converting Two’s Complement to Decimal 1.  If leading bit is one, take two’s

complement to get a positive number.

2.  Add powers of 2 that have “1” in the corresponding bit positions.

3.  If original number was negative, add a minus sign.

n 2n

0 1 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 256 9 512

10 1024

X = 01101000two = 26+25+23 = 64+32+8 = 104ten

Assuming 8-bit 2’s complement numbers.

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More Examples n 2n

0 1 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 256 9 512

10 1024

Assuming 8-bit 2’s complement numbers.

X = 00100111two = 25+22+21+20 = 32+4+2+1 = 39ten

X = 11100110two -X = 00011010

= 24+23+21 = 16+8+2 = 26ten

X = -26ten

Two’s Complement Signed Integers q  Most significant bit is sign bit and has weight –2n-1.

q  Range of an n-bit number: -2n-1 through 2n-1 – 1. - The most negative number (-2n-1) has no positive counterpart.

-23 22 21 20 0 0 0 0 0 0 0 0 1 1 0 0 1 0 2 0 0 1 1 3 0 1 0 0 4 0 1 0 1 5 0 1 1 0 6 0 1 1 1 7

-23 22 21 20 1 0 0 0 -8 1 0 0 1 -7 1 0 1 0 -6 1 0 1 1 -5 1 1 0 0 -4 1 1 0 1 -3 1 1 1 0 -2 1 1 1 1 -1

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Number  Representa3ons  Review  

q  Legmost  bit  zero  indicates    

q  Sign  Magnitude:        - nega1ve  values  

q  One’s  Complement:        - nega1ve  values  

q  Two’s  complement:        - nega1ve  values  

positive number

most significant bit 1 has weight 0

are the 1s complement of positive values(flip every bit)

are the 2s complement of positive valuesmost significant bit 1 has weight –2n-1

Fill  in  the  Table  Bit

PatternValue

(Sign Magnitude)Value

(One’s Complement)Value

(Two’s Complement)

000

001

010

011

100

101

110

111

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Ques3on  q  What  value  does  10011001  represent?    

ASCII  American  Standard  Code  for    

Informa3on  Interchange  

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The ASCII CodeAmerican Standard Code for Information Interchange 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 NUL SOH STX ETX EOT ENQ ACK BEL BS HT LF VT FF CR SO SI

16 DLE DC1 DC2 DC3 DC4 NAK SYN ETB CAN EM SUB ESC FS GS RS US

32 SP ! " # $ % & ' ( ) * + , - . /

48 0 1 2 3 4 5 6 7 8 9 : ; < = > ?

64 @ A B C D E F G H I J K L M N O

80 P Q R S T U V W X Y Z [ \ ] ^ _

96 ` a b c d e f g h i j k l m n o

112 p q r s t u v w x y z { | } ~ DEL

Lower case: 97-122 and upper case: 65-90E.g., ‘a’ is 97 and ‘A’ is 65 (i.e., 32 apart)

char Constantsq  C has char constants (sort of)

q  Examples

Constant Binary Representation (assuming ASCII)

Note

'a' 01100001 letter

'0' 00110000 digit

'\x61' 01100001 hexadecimal form

Use single quotes for char constantUse double quotes for string constant

* Technically 'a' is of type int; automatically truncated to type char when appropriate

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More char Constants

Constant Binary Representation (assuming ASCII)

Note

'\a' 00000111 alert (bell)

'\b' 00001000 backspace

'\f' 00001100 form feed

'\n' 00001010 newline

'\r' 00001101 carriage return

'\t' 00001001 horizontal tab

'\v' 00001011 vertical tab

'\\' 01011100 backslash

'\?' 00111111 question mark

'\'' 00100111 single quote

'\"' 00100010 double quote

'\0' 00000000 null

•  Escape characters

Usedoften

Interesting Properties of ASCII Code q  What is relationship between a decimal digit ('0', '1', …)

and its ASCII code?

q  What is the difference between an upper-case letter ('A', 'B', …) and its lower-case equivalent ('a', 'b', …)?

q  Given two ASCII characters, how do we tell which comes first in alphabetical order?

q  Are 128 characters enough? (http://www.unicode.org/)

