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 IIShadowingOverridingHow to override a methodWhy override a method?EqualityThe equals methodCalling an overridden methodSummaryThe EndJan 15, 2019Polymorphism2SignaturesIn any programming language, a signature is what distinguishes one function or method from anotherIn C, every function has to have a different nameIn Java, two methods have to differ in their names or in the number or types of their parametersfoo(int i) and foo(int i, int j) are differentfoo(int i) and foo(int k) are the samefoo(int i, double d) and foo(double d, int i) are differentIn C++, the signature also includes the return typeBut not in Java!3PolymorphismPolymorphism 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 classThere are two kinds of polymorphism:OverloadingTwo or more methods with different signaturesOverridingReplacing 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 thingExample: 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 nameSo 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 methodsYou 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 assignmentsWidening is legalNarrowing 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 callsLegal because parameter transmission is equivalent to assignmentmyPrint(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 callsIllegal because parameter transmission is equivalent to assignmentmyPrint(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 methodclass 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 IYou can “overload” constructors as well as methods:Counter() { count = 0;}Counter(int start) { count = start;}13Multiple constructors IIOne constructor can “call” another constructor in the same class, but there are special rulesYou call the other constructor with the keyword thisThe call must be the very first thing the constructor doesPoint(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 IThe very first thing any constructor does, automatically, is call the default constructor for its superclassclass Foo extends Bar { Foo() { // constructor super(); // invisible call to superclass constructor ...You can replace this with a call to a specific superclass constructorUse the keyword superThis must be the very first thing the constructor doesclass Foo extends Bar { Foo(String name) { // constructor super(name, 5); // explicit call to superclass constructor ...15Superclass construction IIUnless you specify otherwise, every constructor calls the default constructor for its superclassclass 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 constructorclass 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 above16ShadowingThis is called shadowing—name in class Dog shadows name in class Animalclass Animal { String name = "Animal"; public static void main(String args[]) {
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