Scala-使用可变HashSet时hashCode的行为等于

Question

CASE#1- I have the following class:

class Bear (val aName: String) { 
   def getName: String = aName
   override def equals (a : Any) : Boolean = { println("Test Equals"); true}      
 }

If I run the following code I get the results below:

import scala.collection.mutable.HashSet 
val bear1 = new Bear("Black") 
val bear2 = new Bear("Black") 
val setBears: HashSet[Bear] = HashSet (bear1,bear2) 
println(setBears) 

res: Set(scalaproj.Bear@7d4991ad, scalaproj.Bear@28d93b30)

CASE#2- However, when I add the hashCode method to the class

class Bear (val aName: String) { 
   def getName: String = aName
   override def equals (a : Any) : Boolean = { println("Test Equals"); true}      
   override def hashCode() = { println("Test Hash");100 }
 }

and run the same code I get the results below:

import scala.collection.mutable.HashSet 
val bear1 = new Bear("Black") 
val bear2 = new Bear("Black") 
val setBears: HashSet[Bear] = HashSet (bear1,bear2) 
println(setBears) 

res: Test Hash
     Test Hash
     Test Equals
     Test Hash
     Set(scalaproj.Bear@64)  

My two questions:

1- CASE #2- Why is it that when the hashCode method is defined it is being called three times - twice when adding "bear1" and one time when adding "bear2"

2- CASE #1 - Why is it that the equals method is not called at all when the hashCode is not defined - eventhough it was called in CASE#2. (Even for a case class the overridden equals method is usually called in all cases)

Answer 1

About question #2:

When two objects have the same hashCode , the equals method will be called to settle the 'tie'. Because in this case the hash is different, equals will not be called

About question #1:

In reality what is happening is that when the HashSet is defined, the hash method for bear1 and bear2 are called, because they evaluate to the same value (namely 100), the equals will also be called.

The extra call to the hashCode method is because of the println(setBears) . It shows the hash for each object in the hashset.

So, when creating the HashSet, it adds up to two hashCode calls (one for each bear) and one equals call to settle the tie