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Other Data Types q  Floating Points -  IEEE representation, to be covered later

q  Text strings -  sequence of characters, terminated with NULL (0)

q  Image - array of pixels

o  monochrome: one bit (1/0 = black/white) o  color: red, green, blue (RGB) components (e.g., 8 bits each) o  other properties: transparency

q  Sound -  sequence of fixed-point numbers

Bit-­‐Level  Operators    

(Common  to  C  and  Java)  

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Used in Networking

Used in Encryption/ Decryption

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Used in Compression

•  Look at the DEFLATE algorithm, for instance -  https://en.wikipedia.org/wiki/DEFLATE -  everything in bits, not bytes

Overview of C bit-level operators

~ Bitwise NOT (“flips” bits) & Bitwise AND ^ Bitwise XOR | Bitwise OR

<< Bitwise left shift (shifts bits to left) >> Bitwise right shift (shifts bits to right)

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Observation 1 – Addition q  2’s comp. addition is just binary addition. - assume all integers have the same number of bits -  for now, assume that sum fits in n-bit 2’s comp. representation

01101000 (104) 11110110 (-10)+ 11110000 (-16) + (-9)

01011000 (88) (-19)

Assuming 8-bit 2’s complement numbers.

Observation 2 – Sign Extension q  To add two numbers, we must represent them

with the same number of bits.

q  If we just pad with zeroes on the left:

q  Instead, replicate the MS bit -- the sign bit:

4-bit 8-bit 0100 (4) 00000100 (still 4) 1100 (-4) 00001100 (12, not -4; NOT GOOD!)

4-bit 8-bit 0100 (4) 00000100 (still 4) 1100 (-4) 11111100 (still -4)

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Bit-Level Operations in C q  Operations &, |, ~, ^ available in C

-  Apply to any “integral” data type o  long, int, short, char

-  View arguments as bit vectors -  Arguments applied bit-wise

q  Examples (char data type) -  ~0x41 --> 0xBE

~010000012 --> 101111102 -  ~0x00 --> 0xFF

~000000002 --> 111111112 -  0x69 & 0x55 --> 0x41

011010012 & 010101012 --> 010000012 -  0x69 | 0x55 --> 0x7D

011010012 | 010101012 --> 011111012

Bitwise  Operator:  NOT  (~)  q  NOT  (~)  is  the  same  as  one’s  complement  - Turns  0  to  1,  and  1  to  0  

q  Examples:  - Assume  that  x  is  an  integer  on  8  bits.  Set  x  =  1.          What  is  the  value  of  ~x?  

- Set  the  least  significant  bit  of  y  to  0:  y = y & _____;

0

1

~x

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Bitwise  Operators:  AND  (&)  and  OR  (I)  Bitwise  AND  (&)  

             

 

Bitwise  OR  (|)  

& 0

1

0 1 |

0

1

0 1

0 0 1 1 0 1 0 1

0 0 0 0 1 1 1 1

___

& 15

___

0 0 1 1 0 1 0 1

0 0 0 0 1 1 1 1

___

| 15

___

Assume  that  x  is  an  integer  variable.  

q  Determine  the  least  significant  bit  of  x:   int bit = x & ________;

q  Set  the  least  significant  bit  of  x  to  1:   x = x | ________;

Bitwise  Operators:  AND  (&)  and  OR  (I)  

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Bitwise  Operator:  XOR  q  XOR  (^)  - 0  if  both  bits  are  the  same  - 1  if  the  two  bits  are  different  

q  For  an  integer  x,  what  is  the  value  of    

 x^x

^

0

1

0 1

Relations Between Operations q  DeMorgan’s Laws - Express & in terms of |, and vice-versa

o  A & B = ~(~A | ~B) •  A and B are true if and only if neither A nor B is false

o  A | B = ~(~A & ~B) •  A or B are true if and only if A and B are not both false

q  Exclusive-Or using Inclusive Or o  A ^ B = (~A & B) | (A & ~B)

•  Exactly one of A and B is true o  A ^ B = (A | B) & ~(A & B)

•  Either A is true, or B is true, but not both

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Bitwise  Operator:  Le`  Shi`  q  Leg  shig:    x  <<  y    (same  as  mul1ply  by  2y)  - Shig  bit-­‐vector  x  leg  y  posi1on  - Fill  blanks  with  0    

0 0 1 1 0 1 0 1 53

53<<2

Bitwise  Operator:  Right  Shi`  q  Right  shig:  x  >>  y    (same  as  divide  by  2y)  - Shig  bit-­‐vector  x  right  y  posi1ons  - Fill  in  blanks  with  sign  bit    (this  is  called  arithme/c  shi0)        

0 0 1 1 0 1 0 1 53

53>>2

1 0 1 1 0 1 0 1 -75

-75>>2

sign extension sign extension

Note: Java also has the unsigned shift operator >>> Not available in C.

