Equality (2.1) Advanced Topics on Classes and Objects. Equality (1) Equality (2.2): Common Error

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1 Advanced Topics on Classes and Objects EECS200 B: Advanced Object Oriented Prograing Fall 208 CHEN-WEI WANG Equality (2.) Iplicitly: Every class is a child/sub class of the Object class. The Object class is the parent/super class of every class. There is a useful accessor ethod that every class inherits fro the Object class: boolean equals(object other) Indicates whether soe other object is equal to this one. The default definition inherited fro Object: boolean equals(object other) { return (this == other); e.g., Say p and p2 are of type PointV without the equals ethod redefined, then p.equals(p2) boils down to (p == p2). Very often when you define new classes, you want to redefine / override the inherited definition of equals. of 55 Equality () Recall that A priitive variable stores a priitive e.g., double d =.5; double d2 =.5; A reference variable stores the address to soe object (rather than storing the object itself) e.g., Point p = new Point(2, ) assigns to p the address of the new Point object e.g., Point p2 = new Point(2, ) assigns to p2 the address of another new Point object The binary operator == ay be applied to copare: Priitive variables: their contents are copared e.g., d == d2 evaluates to true Reference variables: the addresses they store are copared (rather than coparing contents of the objects they refer to) e.g., p == p2 evaluates to false because p and p2 are addresses of different objects, even if their contents are identical. 2 of 55 Equality (2.2): Coon Error int i = 0; int j = 2; boolean saevalue = i.equals(j); Copilation Error: the equals ethod is only applicable to reference types. Fix: write i == j instead. 4 of 55

2 Equality () class PointV { double x; double y; PointV(double x, double y) { this.x = x; this.y = y; PointV p = new PointV(2, ); 2 PointV p2 = new PointV(2, ); Syste.out.println( p == p2 ); /* false */ 4 Syste.out.println( p.equals(p2) ); /* false */ At L4, given that the equals ethod is not explicitly redefined/overridden in class PointV, the default version inherited fro class Object is called. Executing p.equals(p2) boils down to (p == p2). If we wish to copare contents of two PointV objects, need to explicitly redefine/override the equals ethod in that class. 5 of 55 Equality (4.) How do we copare contents rather than addresses? Define the accessor ethod equals, e.g., class PointV2 { double x; double y; public boolean equals (Object obj) { if(this == obj) { return true; if(obj == ) { return false; if(this.getclass()!= obj.getclass()) { return false; PointV2 other = (PointV2) obj; return this.x == other.x && this.y == other.y; String s = "(2, )"; PointV2 p = new PointV2(2, ); PointV2 p2 = new PointV2(2, ); Syste.out.println(p. equals (p)); /* true */ Syste.out.println(p.equals()); /* false */ Syste.out.println(p.equals(s)); /* false */ Syste.out.println(p == p2); /* false */ Syste.out.println(p. equals (p2)); /* true */ of 55 Requireents of equals Given that reference variables x, y, z are not : x.equals() Reflexive : Syetric Transitive of 55 x.equals(x) x.equals(y) y.equals(x) x.equals(y) y.equals(z) x.equals(z) API of equals Inappropriate Def. of equals using hashcode Equality (4.2) When aking a ethod call p.equals(o): Variable p is declared of type PointV2 Variable o can be declared of any type (e.g., PointV2, String) We define p and o as equal if: Either p and o refer to the sae object; Or: o is not. p and o at runtie point to objects of the sae type. The x and y coordinates are the sae. Q: In the equals ethod of Point, why is there no such a line: class PointV2 { boolean equals (Object obj) { if(this == ) { return false; A: If this was, a NullPointerException would have occurred and prevent the body of equals fro being executed. 8 of 55

