Test Class A. public class A { private int x, y; public int getx() { return x; } public int gety() { return y; }

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Hannah Bryant
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1 Test Class A Two simple exampel classes public class A { private int x, y; public int getx() { return x; public int gety() { return y; public A( int ax, int ay ) { x = ax; y = ay; public String tostring() { return "<" + x + "," + y + ">"; Some Java Concepts 1

2 Test Class B Two simple exampel classes public class B extends A { private int x, y; public int getx() { return x; public int gety() { return y; public B( int ax, int ay, int bx, int by ) { super( ax, ay); x = bx; y = by; public String tostring() { return "<" + super.getx() + "," + super.gety() + "," + x + "," + y + ">"; Some Java Concepts 2

3 Using HashSet First attempt using HashSet import java.util.*; public class CheckHash { public static void main( String[] args ) { Set<A> aset = new HashSet<A>(); A a1 = new A( 1, 2 ); A a2 = new A( 2, 3 ); A a3 = new A( 1, 2 ); aset.add( a1 ); aset.add( a2 ); aset.add( a3 ); System.out.println( aset ); > java CheckHash [<2,3>, <1,2>, <1,2>] Some Java Concepts 3

4 Using HashSet Second attempt using HashSet public class A {... public int hashcode() { return 7 * x + 11 * y; public class B extends A {... public int hashcode() { return super.hashcode() + 13 * x + 17 * y; > java CheckHash [<1,2>, <1,2>, <2,3>] Some Java Concepts 4

5 Using HashSet Third attempt using HashSet public class A {... public boolean equals( Object o ) { if ( o!= null && getclass() == o.getclass() ) { A a = (A) o; return x == a.x && y == a.y; else return false; public class B extends A {... public boolean equals( Object o ) { if ( super.equals( o ) ) { B b = (B) o; return x == b.x && y == b.y; else return false; Some Java Concepts 5

6 Using HashSet Third attempt using HashSet > java CheckHash [<1,2>, <2,3>] Conclusion When defining new types, make sure that equals och hashcode are properly implemented. The default values does not work! The new definitions must satisfy that: x.equals( y ) implies x.hashcode() == y.hashcode() Within hashcode: choose different primes to multiply the non-pointer fields with. For the pointer fields, use the hashcode of the object that pointer refers to. Some Java Concepts 6

7 Check Inconsistancy Check Inconsistancy public static void archeck( A[] aa, A a ) { aa[0] = a; public static void licheck( List<A> al, A a){ al.add( a ); public static void main( String[] args ) { A[] aa = new A[5]; B[] ba = new B[5]; List<A> al = new ArrayList<A>(); List bl = new ArrayList(); A a = new A( 1, 2); B b = new B( 1, 2, 3, 4); archeck( aa, a); archeck( aa, b); archeck( ba, a); licheck( al, a); licheck( al, b); licheck( bl, a); Some Java Concepts 7

8 Compile and Run Check Inconsistancy > javac CheckInconsistancy.java CheckInconsistancy.java:29: listcheck(java.util.list<a>,a) in CheckInconsistancy cannot be applied to (java.util.list,a) listcheck( bl, a); ^ 1 error With this line taken away. > javac CheckInconsistancy.java > java CheckInconsistancy Exception in thread "main" java.lang.arraystoreexception: A at CheckInconsistancy.arrayCheck(CheckInconsistancy.java:6) at CheckInconsistancy.main(CheckInconsistancy.java:25) Some Java Concepts 8

9 Conclusion Check Inconsistancy Since B is a subtype of A, B[] is a subtype of A[] But, eventhough B is a subtype of A, List is NOT a subtype of List<A> So, the generic type array behaves diffrently from the new generic types (introduced from Java 5). The new view of this was introduced in order to avoid runtime checking of types, which is one of the main reasons why the new generic was introduced. Does this influence anything more? Some Java Concepts 9

