Java generics

Agenda
 Generic Methods
 Generic Classes
 Generics and Inheritance
 Erasure
1

Stack interfaces
interface StringStack{
void push(String s);
String pop();
}
interface IntegerStack{
void push(Integer s);
Integer pop();
}
interface StudentStack{...
3

Sort Method
static void sort(Integer[] array) {
// ...
}
static void sort(Double[] array) {
// ...
}
static void sort(String[] array) {
// ...
}
static void sort(Student[] array){
// ...
}
4

The Problem
 What is wrong with these examples?
 Code redundancy
 No effective code reuse
 Solution?
 Using Object class
 Compile-time type safety
5

The Solution
 Generic types and methods
 Methods with similar implementation
 Applicable for different parameters
6

Generic Methods
 Declaring a method which accepts different parameter
types
 For each method invocation, the compiler searches the
appropriate method
 If the compiler does not find a method, it looks for a
compatible generic method
7
Type Parameter
It says: In this method, E is not a regular type,
it is a generic one

Benefits of Generics
public static < E extends Number>
void printArray( E[] inputArray ){…}
 Restricting possible types
 Compile-time type checking
 printArray(stringArray) brings
 Compiler Error
 or exception?
9

Type parameter as the Return Type
10

Stack Generic Interface
interface Stack<T>{
void push(T s);
T pop();
}
Stack<String> stringStack = new ...
stringStack.push(“salam”);
String s = stringStack.pop();
11

public class Stack<E > {
private E[] elements ;
private final int size; // number of elements in the stack
private int top; // location of the top element
public void push(E pushValue) {
if (top == size - 1) // if stack is full
throw new FullStackException();
elements[++top] = pushValue;
}
public E pop() {
if (top == -1) // if stack is empty
throw new EmptyStackException();
return elements[top--];
}
public Stack() {
size = 10;
top = -1;
elements = (E[]) new Object[size];
}
}
12
A note, later….

Using Stack Class
Stack<String> stack1 = new Stack<String>();
stack1.push("first");
stack1.push("second");
System.out.println(stack1.pop());
Stack<Integer> stack2 = new Stack<Integer>();
stack2.push(1);
stack2.push(2);
13

Compile-time Type Checking
Stack<String> stack1 = new Stack<String>();
stack1.push(new Integer(2));
 Compile-time error
14

public class Stack<E extends Student> {
private E[] elements ;
private final int size; // number of elements in the stack
private int top; // location of the top element
public void push(E pushValue) {
if (top == size - 1) // if stack is full
throw new FullStackException();
elements[++top] = pushValue;
}
public E pop() {
if (top == -1) // if stack is empty
throw new EmptyStackException();
return elements[top--];
}
public Stack() {
size = 10;
top = -1;
elements = (E[]) new Student[size];
}
} 15
A note, later….

Raw Types
 Generic classes and methods can be used without type
parameter
 Stack<String> s = new Stack<String>();
 String as type parameter
 s.push(“salam”);
 s.push(new Integer(12));  Compiler Error
 Stack objectStack = new Stack();
 no type parameter
 s.push(“salam”);
 s.push(new Integer(12));
 s.push(new Student(“Ali Alavi”));
16

No Generics in Runtime
 Generics is a compile-time aspect
 In runtime, there is no generic information
 All generic classes and methods are translated with
raw types
 Byte code has no information about generics
 Only raw types in byte code
 This mechanism is named erasure
17

Erasure
 When the compiler translates generic method into
Java bytecodes
 It removes the type parameter section
 It replaces the type parameters with actual types.
 This process is known as erasure
18

Erasure Example (1)
class Stack<T>{
void push(T s){...}
T pop() {...}
}
 Is translated to
class Stack {
void push(Object s){...}
Object pop() {...}
}
19

Erasure Example (2)
 Translated to
20

What Happens if…
public static <E extends Number> void f(E i){
}
public static void f(Number i){
}
 Compiler Error : Method f(Number) has the same
erasure f(Number) as another method in this type
21

Generics and Inheritance
 A non-generic class can be inherited by a non-generic class
 As we saw before learning generics
 A generic class can be inherited from a non-generic class
 Adding generality to classes
 A non-generic class can be inherited from a generic class
 Removing generality
 A generic class can be inherited by a generic class
22

class GenericList<T> extends Object{
public void add(T t){...}
public T get(int i) {...}
public void remove(int i) {...}
}
class GenericNumericList<T extends Number>
extends GenericList<T>{
}
class NonZeroIntegerList
extends GenericList<Integer>{
public void add(Integer t) {
if(t==null || t==0)
throw new RuntimeException(“Bad value");
super.add(t);
}
}
23