3 Equality (4.) class PointV2 { 2 boolean equals (Object obj) {... if(this.getclass()!= obj.getclass()) { return false; 4 PointV2 other = (PointV2) obj; 5 return this.x == other.x && this.y == other.y; Object obj at L2 declares a paraeter obj of type Object. PointV2 other at L4 declares a variable p of type PointV2. We call such types declared at copile tie as static type. The list of applicable attributes/ethods that we ay call on a variable depends on its static type. e.g., We ay only call the sall list of ethods defined in Object class on obj, which does not include x and y (specific to Point). If we are SURE that an object s actual type is different fro its static type, then we can cast it. e.g., Given that this.getclass() == obj.getclass(), we are sure that obj is also a Point, so we can cast it to Point. Such cast allows ore attributes/ethods to be called upon 9 of 55 (Point) obj at L5. Equality (.) Exercise: Persons are equal if naes and easures are equal. class Person { 2 String firstnae; String lastnae; double weight; double height; boolean equals (Object obj) { 4 if(this == obj) { return true; 5 if(obj == this.getclass()!= obj.getclass()) { return false; Person other = (Person) obj; 8 return 9 this.weight == other.weight && this.height == other.height 0 && this.firstnae. equals (other.firstnae) && this.lastnae. equals (other.lastnae); Q: At L5, will we get NullPointerException if obj is Null? A: No Short-Circuit Effect of obj is, then obj == evaluates to true no need to evaluate the RHS The left operand obj == acts as a guard constraint for the right operand this.getclass()!= obj.getclass(). of 55 Equality (5) Two notions of equality for variables of reference types: Reference Equality : use == to copare addresses Object Equality : define equals ethod to copare contents PointV2 p = new PointV2(, 4); 2 PointV2 p2 = new PointV2(, 4); PointV2 p = new PointV2(4, 5); 4 Syste.out.println(p == p); /* true */ 5 Syste.out.println(p.equals(p)); /* true */ Syste.out.println(p == p2); /* false */ Syste.out.println(p.equals(p2)); /* true */ 8 Syste.out.println(p2 == p); /* false */ 9 Syste.out.println(p2.equals(p)); /* false */ Being reference-equal iplies being object-equal. Being object-equal does not iply being reference-equal. 0 of 55 Equality (.2) Exercise: Persons are equal if naes and easures are equal. class Person { 2 String firstnae; String lastnae; double weight; double height; boolean equals (Object obj) { 4 if(this == obj) { return true; 5 if(obj == this.getclass()!= obj.getclass()) { return false; Person other = (Person) obj; 8 return 9 this.weight == other.weight && this.height == other.height 0 && this.firstnae. equals (other.firstnae) && this.lastnae. equals (other.lastnae); Q: At L5, if swapping the order of two operands of disjunction: this.getclass()!= obj.getclass() obj == Will we get NullPointerException if obj is Null? A: Yes Evaluation of operands is fro left to right. 2 of 55

4 Equality (.) Exercise: Persons are equal if naes and easures are equal. class Person { 2 String firstnae; String lastnae; double weight; double height; boolean equals (Object obj) { 4 if(this == obj) { return true; 5 if(obj == this.getclass()!= obj.getclass()) { return false; Person other = (Person) obj; 8 return 9 this.weight == other.weight && this.height == other.height 0 && this.firstnae. equals (other.firstnae) && this.lastnae. equals (other.lastnae); L0 & L call equals ethod defined in the String class. When defining equals ethod for your own class, reuse equals ethods defined in other classes wherever possible. of 55 Equality in JUnit (.) assertsae(obj, obj2) Passes if obj and obj2 are references to the sae object asserttrue(obj == obj2) assertfalse(obj!= obj2) PointV p = new PointV(, 4); PointV p2 = new PointV(, 4); PointV p = p; assertsae(p, p); /* pass */ assertsae(p2, p); /* fail */ assertequals(exp, exp2) exp == exp2 if exp and exp2 are priitive type 5 of 55 int i = 0; int j = 20; assertequals(i, j); /* fail */ exp.equals(exp2) if exp and exp2 are reference type Q: What if equals is not explicitly defined in obj s declared type? A: assertsae(obj, obj2) PointV2 p4 = new PointV2(, 4); PointV2 p5 = new PointV2(, 4); assertequals(p4, p5); /* pass */ assertequals(p, p2); /* fail different PointV objects */ assertequals(p4, p2); /* fail different types */ Equality (.4) Person collectors are equal if containing equal lists of persons. class PersonCollector { Person[] persons; int nop; /* nuber of persons */ public PersonCollector() {... public void addperson(person p) {... Redefine/Override the equals ethod in PersonCollector. boolean equals (Object obj) { 2 if(this == obj) { return true; if(obj == this.getclass()!= obj.getclass()) { 4 return false; 5 PersonCollector other = (PersonCollector) obj; boolean equal = false; if(this.nop == other.nop) { 8 equal = true; 9 for(int i = 0; equal && i < this.nop; i ++) { 0 equal = this.persons[i].equals(other.persons[i]); return equal; 2 4 of 55 Equality in JUnit (.2) public void testequalityofpointv() { PointV p = new PointV(, 4); PointV p2 = new PointV(, 4); assertfalse(p == p2); assertfalse(p2 == p); /* assertsae(p, p2); assertsae(p2, p); */ /* both fail */ assertfalse(p.equals(p2)); assertfalse(p2.equals(p)); asserttrue(p.x == p2.x && p2.y == p2.y); public void testequalityofpointv2() { PointV2 p = new PointV2(, 4); PointV2 p4 = new PointV2(, 4); assertfalse(p == p4); assertfalse(p4 == p); /* assertsae(p, p4); assertsae(p4, p4); */ /* both fail */ asserttrue(p.equals(p4)); asserttrue(p4.equals(p)); assertequals(p, p4); assertequals(p4, p); public void testequalityofpointvandpointv2() { PointV p = new PointV(, 4); PointV2 p2 = new PointV2(, 4); /* These two assertions do not copile because p and p2 are of different types. */ /* assertfalse(p == p2); assertfalse(p2 == p); */ /* assertsae can take objects of different types and fail. */ /* assertsae(p, p2); */ /* copiles, but fails */ /* assertsae(p2, p); */ /* copiles, but fails */ /* version of equals fro Object is called */ assertfalse(p.equals(p2)); /* version of equals fro PointP2 is called */ assertfalse(p2.equals(p)); of 55

