PolymorphismSignaturesSlide 3OverloadingWhy overload a method?DRY (Don’t Repeat Yourself)Another reason to overload methodsLegal assignmentsLegal method callsIllegal method callsJava uses the most specific methodMultiple constructors IMultiple constructors IISuperclass construction ISuperclass construction IIShadowingAn aside: NamespacesOverridingHow to override a methodWhy override a method?More about toString()EqualityThe equals methodCalling an overridden methodSummaryThe EndJan 14, 2019Polymorphism2SignaturesIn any programming language, a signature is what distinguishes one function or method from anotherIn C, every function has to have a different nameIn Java, two methods have to differ in their names or in the number or types of their parametersfoo(int i) and foo(int i, int j) are differentfoo(int i) and foo(int k) are the samefoo(int i, double d) and foo(double d, int i) are differentIn C++, the signature also includes the return typeBut not in Java!3PolymorphismPolymorphism means many (poly) shapes (morph)In Java, polymorphism refers to the fact that you can have multiple methods with the same name in the same classThere are two kinds of polymorphism:OverloadingTwo or more methods with different signaturesOverridingReplacing an inherited method with another having the same signature4Overloadingclass Test { public static void main(String args[]) { myPrint(5); myPrint(5.0); } static void myPrint(int i) { System.out.println("int i = " + i); } static void myPrint(double d) { // same name, different parameters System.out.println("double d = " + d); }}int i = 5double d = 5.05Why overload a method?So you can use the same names for methods that do essentially the same thingExample: println(int), println(double), println(boolean), println(String), etc.So you can supply defaults for the parameters: int increment(int amount) { count = count + amount; return count;} int increment() { return increment(1);}Notice that one method can call another of the same nameSo you can supply additional information: void printResults() { System.out.println("total = " + total + ", average = " + average);} void printResult(String message) { System.out.println(message + ": "); printResults();}6DRY (Don’t Repeat Yourself)When you overload a method with another, very similar method, only one of them should do most of the work: void debug() { System.out.println("first = " + first + ", last = " + last); for (int i = first; i <= last; i++) { System.out.print(dictionary[i] + " "); } System.out.println();} void debug(String s) { System.out.println("At checkpoint " + s + ":"); debug();}7Another reason to overload methodsYou may want to do “the same thing” with different kinds of data:class Student extends Person { ... void printInformation() { printPersonalInformation(); printGrades(); }}class Professor extends Person() { ... void printInformation() { printPersonalInformation(); printResearchInterests(); }}Java’s print and println methods are heavily overloaded8Legal assignmentsWidening is legalNarrowing is illegal (unless you cast)class Test { public static void main(String args[]) { double d; int i; d = 5; // legal i = 3.5; // illegal i = (int) 3.5; // legal }}9Legal method callsLegal because parameter transmission is equivalent to assignmentmyPrint(5) is like double d = 5; System.out.println(d);class Test { public static void main(String args[]) { myPrint(5); } static void myPrint(double d) { System.out.println(d); }}5.010Illegal method callsIllegal because parameter transmission is equivalent to assignmentmyPrint(5.0) is like int i = 5.0; System.out.println(i);class Test { public static void main(String args[]) { myPrint(5.0); } static void myPrint(int i) { System.out.println(i); }}myPrint(int) in Test cannot be applied to (double)11Java uses the most specific methodclass Test { public static void main(String args[]) { myPrint(5); myPrint(5.0); } static void myPrint(double d) { System.out.println("double: " + d); } static void myPrint(int i) { System.out.println("int: " + i); }}int:5double: 5.012Multiple constructors IYou can “overload” constructors as well as methods:Counter() { count = 0;}Counter(int start) { count = start;}13Multiple constructors IIOne constructor can “call” another constructor in the same class, but there are special rulesYou call the other constructor with the keyword thisThe call must be the very first thing the constructor doesPoint(int x, int y) { this.x = x; this.y = y; sum = x + y;}Point() { this(0, 0);}A common reason for overloading constructors is (as above) to provide default values for missing parameters14Superclass construction IThe very first thing any constructor does, automatically, is call the default constructor for its superclassclass Foo extends Bar { Foo() { // constructor super(); // invisible call to superclass constructor ...You can replace this with a call to a specific superclass constructorUse the keyword superThis must be the very first thing the constructor doesclass Foo extends Bar { Foo(String name) { // constructor super(name, 5); // explicit call to superclass constructor ...15Superclass construction IIUnless you specify otherwise, every constructor calls the default constructor for its superclassclass Foo extends Bar { Foo() { // constructor super(); // invi sible call to superclass constructor ...You can use this(...) to call another constructor in the same class:class Foo extends Bar { Foo(String message) { // constructor this(message, 0, 0); // your explici t call to another constructor ...You can use super(...) to call a specific superclass constructorclass Foo extends Bar { Foo(String name) { // constructor super(name, 5); // your explicit call to some superclass constructor ...Since the call to another constructor must be the very first thing you do in the constructor, you can only do one of the above16ShadowingThis is called shadowing—name in class Dog shadows name in class Animalclass Animal { String name = "Animal"; public static void