Why does this result in 1?

up vote
6
down vote

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I'm trying to figure out why the output of this code results in 1?

The code is as follows:

importjava.util.HashSet;
importjava.util.Set;

public class Test 
 public static void main(String args) 
 Set<Foo> myFooSet= newHashSet<Foo(); 
 myFooSet.add(new Foo(2));
 myFooSet.add(new Foo(1));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(2));
 System.out.print(myFooSet.size());
 

class Foo 
 Integer code;
 Foo(Integer c) 
 code= c;
 
 public boolean equals(Foo f) 
 return false;
 
 public boolean equals(Object f) 
 return true;
 
 public int hashCode() 
 return 17;

I was wondering why does this code result in 1?

some extraa details

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

asked 42 mins ago

Fredrick Norman

391

New contributor

3

Because the hashcode is the same and equals returns true so all 5 elemens are considered equal and replace each other (as you surely know sets don't allow duplicates). Read up on equals() and hashCode() for more information.
â€“Â Thomas
40 mins ago

1

@Thomas actually they don't replace each other. Only the first element is added.
â€“Â Eran
36 mins ago

@Eran ah yes, you're right. I should reread the JavaDoc ("If this set already contains the element, the call leaves the set unchanged and returns false.") - d'oh.
â€“Â Thomas
35 mins ago

add a commentÂ |Â

up vote
6
down vote

favorite

I'm trying to figure out why the output of this code results in 1?

The code is as follows:

importjava.util.HashSet;
importjava.util.Set;

public class Test 
 public static void main(String args) 
 Set<Foo> myFooSet= newHashSet<Foo(); 
 myFooSet.add(new Foo(2));
 myFooSet.add(new Foo(1));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(2));
 System.out.print(myFooSet.size());
 

class Foo 
 Integer code;
 Foo(Integer c) 
 code= c;
 
 public boolean equals(Foo f) 
 return false;
 
 public boolean equals(Object f) 
 return true;
 
 public int hashCode() 
 return 17;

I was wondering why does this code result in 1?

some extraa details

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

asked 42 mins ago

Fredrick Norman

391

New contributor

3

Because the hashcode is the same and equals returns true so all 5 elemens are considered equal and replace each other (as you surely know sets don't allow duplicates). Read up on equals() and hashCode() for more information.
â€“Â Thomas
40 mins ago

1

@Thomas actually they don't replace each other. Only the first element is added.
â€“Â Eran
36 mins ago

@Eran ah yes, you're right. I should reread the JavaDoc ("If this set already contains the element, the call leaves the set unchanged and returns false.") - d'oh.
â€“Â Thomas
35 mins ago

add a commentÂ |Â

up vote
6
down vote

favorite

I'm trying to figure out why the output of this code results in 1?

The code is as follows:

importjava.util.HashSet;
importjava.util.Set;

public class Test 
 public static void main(String args) 
 Set<Foo> myFooSet= newHashSet<Foo(); 
 myFooSet.add(new Foo(2));
 myFooSet.add(new Foo(1));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(2));
 System.out.print(myFooSet.size());
 

class Foo 
 Integer code;
 Foo(Integer c) 
 code= c;
 
 public boolean equals(Foo f) 
 return false;
 
 public boolean equals(Object f) 
 return true;
 
 public int hashCode() 
 return 17;

I was wondering why does this code result in 1?

some extraa details

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

asked 42 mins ago

Fredrick Norman

391

New contributor

I'm trying to figure out why the output of this code results in 1?

The code is as follows:

importjava.util.HashSet;
importjava.util.Set;

public class Test 
 public static void main(String args) 
 Set<Foo> myFooSet= newHashSet<Foo(); 
 myFooSet.add(new Foo(2));
 myFooSet.add(new Foo(1));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(3));
 myFooSet.add(new Foo(2));
 System.out.print(myFooSet.size());
 

class Foo 
 Integer code;
 Foo(Integer c) 
 code= c;
 
 public boolean equals(Foo f) 
 return false;
 
 public boolean equals(Object f) 
 return true;
 
 public int hashCode() 
 return 17;

I was wondering why does this code result in 1?

some extraa details

java hashmap hashset

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

asked 42 mins ago

Fredrick Norman

391

New contributor

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

asked 42 mins ago

Fredrick Norman

391

New contributor

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

edited 41 mins ago

Himanshu Bhardwaj

3,24321030

asked 42 mins ago

Fredrick Norman

391

New contributor

asked 42 mins ago

Fredrick Norman

391

asked 42 mins ago

Fredrick Norman

391

New contributor

Fredrick Norman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

3

Because the hashcode is the same and equals returns true so all 5 elemens are considered equal and replace each other (as you surely know sets don't allow duplicates). Read up on equals() and hashCode() for more information.
â€“Â Thomas
40 mins ago

1

@Thomas actually they don't replace each other. Only the first element is added.
â€“Â Eran
36 mins ago

@Eran ah yes, you're right. I should reread the JavaDoc ("If this set already contains the element, the call leaves the set unchanged and returns false.") - d'oh.
â€“Â Thomas
35 mins ago

add a commentÂ |Â

3

Because the hashcode is the same and equals returns true so all 5 elemens are considered equal and replace each other (as you surely know sets don't allow duplicates). Read up on equals() and hashCode() for more information.
â€“Â Thomas
40 mins ago

1

@Thomas actually they don't replace each other. Only the first element is added.
â€“Â Eran
36 mins ago

@Eran ah yes, you're right. I should reread the JavaDoc ("If this set already contains the element, the call leaves the set unchanged and returns false.") - d'oh.
â€“Â Thomas
35 mins ago

Because the hashcode is the same and equals returns true so all 5 elemens are considered equal and replace each other (as you surely know sets don't allow duplicates). Read up on equals() and hashCode() for more information.
â€“Â Thomas
40 mins ago

@Thomas actually they don't replace each other. Only the first element is added.
â€“Â Eran
36 mins ago

@Eran ah yes, you're right. I should reread the JavaDoc ("If this set already contains the element, the call leaves the set unchanged and returns false.") - d'oh.
â€“Â Thomas
35 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
9
down vote

Your defined public boolean equals(Object f) and public int hashCode() methods in the Foo class that basically say all Foo instances are equal to each other, so only one instance of Foo can be added to any HashSet.