5 Equality in JUnit (.) public void testpersoncollector() { Person p = new Person("A", "a", 80,.8); Person p2 = new Person("A", "a", 80,.8); Person p = new Person("B", "b", 200, 2.); Person p4 = p; assertfalse(p == p2); asserttrue(p.equals(p2)); asserttrue(p == p4); asserttrue(p.equals(p4)); PersonCollector pc = new PersonCollector(); PersonCollector pc2 = new PersonCollector(); assertfalse(pc == pc2); asserttrue(pc.equals(pc2)); pc.addperson(p); assertfalse(pc.equals(pc2)); pc2.addperson(p2); assertfalse(pc.persons[0] == pc2.persons[0]); asserttrue(pc.persons[0].equals(pc2.persons[0])); asserttrue(pc.equals(pc2)); pc.addperson(p); pc2.addperson(p4); asserttrue(pc.persons[] == pc2.persons[]); asserttrue(pc.persons[].equals(pc2.persons[])); asserttrue(pc.equals(pc2)); pc.addperson(new Person("A", "a", 5,.5)); pc2.addperson(new Person("A", "a", 5,.55)); assertfalse(pc.persons[2] == pc2.persons[2]); assertfalse(pc.persons[2].equals(pc2.persons[2])); assertfalse(pc.equals(pc2)); of 55 Why Ordering Between Objects? (2) class Eployee { int id; double salary; Eployee(int id) { this.id = id; void setsalary(double salary) { this.salary = salary; 2 public void testuncoparableeployees() { Eployee alan = new Eployee(2); 4 Eployee ark = new Eployee(); 5 Eployee to = new Eployee(); Eployee[] es = {alan, ark, to; Arrays.sort(es); 8 Eployee[] expected = {to, alan, ark; 9 assertarrayequals(expected, es); L8 triggers a java.lang.classcastexception: Eployee cannot be cast to java.lang.coparable Arrays.sort expects an array whose eleent type defines a precise ordering of its instances/objects. 9 of 55 Why Ordering Between Objects? () Each eployee has their nuerical id and salary. e.g., (alan, 2, ), (ark,, 450.), (to,, 450.) Proble: To facilitate an annual review on their statuses, we want to arrange the so that ones with saller id s coe before ones with larger id s.s e.g., to, alan, ark Even better, arrange the so that ones with larger salaries coe first; only copare id s for eployees with equal salaries. e.g., alan, to, ark Solution : Define ordering of Eployee objects. [ Coparable interface, copareto ethod ] Use the library ethod Arrays.sort. 8 of 55 Defining Ordering Between Objects (.) class CEployee ipleents Coparable <CEployee> {... /* attributes, constructor, utator siilar to Eployee public int copareto(ceployee e) { return this.id - e.id; Given two CEployee objects ce and ce2: ce.copareto(ce2) > 0 [ ce is greater than ce2 ] ce.copareto(ce2) == 0 [ ce is equal to ce2 ] ce.copareto(ce2) < 0 [ ce is saller than ce2 ] Say ces is an array of CEployee (CEployee[] ces), calling Arrays.sort(ces) re-arranges ces, so that: ces[0] CEployee object 20 of 55 ces[] CEployee object... ces[ces.length - ] CEployee object