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Bitwise  Operators  q  Print  all  the  bits  of  a  character  from  right  to  leg  (star1ng  with  the  least  significant  bit):  

0 0 1 1 0 1 0 1 char c = 0xB5; void ReversePrintBits(char c) { /* add code here */ }

Bitwise  Operators  q  Print  all  the  bits  of  a  character  from  leg  to  right  (star1ng  with  the  most  significant  bit):  

0 0 1 1 0 1 0 1 char c = 0xB5; void PrintBits(char c) { /* add code here */ }

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q  Used  to  change  or  query  one  or  more  bits  in  a  variable.  

q  The  bitmask  indicates  which  bits  are  to  be  affected.  

q  Common  opera1ons:  - Set  one  or  more  bits  (set  to  1)  - Clear  one  or  more  bits  (set  to  zero)  

- Read  one  or  more  bits  

q  Examples:  - Set  bit  2  of  x  (bit  0  is  least  significant):  x = x ________ - Clear  bit  3  of  x:   x = x ____________

- Read  bit  4  of  x:   bit = x ____________

Bitmasks  

q  Set  the  least  significant  byte  of  x  to  FF:  

x = x ________;  

q  Clear  the  least  significant  byte  of  x:  

x = x ________;

q  Read  the  least  significant  byte  of  x:  

byte = x ________;

Bitmasks  

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q  Set  to  1  bits  2,  4  and  7  of  x  (0  is  least  significant):  

x = x ________;  

q  Clear  bits  3,  4  and  5  of  x:  

x = x ________;

Bitmasks  

Contrast:    Logical  Operators  

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Familiar  Operators  (common  to  C,  Java)  Category OperatorsArithmetic ++expr --expr expr++ expr--

expr1*expr2 expr1/expr2 expr1%expr2 expr1+expr2 expr1-expr2

Assignment expr1=expr2 expr1*=expr2 expr1/=expr2 expr1%=expr2 expr1+=expr2 expr1-=expr2

Relational expr1<expr2 expr1<=expr2 expr1>expr2 expr1>=expr2 expr1==expr2 expr1!=expr2

Logical !expr expr1&&expr2 expr1||expr2

Function Call func(paramlist)

Cast (type)expr

Conditional expr1?expr2:expr3

Logical  Operators  

q  Always  return  0  or  1  

q  View  0  as  “False”  

q  Anything  nonzero  as  “True”  

q  Examples  (char  data  type)  

- !0x41 --> _____ - !0x00 --> _____ - !!0x41 --> _____

- 0x69 && 0x55 --> _____ - 0x69 || 0x55 --> _____

&& (Logical AND), || (Logical OR), ! (Logical NOT)

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What  is  the  Output?  int a = 0x43, b = 0x21;

printf("a | b = %x\n", a | b); _________

Printf("a || b = %x\n", a || b); _________

printf("a & b = %x\n", a & b); _________

printf("a && b = %x\n", a && b); _________

Mental  Exercise  /*      *      isNotEqual  -­‐  return  0  if  x  ==  y,  and  1  otherwise      *      Examples:  isNotEqual(5,5)  =  0,  isNotEqual(4,5)  =  1  */    int isNotEqual (int x, int y) { return ______________; }

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Limits  of  the  Machine:    

How  much  memory  space  for  data?  

Storage  Units  

1 bit = smallest unit of memory

1 byte = 8 bits

4 bytes = 1 word (system dependent)

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Words  q  On  most  machines,  bytes  are  assembled  into  larger  structures  called  “words”,  where  a  word  is  usually  defined  to  be  the  number  of  bits  the  processor  can  operate  on  at  one  1me.    

q  Some  machines  use  four-­‐byte  words  (32  bits),  while  some  others  use  8-­‐byte  words  (64  bits)  and  some  machines  use  less  conven1onal  sizes.  

0000 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011

32-bitWords Bytes Addr.