10 Check generic types CheckGenerics, first attempt public class CheckGeneric<T> { T elem; T[] elema; List<T>[] elemli; public CheckGeneric( T elem1, T elem2 ) { if ( elem1.compareto(elem2) < 0 ) elem = elem1; else elem = elem2; elema = new T[5]; elemli = new List<T>[5]; Some Java Concepts 10

11 CheckGenerics, compile Check generic types > javac CheckGeneric.java CheckGeneric.java:9: cannot find symbol symbol : method compareto(t) location: class java.lang.object if ( elem1.compareto(elem2) < 0 ) ^ CheckGeneric.java:13: generic array creation elema = new T[5]; ^ CheckGeneric.java:14: generic array creation elemli = new List<T>[5]; ^ 3 errors Some Java Concepts 11

12 CheckGenerics, second attempt Check generic types public class CheckGeneric<T extends Comparable<T>> {... public CheckGeneric( T elem1, T elem2 ) { if ( elem1.compareto(elem2) < 0 ) elem = elem1; else elem = elem2; elema = (T[]) new Object[5]; elemli = (List<T>[]) new Object[5]; Some Java Concepts 12

13 CheckGenerics, compile > javac CheckGeneric.java Note: CheckGeneric.java uses unchecked or unsafe operations. Note: Recompile with -Xlint:unchecked for details. > javac -Xlint:unchecked CheckGeneric.java CheckGeneric.java:13: warning: [unchecked] unchecked cast found : java.lang.object[] required: T[] elema = (T[]) new Object[5]; ^ CheckGeneric.java:14: warning: [unchecked] unchecked cast found : java.lang.object[] required: java.util.list<t>[] elemli = (List<T>[]) new Object[5]; ^ 2 warnings Check generic types Some Java Concepts 13

14 CheckGenerics Check generic types public class A implements Comparable<A> {... public int compareto( A a ) { int diff = x - a.x; return diff == 0? y - a.y : diff; Some Java Concepts 14

15 CheckGenerics, test program Check generic types public static void main( String[] args ) { A a1 = new A(1,2), a2 = new A(2,1); CheckGeneric<A> cga = new CheckGeneric<A>(a1,a2); B b1 = new B(1,2,3,4), b2 = new B(4,3,2,1); if (b1.equals(b2) ) System.out.println("Yes"); else System.out.println("No"); CheckGeneric cgb = new CheckGeneric(b1,b2); Some Java Concepts 15

16 CheckGenerics Check generic types > javac CheckGeneric.java CheckGeneric.java:26: type parameter B is not within its bound CheckGeneric cgb = new CheckGeneric(b1,b2); ^ CheckGeneric.java:26: type parameter B is not within its bound CheckGeneric cgb = new CheckGeneric(b1,b2); ^ Note: CheckGeneric.java uses unchecked or unsafe operations. Note: Recompile with -Xlint:unchecked for details. 2 errors Some Java Concepts 16

17 What is? Generic contra arrays The questionmark is usually called wild card. Among other things, it is used to refer to a type of which we do not need to know which particular one we have. We have the classes A and B, and that B extends A. We have learned that for the generic type constructor [], i.e. array, that B[] is a subclass of A[]. We also learned that Set is not a subclass Set<A>. In stead, we have have: Set<?> is a direct supertype to all Set<T>, where T is some concrete typ, e.g. A, B or Integer. E.g. we can declare a variable of type Set<?> and let it refer to any Set Set<?> unknownset = new HashSet<A>();... unknownset = new HashSet(); List<?>[] av = new List<?> [5]; Some Java Concepts 17

18 CheckGenerics Check generic types Furthermore we can let? range only over all superclasses of a type by the super operator. This will solve our earlier problem! public class CheckGeneric<T extends Comparable<? super T >> { Now everything works nice! Some Java Concepts 18

Rules and syntax for inheritance. The boring stuff

Rules and syntax for inheritance The boring stuff The compiler adds a call to super() Unless you explicitly call the constructor of the superclass, using super(), the compiler will add such a call for