6 Defining Ordering Between Objects (.2) public void testcoparableeployees_() { /* * CEployee ipleents the Coparable interface. * Method copareto copares id s only. */ CEployee alan = new CEployee(2); CEployee ark = new CEployee(); CEployee to = new CEployee(); alan.setsalary(4500.4); ark.setsalary(450.); to.setsalary(450.); CEployee[] es = {alan, ark, to; /* When coparing eployees, * their salaries are irrelevant. */ Arrays.sort(es); CEployee[] expected = {to, alan, ark; assertarrayequals(expected, es); 2 of 55 Defining Ordering Between Objects (2.2) Alternatively, we can use extra if stateents to express the logic ore clearly. class CEployee2 ipleents Coparable <CEployee2> { 2... /* attributes, constructor, utator siilar to Eployee 4 public int copareto(ceployee2 other) { 5 if(this.salary > other.salary) { return -; 8 else if (this.salary < other.salary) { 9 return ; 0 else { /* equal salaries */ 2 return this.id - other.id; 4 2 of 55 Defining Ordering Between Objects (2.) Let s now ake the coparison ore sophisticated: Eployees with higher salaries coe before those with lower salaries. When two eployees have sae salary, whoever with lower id coes first. class CEployee2 ipleents Coparable <CEployee2> { 2... /* attributes, constructor, utator siilar to Eployee 4 public int copareto(ceployee2 other) { 5 int salarydiff = Double.copare(this.salary, other.salary); int iddiff = this.id - other.id; if(salarydiff!= 0) { return - salarydiff; 8 else { return iddiff; L5: Double.copare(d, d2) returns -(d < d2), 0 (d == d2), or + (d > d2). L: Why inverting the sign of salarydiff? this.salary > other.salary Double.copare(this.salary, other.salary) > 0 But we should consider eployee with higher salary as saller. We want that eployee to coe before the other one! 22 of 55 Defining Ordering Between Objects (2.) 2 public void testcoparableeployees_2() { /* 4 * CEployee2 ipleents the Coparable interface. 5 * Method copareto first copares salaries, then * copares id s for eployees with equal salaries. */ 8 CEployee2 alan = new CEployee2(2); 9 CEployee2 ark = new CEployee2(); 0 CEployee2 to = new CEployee2(); alan.setsalary(4500.4); 2 ark.setsalary(450.); to.setsalary(450.); 4 CEployee2[] es = {alan, ark, to; 5 Arrays.sort(es); CEployee2[] expected = {alan, to, ark; assertarrayequals(expected, es); 8 24 of 55

7 Defining Ordering Between Objects () When you have your class C ipleent the interface Coparable<C>, you should design the copareto ethod, such that given objects c, c2, c of type C: Asyetric : (c.copareto(c2) < 0 c2.copareto(c) < 0) (c.copareto(c2) > 0 c2.copareto(c) > 0) We don t have c < c2 and c2 < c at the sae tie! Transitive : c.copareto(c2) < 0 c2.copareto(c) < 0 c.copareto(c) < 0 c.copareto(c2) > 0 c2.copareto(c) > 0 c.copareto(c) > 0 We have c < c2 c2 < c c < c Q. How would you define the copareto ethod for the Player class of a rock-paper-scissor gae? [Hint: Transitivity] 25 of 55 Hashing: Arrays are Maps Each array entry is a pair: an object and its nuerical index. e.g., say String[] a = {"A", "B", "C", how any? : (0, "A"), (, "B"), (2, "C") Search s are the set of nuerical index s. The set of index s are unique [e.g., 0.. (a.length )] Given a valid index i, we can Uniquely deterines where the object is [(i + ) th ite] Efficiently retrieves that object [a[i] fast eory access] Maps in general ay have non-nuerical s: Student ID [student record] Social Security Nuber [resident record] Passport Nuber [citizen record] Residential Address [household record] Media Access Control (MAC) Address [PC/Laptop record] Web URL [web page]... 2 of 55 Hashing: What is a Map? A ap (a.k.a. table or dictionary) stores a collection of. entry Map ENTRY (SEARCH) KEY Each entry is a pair: a and its (search). VALUE D 25 C F 4 Z A 9 C Q Each search : Uniquely identifies an object in the ap Should be used to efficiently retrieve the associated Search s ust be unique (i.e., do not contain duplicates). 2 of 55 Hashing: Naive Ipleentation of Map Proble: Support the construction of this siple ap: ENTRY (SEARCH) KEY VALUE D 25 C F 4 Z A 9 C Q Let s just assue that the axiu ap capacity is 00. Naive Solution: Let s understand the expected runtie structures before seeing the Java code! 28 of 55