0012 0013 0014

64-bitWords

Addr =??

Addr =??

Addr =??

Addr =??

Addr =??

Addr =??

0000

0004

0008

0012

0000

0008

The  sizeof  Operator  

q Unique  among  operators:    evaluated  at  compile-­‐1me  

q Evaluates  to  type  size_t;  on  tanner,  same  as  unsigned  int  

q Examples  

Category Operators

sizeof sizeof(type) sizeof(expr)

int i = 10; double d = 100.0; … … sizeof(int) … /* On tanner, evaluates to 4 */ … sizeof(i) … /* On tanner, evaluates to 4 */ … sizeof(double)… /* On tanner, evaluates to 8 */ … sizeof(d) … /* On tanner, evaluates to 8 */ … sizeof(d + 200.0) … /* On tanner, evaluates to 8 */

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Determining  Data  Sizes  q  To  determine  data  sizes  on  your  computer  

q  Output  on  tanner  

#include <stdio.h> int main() { printf("char: %d\n", (int)sizeof(char)); printf("short: %d\n", (int)sizeof(short)); printf("int: %d\n", (int)sizeof(int)); printf("long: %d\n", (int)sizeof(long)); printf("float: %d\n", _________________); printf("double: %d\n", _________________); printf("long double: %d\n", _________________); return 0; }

char: 1 short: 2 int: 4 long: 4 float: 4 double: 8 long double: 16

Overflow:  Running  Out  of  Room  q  Adding  two  large  integers  together  - Sum  might  be  too  large  to  store  in  available  bits  - What  happens?  

 

q  We have overflow if: -  signs of both operands are the same, and -  sign of sum is different.

01000 (8) 11000 (-8) + 01001 (9) + 10111 (-9)

10001 (-15) 01111 (+15) Assuming 5-bit 2’s complement numbers.

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Overflow  q  Unsigned  integers  - All  arithme1c  is  “modulo”  arithme1c  - Sum  would  just  wrap  around  

q  Signed  integers  - Can  get  nonsense  values  - Example  with  16-­‐bit  integers    

o  Sum:  10000+20000+30000    o  Result:  -­‐5536  

Try  It  Out  q  Write  a  program  that  computes  the  sum  

 

10000+20000+30000  

   Use  only  short  int  variables  in  your  code:  

short int a = 10000; short int b = 20000; short int c = 20000; short int sum = a + b + c; printf("sum = %d\n", sum);                  

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Exercise  q  Assume  only  four  bits  are  available  for  represen1ng  integers,  and  signed  integers  are  represented  in  2’s  complement.  

q  Compute  the  value  of  the  expression  7  +  7  

short int x = 15213; unsigned short int ux = (unsigned short) x; short int y = -15213; unsigned short int uy = (unsigned short) y;

Casting Signed to Unsigned q  C Allows Conversions from Signed to Unsigned

q  Resulting Value - No change in bit representation - Nonnegative values unchanged

o  ux = 15213 - Negative values change into (large) positive values

o  uy = 50323

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Try  It  Out  q  C  code:  

char a = 0xFF; unsigned char b = 0xFF; printf("a = %d\n", a); printf("b = %d\n", b);

Int to Char? Try It Out …#include <stdio.h>

int main() { char c = 0x81; int i; i = c; printf(" integer = %x\n character = %x\n", i, c); i = 0x87654321; c = i; printf(" integer = %d\n character code = %d\n", i, c); return 0; }

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C  vs.  Java:  Cast  Conversions  q  Java:  demo1ons  are  not  automa1c  C:  demo1ons  are  automa1c  

int i; char c; … i = c; /* Implicit promotion */ /* Sign extension in Java and C */ c = i; /* Implicit demotion */ /* Java: Compiletime error */ /* C: OK; truncation */ c = (char)i; /* Explicit demotion */ /* Truncation in Java and C */

What  did  we  learn?  q  Computer  represents  everything  in  binary  - Integers,  floa1ng-­‐point  numbers,  characters,  …  - Pixels,  sounds,  colors,  etc.  

q  Memory  is  bytes,  words,  endian,  two’s  complement,  ASCII,  conver1ng  between  hex-­‐binary-­‐decimal,  binary  opera1ons,  sign  extension,  limits  of  machine,  overflow,  cast  conversions