Some Notes
 We can also create generic interfaces
interface Stack<T>{
void push(T s);
T pop();
}
 No primitives as type parameters
24

Multiple Type Parameters
class MultipleType<T,K>{
private T t;
public T getT() {
return t;
}
public void setT(T t) {
this.t = t;
}
public void doSomthing(K k, T t){…}
}
MultipleType<String, Integer> multiple =
new MultipleType<String, Integer>();
multiple.doSomthing(5, "123");
25

Note
 You can not instantiate generic classes
class Stack<T>{
T ref = new T();
}
 Syntax Error: Cannot instantiate the type T
 Why?
26

Note (2)
 You can not instantiate generic classes
class Stack<T>{
T[] elements = new T[size];
}
 Syntax Error: Cannot instantiate the type T
 Why?
27

Note (3)
 You cannot create a generic array
class Box<T> {
final T x;
Box(T x) {
this.x = x;
}
}
 Then, this line brings a compile error:
Box<String>[] bsa = new Box<String>[3];
 Why?
28
Syntax Error:
Cannot create a generic array of Box<String>

Reason
 Operations such as instanceof and new are runtime
operations
 They use a type at runtime
 With erasure type information is removed at runtime
 So these operations are Meaningless
 Although, they may be possible
29
T ref = new T();  impossible
• which constructor?
T[] elements = new T[size];  Meaningless
Box<String>[] bsa = new Box<String>[3];  Meaningless

Generics and Java 7
 Older versions:
ArrayList<String> list = new ArrayList<String>();
 With Java 7:
ArrayList<String> list = new ArrayList<>();
 Type information after new are ignored.
List<Map<Long, Set<Integer>>> list = new ArrayList<>();
30

Further Reading
 Wildcards as type parameters
 Java generics vs. C++ templates
 Erasure is different in these languages
 Type Argument inference
 More on erasure
 TIJ is so better than Deitel in generics chapter
 More Depth
31

Wow!!!
public static void wow(ArrayList<String> list) {
Method method = list.getClass().getMethod("add",
Object.class);
method.invoke(list, new Integer(2));
}
public static void main(String args[]) {
ArrayList<String> s = new ArrayList<String>();
wow(s);
for (Object string : s) {
System.out.println(string);
}
}
32

A Note on Inheritance
class A{
public Object f(Object o){
return new Object();
}
}
class B extends A{
return new String(“hello");
}
}
 B.f() overrides A.f()
33

class A{
}
}
class B extends A{
public String f(Object o){
return new String("salam");
}
}
 B.f() overrides A.f()
34

class A{
}
}
class B extends A{
public Object f(String o){
return new String("salam");
}
}
 B.f() is overloading A.f()
 B.f() does not override A.f()
35

Pair class (Quiz)
 Pair<T, K>
 equals
 toString
36

class Pair<T,K>{
private T first;
private K second;
public Pair(T t, K k) {
this.first = t;
this.second = k;
}
public T getFirst() {
return first;
}
public K getSecond() {
return second;
}
public String toString() {
return "[" + second + ", " + first + "]";
}
}
37

Pair<Integer, String> pair1 =
new Pair<Integer, String>(4, "Ali");
Integer i = pair1.getFirst();
String s = pair1.getSecond();
Pair<String, Boolean> pair2 =
new Pair<String, Boolean>("salam", true);
String ss = pair2.getFirst();
Boolean bb = pair2.getSecond();
38

equals() method
public boolean equals(Pair<T,K> pair) {
return
pair.first.equals(first)
&&
pair.second.equals(second);
}
 What is wrong with this implementation?
39

boolean equals(Pair<T,K> pair)
 It should check for nullity of pair
 It should check for nullity of pair.first and pair.second
 It should check for nullity of this.first and this.second
 This method does not override equals()
 It is overloading it
 Correct signature:
boolean equals(Object pair)
 What if parameter is not a Pair?
40

Type Checking
public boolean equals(Object o) {
Pair<T, K> pair = null;
try{
pair = (Pair<T, K>) o;
}catch(ClassCastException e){
return false;
}
return pair.first.equals(first) &&
pair.second.equals(second);
}
41

Java generics

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