8 Hashing: Naive Ipleentation of Map (0) After executing = new () : Attribute. initialized as an array of 00 slots. Attribute. is 0, eaning: Current nuber of stored in the ap is 0. Index for storing the next new entry is 0. Hashing: Naive Ipleentation of Map (2) After executing.put(new (25, "C")) : Attribute. has 98 slots. Attribute. is 2, eaning: Current nuber of stored in the ap is 2. Index for storing the next new entry is [0].[] 25 D 29 of 55 of 55 Hashing: Naive Ipleentation of Map () After executing.put(new (, "D")) : Attribute. has slots. Attribute. is, eaning: Current nuber of stored in the ap is. Index for storing the next new entry is. Hashing: Naive Ipleentation of Map () After executing.put(new (, "F")) : Attribute. has 9 slots. Attribute. is, eaning: Current nuber of stored in the ap is. Index for storing the next new entry is [0].[0].[].[2] 25 D D F 0 of 55 2 of 55

9 Hashing: Naive Ipleentation of Map (4) After executing.put(new (4, "Z")) : Attribute. has 9 slots. Attribute. is 4, eaning: Current nuber of stored in the ap is 4. Index for storing the next new entry is 4. Hashing: Naive Ipleentation of Map () After executing.put(new (9, "C")) : Attribute. has 94 slots. Attribute. is, eaning: Current nuber of stored in the ap is. Index for storing the next new entry is [0].[].[2].[].[0].[].[2].[].[4].[5] D F Z D F Z A of 55 5 of 55 Hashing: Naive Ipleentation of Map (5) After executing.put(new (, "A")) : Attribute. has 95 slots. Attribute. is 5, eaning: Current nuber of stored in the ap is 5. Index for storing the next new entry is 5. Hashing: Naive Ipleentation of Map () After executing.put(new (, "Q")) : Attribute. has 9 slots. Attribute. is, eaning: Current nuber of stored in the ap is. Index for storing the next new entry is [0].[].[2].[].[4].[0].[].[2].[].[4].[5].[] D 25 F 4 Z A D 25 F 4 Z A 9 Q 4 of 55 of 55

10 Hashing: Naive Ipleentation of Map (8.) public class { private int ; private String ; public (int, String ) { this. = ; this. = ; /* Getters and Setters for and */ of 55 Hashing: Naive Ipleentation of Map (8.) public void test() { = new (); asserttrue(.size() == 0);.put(, "D");.put(25, "C");.put(, "F");.put(4, "Z");.put(, "A");.put(9, "C");.put(, "Q"); asserttrue(.size() == ); /* inquiries of existing */ asserttrue(.get().equals("d")); asserttrue(.get().equals("q")); /* inquiry of non-existing */ asserttrue(.get() == ); 9 of 55 Hashing: Naive Ipleentation of Map (8.2) public class { private final int MAX_CAPCAITY = 00; private [] ; private int ; /* nuber of */ public () { = new [MAX_CAPCAITY]; = 0; public int size() { return ; public void put(int, String ) { e = new (, ); [] = e; ++; 8 of 55 Required Reading: Point and PointCollector Hashing: Naive Ipleentation of Map (8.4) public class { private final int MAX_CAPCAITY = 00; public String get (int ) { for(int i = 0; i < ; i ++) { e = [i]; int k = e.getkey(); if(k == ) { return e.getvalue(); return ; Say is: {(, D), (25, C), (, F), (4, Z), (, A), (9, C), (, Q),,... How efficient is.get()? [ iteration ] How efficient is.get()? [ iterations ] If is full, worst case of.get(k)? [ 00 iterations ] If with 0, worst case of.get(k)? [ 0 iterations ] get s worst-case perforance is linear on size of.! A uch faster (and correct) solution is possible! 40 of 55

