1. OOP in Java : © W. Milner 2005 : Slide 1 OOP in Java

2. OOP in Java : © W. Milner 2005 : Slide 2 Course outline 1.Getting started, primitive data types and control structures 2.Classes and objects 3.Extending classes 4.Using some standard packages 5.OOP revisited Sessions 1 to 3 introduce and sketch out the ideas of OOP. Session 5 deals with these ideas in closer detail.

3. OOP in Java : © W. Milner 2005 : Slide 3 Recommended text  The Java Programming Language Third Edition by Arnold, Gosling and Holmes – Addison Wesley  Gosling created the Java language  In the spirit of The C Programming Language by K & R, and the C++ Programming Language by Stroustrup  If you’ve never written a computer program before, it won’t make much sense

4. OOP in Java : © W. Milner 2005 : Slide 4 What you need to know already  What compiling and iterpreting mean  Some knowledge of C –Much Java syntax is like C, so –We’ll say ‘it’s like C except that..’

5. OOP in Java : © W. Milner 2005 : Slide 5 Getting started  Just follow these steps – explanations later  Look at java.sun.com for what is available  You need J2SE = Java 2 Standard Edition  Download and install the JDK – currently 5.0 Update 5  This will install software needed in C:\Program files\java  Create a directory where you will save your Java programs – call the folder JavaProgs

6. OOP in Java : © W. Milner 2005 : Slide 6 Keeping Going  Start NotePad and enter the program on the next slide.  Save it with the filename “First.java” – must be this name. Put quotes around it – make sure Notepad does not put.txt on the end.  Note CAPITAL F.  Save it in your JavaProgs folder:

7. OOP in Java : © W. Milner 2005 : Slide 7 The first program public class First { public static void main(String[] args) { System.out.println("Hello world"); }

8. OOP in Java : © W. Milner 2005 : Slide 8 Then..  Switch to the command prompt (Start Programs Accessories Command Prompt) and navigate to your JavaProgs folder  Use cd to change directory. Cd.. Moves ‘up’ a level.  You should have something like: 

9. OOP in Java : © W. Milner 2005 : Slide 9 Compile it  The compiler is called javac. You’ll need to include the path to it, so type in:   After this, get this line back using the up and down arrow keys.  Then run it like this:

10. OOP in Java : © W. Milner 2005 : Slide 10 Explanations..  All Java source code files are called Something.java  You give this to the javac compiler  This produces a file called Something.class  This is in bytecode  This can be interpreted by the java interpreter

11. OOP in Java : © W. Milner 2005 : Slide 11 More explanations  Each source code file defines a class  The name of the class it defines (First in our case) must match the filename – hence First.java  This is compiled to a bytecode file named First.class  Java is case-sensitive (like C)  Classes should start with a capital letter.  So the file must be called First.java and not first.java

12. OOP in Java : © W. Milner 2005 : Slide 12 Exercise  Change the file First.java so it defines a class called Second – and save it as Second.java  Change it so it outputs ‘my second program’ instead of ‘Hello world’  Compile and run it.

13. OOP in Java : © W. Milner 2005 : Slide 13 Features of Java  Java is a general purpose high level language  Core Java is well-defined and stable  Versions are useful in many situations – desktop, server, embedded, mobile etc  Java is a trademark of Sun Microsystems  Java is cross -platform

14. OOP in Java : © W. Milner 2005 : Slide 14 More features  Java is a pure object-oriented language.  No functions, no global variables  Unlike C++, which is C with objects  Java is designed to make it hard to write programs which will crash  No pointers  Compiler will not compile doubtful code  Programmer must write code that ‘catches exceptions’  Run-time checks eg array bounds exceptions  Slower than C or C++ (not much), but less chance of crash

15. OOP in Java : © W. Milner 2005 : Slide 15 Types  Recall data type from C – int, char double etc  Java has 2 kinds of types: Primitive types Reference types

16. OOP in Java : © W. Milner 2005 : Slide 16 Primitive types  These are simple types like char, double, int  Similar to C  Main difference – the sizes of these types are defined eg an int is 4 bytes  Each hardware platform has its own 'virtual machine'  Which all look the same  So can all have the same data sizes  All chars use UNICODE character set - so characters are 2 bytes long

17. OOP in Java : © W. Milner 2005 : Slide 17 Reference types  Reference type variables are objects  Objects belong to classes  Objects made by a constructor

18. OOP in Java : © W. Milner 2005 : Slide 18 Primitive types first  We will look at primitive types first  and control structures (loops and ifs)  Then look at classes and objects

19. OOP in Java : © W. Milner 2005 : Slide 19 Program format public class Testing { public static void main(String[] args) { // find the area of a circle.. double radius = 5.0; double area; area = 3.1416 * radius * radius; System.out.println("Area = " + area); } Use this format to start with In file Testing.java Code goes here Explain rest later

