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    <p>Generally Scala is immutable, you usually have to handle mutable cases yourself. Also case classes are immutable by nature and their copy method is generated by the compiler. So yes, they would share the same object reference. This is one of the reasons immutability is nice.</p> <p>Your second question is a bit different. In that case the classes are constructed one after the other in seperate contexts.</p> <p>It's also a good idea to check what's being compiled:</p> <pre><code>&gt;scalac -print test.scala [[syntax trees at end of cleanup]] // test.scala package test { class Foo extends Object { private[this] val list: List = _; &lt;stable&gt; &lt;accessor&gt; def list(): List = Foo.this.list; def &lt;init&gt;(): b.Foo = { Foo.super.&lt;init&gt;(); Foo.this.list = immutable.this.List.apply(scala.this.Predef.wrapIntArray(Array[Int]{1, 2, 3, 4, 5, 6})); () } } } </code></pre> <p>From this we can see Scala creating a new list each time. Changing this to <code>var</code> won't change anything, we can check:</p> <pre><code>&gt;scalac -print test.scala [[syntax trees at end of cleanup]] // test.scala package test { class Foo extends Object { private[this] var list: List = _; &lt;accessor&gt; def list(): List = Foo.this.list; &lt;accessor&gt; def list_=(x$1: List): Unit = Foo.this.list = x$1; def &lt;init&gt;(): b.Foo = { Foo.super.&lt;init&gt;(); Foo.this.list = immutable.this.List.apply(scala.this.Predef.wrapIntArray(Array[Int]{1, 2, 3, 4, 5, 6})); () } } } </code></pre> <p>It only generated a setter method for it (<code>def list_=(x$1: List)</code>).</p> <p>If you would like to reference the same list in both cases then initialize it with an object's default list:</p> <pre><code>object Foo { val DefaultList = List(1,2,3,4,5,6) } class Foo { var list = Foo.DefaultList } </code></pre> <p>Which compiles to the following:</p> <pre><code>&gt;scalac -print test.scala [[syntax trees at end of cleanup]] // test.scala package test { object Foo extends Object { private[this] val DefaultList: List = _; &lt;stable&gt; &lt;accessor&gt; def DefaultList(): List = Foo.this.DefaultList; def &lt;init&gt;(): test.Foo.type = { Foo.super.&lt;init&gt;(); Foo.this.DefaultList = immutable.this.List.apply(scala.this.Predef.wrapIntArray(Array[Int]{1, 2, 3, 4, 5, 6})); () } }; class Foo extends Object { private[this] var list: List = _; &lt;accessor&gt; def list(): List = Foo.this.list; &lt;accessor&gt; def list_=(x$1: List): Unit = Foo.this.list = x$1; def &lt;init&gt;(): test.Foo = { Foo.super.&lt;init&gt;(); Foo.this.list = Foo.DefaultList(); () } } } </code></pre> <p>As you can see the list is only created once and then the reference through the <code>def DefaultList(): List</code> assigned on each Foo class instantiation.</p>
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    1. POHow costly is it to create(or copy) an object?
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    1. USAkos Krivachy
    1. USAkos Krivachy
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    1. COThank for your answer! But I'm not sure what you meant by "same object reference". If the case class has 4 mix-in traits and echo of them has 5 fields, does it have to copy 4*5=20 references when copying? Or only 4 references?(more precisely, when a class mix-in a trait, does it have all fields from the trait, or just a reference to a dummy instance of the trait?)
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      1. USLai Yu-Hsuan
    2. COOk, I see you answered your question :)
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