primary constructor, values with underscore

Question

A try this (in Scala 2.10.3) :

scala> class A(var a: Int = _)
<console>:1: error: unbound placeholder parameter
       class A(var a: Int = _)
                            ^

What is the problem?, the underscore for default value-type can not be used in a primary constructor?

Revan · Accepted Answer · 2013-12-02 19:50:00Z

2

The underscore is used for placeholder of the default value of the type (see Scala language specification). I am not really sure why you would want to do this type of initialization instead of: class A(var a: Int = 0) as it is likely to be more concise and other coders will know immediately the default value.

answered Dec 2, 2013 at 19:50

Revan

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5 Comments

user3058661 Over a year ago

Only testing for study the language, my question is, if a try this: scala> class A { var a: Int = _ } defined class A, then there is no problem with it, because the value of a constructor is a field for the class or not?.

Revan Over a year ago

When using curly braces, you're defining the body of the class. With { var a: Int _ } you've created an instance variable, a which is initialized to the default value of Int. When doing class A(var a: Int = _) you're asking scala to create an new class A which accepts an Int in it's constructor argument.

user3058661 Over a year ago

Ok, I thought the value in a constructor was created like a field so this would allow use underscore for default value :( , Thanks for you time.

Revan Over a year ago

You are correct, scala will provide a getter/setter for that field. For instance, class A(val myInt: Int) can later be accessed externally via: val a = new A(4); println(a.myInt). Without the val/var in the constructor argument that parameter will not be a field of the class.

Shannon Over a year ago

Seems inconsistent that Scala programmers should only sometimes use "_" to represent the default value of a type. Anybody know why Scala language designers decided to do it this way?

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