Need some help with Scala's instance variables

Question

Assume this Java code:

public class A {
   public A(String g) {
      x += g.length();
   }

   private int x = 0;
}

If I create an instance of A, like this:

A a = new A("geo");

after this call, the value of x will be 3. What am I doing wrong in my Scala code?

class A(val g:String) {
  x += g.length
  var x:Int = 0
}

object x extends Application {
  val x = new A("geo")
  println(x.x)
}

This prints 0. I assumed that when the compiler reaches the var x:Int = 0, the body of the main constructor has ended. Am I wrong? How else could you declare instance variables in Scala ( assuming I don't want them in my constructor ) ?

Every line of code that is not inside a def (method body) is constructor code. Val and var initialization code is included in that and the constructor is the top-to-bottom accumulation of all that code. Note that this includes things like method calls interspersed between val, var and def definitions. — Randall Schulz
– Randall Schulz, Commented Dec 16, 2009 at 21:37

Mitch Blevins · Accepted Answer · 2009-12-16 22:47:42Z

7

Keep in mind that your code translates into something similar (but not exactly) to this:

public class A {

  private final String g;
  private int x;

  public A(String g) {
    this.g = g;
    x_$eq(x() + g.length());
    x = 0;
  }

  public void x_$eq(int arg0) {
    x = arg0;
  }

  public int x() {
    return x;
  }

  public String g() {
    return g;
  }
}

But vars defined in (non-constructor) methods get translated into actual local variables.

Not sure if this explains the reasoning so much as highlights one of the differences.

EDIT - Changed "translation" from scala to java for clarity and ability to more accurately represent what is happening.

edited Dec 16, 2009 at 22:47

answered Dec 16, 2009 at 21:34

Mitch Blevins

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3 Comments

Geo Over a year ago

So, why wasn't _x 3? That's how I'm reading the code representation.

Geo Over a year ago

But how is x treated? As a local variable? Is the compiler declaring it?

faran Over a year ago

The var x is in instance variable in all cases. With the declaration var x: Int = 0, the variable x is a private variable with public x() and x_=() methods automatically created.

Daniel C. Sobral · Accepted Answer · 2009-12-17 11:19:42Z

7

Your confusion results from a misunderstanding of how constructors in Scala works. Specifically, let's translate the Scala code you posted into Java:

class A(val g:String) {
  x += g.length
  var x:Int = 0
}

becomes

public class A {
   public A(String g) {
      x += g.length();
      x = 0;
   }
   private int x;
}

The reason is simple. The whole body of a class in Scala is the primary constructor for that class. That means the statements in it, and initializing val and var are statements, will be executed in the order they are found.

PS: Here is the actual, true rendition of that code.

Scala 2.7

C:\Users\Daniel\Documents\Scala\Programas> scalac -print A.scala
[[syntax trees at end of cleanup]]// Scala source: A.scala
package <empty> {
  class A extends java.lang.Object with ScalaObject {
    @remote def $tag(): Int = scala.ScalaObject$class.$tag(A.this);
    <paramaccessor> private[this] val g: java.lang.String = _;
    <stable> <accessor> <paramaccessor> def g(): java.lang.String = A.this.g;
    private[this] var x: Int = _;
    <accessor> def x(): Int = A.this.x;
    <accessor> def x_=(x$1: Int): Unit = A.this.x = x$1;
    def this(g: java.lang.String): A = {
      A.this.g = g;
      A.super.this();
      A.this.x_=(A.this.x().+(g.length()));
      A.this.x = 0;
      ()
    }
  }
}

Scala 2.8

C:\Users\Daniel\Documents\Scala\Programas>scalac -print A.scala
[[syntax trees at end of cleanup]]// Scala source: A.scala
package <empty> {
  class A extends java.lang.Object with ScalaObject {
    <paramaccessor> private[this] val g: java.lang.String = _;
    <stable> <accessor> <paramaccessor> def g(): java.lang.String = A.this.g;
    private[this] var x: Int = _;
    <accessor> def x(): Int = A.this.x;
    <accessor> def x_=(x$1: Int): Unit = A.this.x = x$1;
    def this(g: java.lang.String): A = {
      A.this.g = g;
      A.super.this();
      A.this.x_=(A.this.x().+(g.length()));
      A.this.x = 0;
      ()
    }
  }
}

edited Dec 17, 2009 at 11:19

answered Dec 16, 2009 at 22:37

Daniel C. Sobral

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3 Comments

Mitch Blevins Over a year ago

Good to see you have as much difficulty as me free-handing java without slipping in scala-isms. You missed a semi-colon and used the non-existent-in-java underscore initializer.

Daniel C. Sobral Over a year ago

Ok, code fixed, and compiled to make sure. Why can't Java be reasonable? :-)

Daniel C. Sobral Over a year ago

Ok, I fixed the code, though I see you offered a superior answer (ie, a more accurate translation into Java).

Jim Barrows · Accepted Answer · 2009-12-16 19:52:11Z

4

Change this:

class A(val g:String) {
  x += g.length
  var x:Int = 0
}

to

class A(val g:String) {
  var x = g.length

}

answered Dec 16, 2009 at 19:52

Jim Barrows

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1 Comment

Alan Krueger Over a year ago

Why not just make it a val? There's no evidence from the sample code that it needs to be modified.

user233238 · Accepted Answer · 2009-12-16 20:07:47Z

4

var x:Int = 0 make this the first line of the constructor

class A(val g:String) {  
   var x:Int = 0  
   x += g.length  
}

edited Dec 16, 2009 at 20:07

answered Dec 16, 2009 at 20:00

user233238

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Comments

Ken Bloom · Accepted Answer · 2009-12-18 02:22:40Z

0

Why does Scala let you reference x before you declared it? In any other scope, this would be illegal.

scala> def foo(g:String) = { x+=1; var x=0; x}
<console>:4: error: forward reference extends over definition of variable x
       def foo(g:String) = { x+=1; var x=0; x}

answered Dec 18, 2009 at 2:22

Ken Bloom

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