20. OOP in Java : © W. Milner 2005 : Slide 20 Variables - declaring and assigning  // starts a line comment  double area declares a variable called area of type double  double radius = 5.0; declares and initializes a variable  variables can be declared anywhere in a block { }  statements end in ; like C and C++ public class Testing { public static void main(String[] args) { // find the area of a circle.. double radius = 5.0; double area; area = 3.1416 * radius * radius; System.out.println("Area = " + area); }

21. OOP in Java : © W. Milner 2005 : Slide 21 Console output  System.out.println("Area = " + area);  This takes a single string argument – but..  The + causes area to be converted to the equivalent string, and  concatenated with the left hand operand  so we get  "Area = 5.72" or whatever  System.out.print stays on same line

22. OOP in Java : © W. Milner 2005 : Slide 22 Primitive data types - numeric Java data type sizes are platform independent All are signed Top four are integer, bottom two are floating point Variables of these types declared like short a,b,c;or initialised when declared double x = 1.502;

23. OOP in Java : © W. Milner 2005 : Slide 23 Numeric data types  Can get overflow eg –int i = 64000; –int n = i * i ;  Specify type of constant like –x = 1000L; // defaults to integer –f = 1.0F; // force float - defaults to double –x = 0x1FF;// hex  Operators + - * /  % is mod = remainder eg 13 % 4 is 1  Short cut – same as C += x+=8; same as x = x + 8; -= x-=8; same as x = x - 8; *= x*=8; same as x = x * 8; /=x/=8;same as x = x / 8; %=x%=8;same as x = x % 8;

24. OOP in Java : © W. Milner 2005 : Slide 24 Overflow Exercise  Use code to produce overflow as in the previous slide  Find out what happens when you compile/run it.

25. OOP in Java : © W. Milner 2005 : Slide 25 Two types of division  float f = 1.0 / 2.0;// floating point  int i = 1 / 2;// i is 0  if both operands are integer, / is integer version - it gives the quotient and discards the remainder  So / is overloaded – different versions, same name

26. OOP in Java : © W. Milner 2005 : Slide 26 Increment and decrement  x++; is the same as x = x + 1;  y--; is the same as y = y - 1;  post-increment is like a = b++; which first assigns b to a then increments b  pre-increment is a = ++b; which first increments b then assigns the new value to a

27. OOP in Java : © W. Milner 2005 : Slide 27 Type casts  Assigning a small type to a larger type is no problem eg  int i; long x; i = 32; x = i; OK because x more bits than i  But reverse gives ‘possible loss of precision’ eg  int i; long x; x = 32; i = x;// gives compile error type cast  Problem solved by a type cast ie i = (int) x;

28. OOP in Java : © W. Milner 2005 : Slide 28 Type cast exercise  Try out  int i; long x; x = 32; i = x; In a program.  Fix it as in the previous slide

29. OOP in Java : © W. Milner 2005 : Slide 29 Char type  char is for a single character, like  char c = ‘A’; note single quotes  c++;makes c = ‘B’  Strings are different - see later Unicode  Java uses Unicode not ASCII - 16 bits per character eg import java.applet.*; import java.awt.*; public class TestApplet extends Applet { public void paint(Graphics g) { char c; Font f = new Font("Arial Unicode MS",Font.PLAIN,20); g.setFont(f); c = '\u098a';// Unicode constant g.drawString("Some Bengali: " + c,10,30 ); }

30. OOP in Java : © W. Milner 2005 : Slide 30 boolean type  If a variable is declared to be boolean, it is restricted to 2 values - true and false  boolean result; result = true;  result = ( x > y ); result is true if x is greater than y  also = == != == not the same as =  && and|| or ! not  result = ( x > y ) && ( y < 5 ); result is true if x is greater than y and y is less than 5  && and || are short-cut operators

31. OOP in Java : © W. Milner 2005 : Slide 31 bitwise operators  & is bitwise AND (like both) | is bitwise OR (like either ) ^ is XOR (like not equal) ~ is NOT  eg if x = 9 1001 and y = 10 1010  x & y is 8 1000  x | y is 11 1011  x ^ y is 3 0011  ~ 0xFFFFFFFE is 1inverting 11111111110

32. OOP in Java : © W. Milner 2005 : Slide 32 bit shift operators  >> is shift right eg if x = 7 or in binary 0000 0111 x >> 1 is 0000 0011  << is shift left so x << 1 is 0000 1110 = 14

33. OOP in Java : © W. Milner 2005 : Slide 33 Bit shift exercise  Try out this code: int i = 6; System.out.println(i&1); i>>=1; System.out.println(i&1); i>>=1; System.out.println(i&1);  Explain what you get

34. OOP in Java : © W. Milner 2005 : Slide 34 precedence  2 * 3 + 4 is 10 not 14  2*(3+4) is 14  Highest++ -- * / % + - >= == != && ||  Lowest= += -= *= /= %=  Use brackets when in doubt