Therefore myFooSet.size() will return 1, regardless of the number of Foo elements you attempted to add to it (as long as you added at least one).

Note: your public boolean equals(Foo f) method is never used by HashSet, since HashSet only uses the equals method declared in Object class - public boolean equals(Object obj) - which you overrode to always return true.

edited 32 mins ago

answered 40 mins ago

Eran

268k35415500

1

The other important detail to make explicit is the equals(Foo) overload is never invoked, because HashMap has no way to know that exists, so equals(Object) is the overload that is invoked. So whilst new Foo(1).equals(new Foo(1)) is false, that's never actually used.
â€“Â Andy Turner
35 mins ago

add a commentÂ |Â

up vote
1
down vote

Adding to @Eran answer, defining equals(Foo f) method isn't overriding the Object.equals(java.lang.Object) method being used when comparing, even if your Object is Foo

public boolean equals(Foo f) {

Isn't called in your code

answered 35 mins ago

user7294900

15.6k92851

add a commentÂ |Â

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2 Answers
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2 Answers
2

active

oldest

votes

up vote
9
down vote

Therefore myFooSet.size() will return 1, regardless of the number of Foo elements you attempted to add to it (as long as you added at least one).

edited 32 mins ago

answered 40 mins ago

Eran

268k35415500

1

The other important detail to make explicit is the equals(Foo) overload is never invoked, because HashMap has no way to know that exists, so equals(Object) is the overload that is invoked. So whilst new Foo(1).equals(new Foo(1)) is false, that's never actually used.
â€“Â Andy Turner
35 mins ago

add a commentÂ |Â

up vote
9
down vote

Therefore myFooSet.size() will return 1, regardless of the number of Foo elements you attempted to add to it (as long as you added at least one).

edited 32 mins ago

answered 40 mins ago

Eran

268k35415500

1

The other important detail to make explicit is the equals(Foo) overload is never invoked, because HashMap has no way to know that exists, so equals(Object) is the overload that is invoked. So whilst new Foo(1).equals(new Foo(1)) is false, that's never actually used.
â€“Â Andy Turner
35 mins ago

add a commentÂ |Â

up vote
9
down vote

Therefore myFooSet.size() will return 1, regardless of the number of Foo elements you attempted to add to it (as long as you added at least one).

edited 32 mins ago

answered 40 mins ago

Eran

268k35415500

Therefore myFooSet.size() will return 1, regardless of the number of Foo elements you attempted to add to it (as long as you added at least one).

edited 32 mins ago

answered 40 mins ago

Eran

268k35415500

edited 32 mins ago

answered 40 mins ago

Eran

268k35415500

answered 40 mins ago

Eran

268k35415500

answered 40 mins ago

Eran

268k35415500

1

The other important detail to make explicit is the equals(Foo) overload is never invoked, because HashMap has no way to know that exists, so equals(Object) is the overload that is invoked. So whilst new Foo(1).equals(new Foo(1)) is false, that's never actually used.
â€“Â Andy Turner
35 mins ago

add a commentÂ |Â

1

The other important detail to make explicit is the equals(Foo) overload is never invoked, because HashMap has no way to know that exists, so equals(Object) is the overload that is invoked. So whilst new Foo(1).equals(new Foo(1)) is false, that's never actually used.
â€“Â Andy Turner
35 mins ago

The other important detail to make explicit is the equals(Foo) overload is never invoked, because HashMap has no way to know that exists, so equals(Object) is the overload that is invoked. So whilst new Foo(1).equals(new Foo(1)) is false, that's never actually used.
â€“Â Andy Turner
35 mins ago

add a commentÂ |Â

up vote
1
down vote

Adding to @Eran answer, defining equals(Foo f) method isn't overriding the Object.equals(java.lang.Object) method being used when comparing, even if your Object is Foo

public boolean equals(Foo f) {

Isn't called in your code

answered 35 mins ago

user7294900

15.6k92851

add a commentÂ |Â

up vote
1
down vote

Adding to @Eran answer, defining equals(Foo f) method isn't overriding the Object.equals(java.lang.Object) method being used when comparing, even if your Object is Foo

public boolean equals(Foo f) {

Isn't called in your code

answered 35 mins ago

user7294900

15.6k92851

add a commentÂ |Â

up vote
1
down vote

Adding to @Eran answer, defining equals(Foo f) method isn't overriding the Object.equals(java.lang.Object) method being used when comparing, even if your Object is Foo

public boolean equals(Foo f) {

Isn't called in your code

answered 35 mins ago

user7294900

15.6k92851

Adding to @Eran answer, defining equals(Foo f) method isn't overriding the Object.equals(java.lang.Object) method being used when comparing, even if your Object is Foo

public boolean equals(Foo f) {

Isn't called in your code

answered 35 mins ago

user7294900

15.6k92851

answered 35 mins ago

user7294900

15.6k92851

answered 35 mins ago

user7294900

15.6k92851

answered 35 mins ago

user7294900

15.6k92851

add a commentÂ |Â

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