11 Hashing: Hash Table () A k 0 hashing hc(k) A.length - A[hc(k)] Given a (nuerical or non-nuerical) search k: Apply a function hc so that hc(k) returns an integer. We call hc(k) the hash code of k. Value of hc(k) denotes a valid index of soe array A. Rather than searching through array A, go directly to A[ hc(k) ] to get the associated. Both coputations are fast: Converting k to hc(k) Indexing into A[ hc(k) ] 4 of 55 Hashing: Hash Table as a Bucket Array (2.2) For illustration, assue A.length is 0 and hc(k) = k%. hc(k) = k% (SEARCH) KEY VALUE D 25 C F 4 Z A 9 C Q (,D) (25,C) (,F) (4,Z) Collision: unequal s have sae hash code (e.g., 25,, 4) When there are ultiple in the sae bucket, we distinguish between the using their unequal s. 4 of 55 (,A) (9,C) (,Q) Hashing: Hash Table as a Bucket Array (2.) For illustration, assue A.length is and hc(k) = k%. hc(k) = k% (SEARCH) KEY VALUE D 25 C F 4 Z A 9 C Q (,D) (25,C) (,F) (4,Z) Collision: unequal s have sae hash code (e.g., 25,, 4) Unavoidable as nuber of, but a good hash function should have sizes of the buckets uniforly distributed. 42 of 55 (,A) (9,C) (,Q) Hashing: Contract of Hash Function Principle of defining a hash function hc: k.equals(k2) hc(k) == hc(k2) Equal s always have the sae hash code. Equivalently, according to contrapositive: hc(k) hc(k2) k.equals(k2) Different hash codes ust be generated fro unequal s. What if k.equals(k2)? hc(k) == hc(k2) [collision e.g., 25 and ] hc(k) hc(k2) [no collision e.g., 25 and ] What if hc(k) == hc(k2)? k.equals(k2) [collision e.g., 25 and ] k.equals(k2) [sound hash function] 44 of 55 inconsistent hashcode and equals

12 Hashing: Defining Hash Function in Java () The Object class (coon super class of all classes) has the ethod for redefining the hash function for your own class: public class IntegerKey { 2 private int k; public IntegerKey(int k) { this.k = k; 5 public int hashcode() { return k % public boolean equals(object obj) { 8 if(this == obj) { return true; 9 if(obj == ) { return false; 0 if(this.getclass()!= obj.getclass()) { return false; IntegerKey other = (IntegerKey) obj; 2 return this.k == other.k; Q: Can we replace L2 by return this.hashcode() == other.hashcode()? A: No When collision happens, s with sae hash code (i.e., in the sae bucket) cannot be distinguished. 45 of 55 Hashing: Using Hash Table in Java public void testhashtable() { Hashtable<IntegerKey, String> table = new Hashtable<>(); IntegerKey k = new IntegerKey(9); IntegerKey k2 = new IntegerKey(9); asserttrue(k.equals(k2)); asserttrue(k.hashcode() == k2.hashcode()); table.put(k, "D"); asserttrue(table.get(k2).equals("d")); 4 of 55 Hashing: Defining Hash Function in Java (2) public void testcustoizedhashfunction() { IntegerKey ik = new IntegerKey(); /* % == */ asserttrue(ik.hashcode() == ); IntegerKey ik9_ = new IntegerKey(9); /* 9 % == */ IntegerKey ik9_2 = new IntegerKey(9); IntegerKey ik = new IntegerKey(); /* % == */ asserttrue(ik9_.hashcode() == ); asserttrue(ik9_2.hashcode() == ); asserttrue(ik.hashcode() == ); asserttrue(ik9_.hashcode() == ik9_2.hashcode()); asserttrue(ik9_.equals(ik9_2)); Hashing: Defining Hash Function in Java () When you are given instructions as to how the hashcode ethod of a class should be defined, override it anually. Otherwise, use Eclipse to generate the equals and hashcode ethods for you. Right click on the class. Select Source. Select Generate hashcode() and equals(). Select the relevant attributes that will be used to copute the hash. asserttrue(ik9_.hashcode() == ik.hashcode()); assertfalse(ik9_.equals(ik)); 4 of of 55