35. OOP in Java : © W. Milner 2005 : Slide 35 Control - if  for example if ( x== 5 ) { y = 2; a++; } else c++;  round brackets around boolean expression  indentation  no then as in Visual Basic  block { } around several steps to do  no block if just one step - if (x<4) a=4;  else can be omitted if not needed

36. OOP in Java : © W. Milner 2005 : Slide 36 if example - validation  for example if ( ( age>0 ) && ( age < 130 ) ) System.out.println(‘age is valid’); else { System.out.println(‘age is invalid’);.. code to deal with error.. }  beware of if ( x==5 ); y = 2;

37. OOP in Java : © W. Milner 2005 : Slide 37 switch - I  used where many alternative actions are possible  example - switch (y) { case 5:a = 7; case 9:b = 3; case 4:a = 8; default:z = 2; }  y can be expression (like x + 4) but must be integral  the 5, 9, 4 etc must be constants  default is optional

38. OOP in Java : © W. Milner 2005 : Slide 38 switch - II  the action ‘falls through’ - when one case is triggered, all the following options execute to the end of the switch break  so often combine with break - example - switch (y) { case 5:a = 7;break; case 9:b = 3;break; case 4:a = 8;break; }  include final break - not essential but good practice, since if add further option do not need to go back and add break to previous

39. OOP in Java : © W. Milner 2005 : Slide 39 Conditional operator ? ;  example x = ( y > 4 ) ? 7 : 3; if y is greater than 4, x becomes 7, and otherwise, it becomes 3  in general a ? b : c b is evaluated if a is true, c if it is not  example int x =9, y = 10, a; a = (x > 9) ? x++ : y++ ;  after this a = 10, y is 11, x is still 9

40. OOP in Java : © W. Milner 2005 : Slide 40 loops - while  loops repeat blocks of code - called iteration  example - output the numbers 3, 6, 9,... 99 x = 3; while ( x<102 ) { System.out.println( x ); x += 3; }  in general, while (boolean expression) statement or block to repeat  need to initialise variables  may loop zero times if false first time  use indentation

41. OOP in Java : © W. Milner 2005 : Slide 41 loops - do while  example - output the numbers 3, 6, 9,... 99 x = 3; do { System.out.println( x ); x += 3; } while ( x<102 )  in general, do statement or block to repeat while (boolean expression)  unlike a while, it will execute the loop at least once

42. OOP in Java : © W. Milner 2005 : Slide 42 loops - for  example - output the numbers 3, 6, 9,... 99 for ( x = 3; x<102; x+=3 ) System.out.println(x);  in general for ( ; ; )  may loop zero times  add up the integers 1 + 2 + 3 +...100 int t = 0; int x; for ( x = 1; x<101; x++) t += x; System.out.println( t );

43. OOP in Java : © W. Milner 2005 : Slide 43 loops - for - II statement list  can use statement list, like int t; int x; for ( x = 1, t = 0; x<101; x++) t+=x; System.out.println(t);  can omit any part, (retain separating ; ) like int t = 0; int x = 1; for ( ; x<101; x++) t+=x; System.out.println(t);  for (; ; ) loops forever

44. OOP in Java : © W. Milner 2005 : Slide 44  can declare variable in for, like int t = 0; for ( int x = 1; x<101; x++) t+=x; System.out.println(t); in this case the scope of the variable is limited to the for statement loops - for - III do not do this - for ( int x = 1; x<101; x++); t+=x;

45. OOP in Java : © W. Milner 2005 : Slide 45 Arrays - I elements index  An array is a set of boxes (elements) each labelled with a number (index)  Arrays are declared and created as int [ ] numbers = new int [ 100 ]; which makes an array of 100 integers called numbers  or do it in 2 steps int [ ] numbers; //declare it numbers = new int [ 100 ]; // create it  or initialise it int [ ] numbers = { 4, 2, 1, 3, 5 };  can have arrays of anything

46. OOP in Java : © W. Milner 2005 : Slide 46 Arrays - II  Array elements referred to like numbers [4] = 18;  Multi-dimensional arrays created and used like int [ ] [ ] table = new int [5] [10]; table[3][4]=7;  Array element numbering starts at 0  so int [ ] numbers = new int [ 100 ]; creates 100 elements, from numbers[0] to numbers[99]  array bounds are checked at compile time and run time

47. OOP in Java : © W. Milner 2005 : Slide 47 Arrays - sorting int [ ] numbers = new int [5]; //.. put some numbers in the array, then...sort them // a bubble sort.. for ( int i = 0; i < 5; i++ ) for ( int j = 0; j < 4-i; j++ ) if ( numbers[ j ] > numbers[ j+1] ) {// swap them int temp; temp = numbers[ j ]; numbers[ j ] = numbers[ j+1 ]; numbers[ j+1 ] = temp; };

48. OOP in Java : © W. Milner 2005 : Slide 48 Array exercise  Declare an array of 100 doubles  Fill the array with random numbers (use Math.random(); )  Print them out