13 Hashing: Defining Hash Function in Java (4..) Caveat : Always ake sure that the hashcode and equals are redefined/overridden to work together consistently. e.g., Consider an alternative version of the IntegerKey class: public class IntegerKey { private int k; public IntegerKey(int k) { this.k = k; /* hashcode() inherited fro Object NOT overridden. public boolean equals(object obj) { if(this == obj) { return true; if(obj == ) { return false; if(this.getclass()!= obj.getclass()) { return false; IntegerKey other = (IntegerKey) obj; return this.k == other.k; 49 of 55 Hashing: Defining Hash Function in Java (4.2) 2 public void testdefaulthashfunction() { IntegerKey ik9_ = new IntegerKey(9); 4 IntegerKey ik9_2 = new IntegerKey(9); 5 asserttrue(ik9_.equals(ik9_2)); asserttrue(ik9_.hashcode()!= ik9_2.hashcode()); 8 public void testhashtable() { 9 Hashtable<IntegerKey, String> table = new Hashtable<>(); 0 IntegerKey k = new IntegerKey(9); IntegerKey k2 = new IntegerKey(9); 2 asserttrue(k.equals(k2)); asserttrue(k.hashcode()!= k2.hashcode()); 4 table.put(k, "D"); 5 asserttrue(table.get(k2) == ); L, 4, 0, : Default version of hashcode, inherited fro Object, returns a distinct integer for every new object, despite its contents. [ Fix: Override hashcode of your classes! ] 5 of 55 Hashing: Defining Hash Function in Java (4..2) public class IntegerKey { private int k; public IntegerKey(int k) { this.k = k; /* hashcode() inherited fro Object NOT overridden. public boolean equals(object obj) { if(this == obj) { return true; if(obj == ) { return false; if(this.getclass()!= obj.getclass()) { return false; IntegerKey other = (IntegerKey) obj; return this.k == other.k; Proble? Default ipleentation of hashcode() fro the Object class: Objects with distinct addresses have distinct hash code s. Violation of the Contract of hashcode(): hc(k) hc(k2) k.equals(k2) What about equal objects with different addresses? 50 of 55 Index () Equality () Equality (2.) Equality (2.2): Coon Error Equality () Requireents of equals Equality (4.) Equality (4.2) Equality (4.) Equality (5) Equality (.) Equality (.2) Equality (.) Equality (.4) Equality in JUnit (.) 52 of 55

14 Index (2) Equality in JUnit (.2) Equality in JUnit (.) Why Ordering Between Objects? () Why Ordering Between Objects? (2) Defining Ordering Between Objects (.) Defining Ordering Between Objects (.2) Defining Ordering Between Objects (2.) Defining Ordering Between Objects (2.2) Defining Ordering Between Objects (2.) Defining Ordering Between Objects () Hashing: What is a Map? Hashing: Arrays are Maps Hashing: Naive Ipleentation of Map Hashing: Naive Ipleentation of Map (0) 5 of 55 Index (4) Hashing: Contract of Hash Function Hashing: Defining Hash Function in Java () Hashing: Defining Hash Function in Java (2) Hashing: Using Hash Table in Java Hashing: Defining Hash Function in Java () Hashing: Defining Hash Function in Java (4..) Hashing: Defining Hash Function in Java (4..2) Hashing: Defining Hash Function in Java (4.2) 55 of 55 Index () Hashing: Naive Ipleentation of Map () Hashing: Naive Ipleentation of Map (2) Hashing: Naive Ipleentation of Map () Hashing: Naive Ipleentation of Map (4) Hashing: Naive Ipleentation of Map (5) Hashing: Naive Ipleentation of Map () Hashing: Naive Ipleentation of Map () Hashing: Naive Ipleentation of Map (8.) Hashing: Naive Ipleentation of Map (8.2) Hashing: Naive Ipleentation of Map (8.) Hashing: Naive Ipleentation of Map (8.4) Hashing: Hash Table () Hashing: Hash Table as a Bucket Array (2.) Hashing: Hash Table as a Bucket Array (2.2) 54 of 55