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Home/ Questions/Q 8426533
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T04:33:22+00:00 2026-06-10T04:33:22+00:00

I’m trying to overload a constructor in a generic scala class but it’s not

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I’m trying to overload a constructor in a generic scala class but it’s not compiling.

Here’s my code:

class V[T](m: Map[T,Double]) {
    def this(dt: Seq[Double]) = this(dt.zipWithIndex.map(_.swap).toMap)
}

And the error messages I get:

ERROR: called constructor's definition must precede calling constructor's definition : line 6

ERROR: overloaded method constructor V with alternatives:   
(dt: Seq[Double])V[T] <and>   (m: Map[T,Double])V[T]  cannot be applied to 
(scala.collection.immutable.Map[Int,Double]) : line 6

As far as I understand constructor overloading in scala, I think I’m following the proper syntax and the restriction that the call to this should precede everything else.

So what am I doing wrong and how can I fix this?

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  1. Editorial Team
    Editorial Team
    2026-06-10T04:33:24+00:00Added an answer on June 10, 2026 at 4:33 am

    With

    def this(dt: Seq[Double]) = this(dt.zipWithIndex.map(_.swap).toMap)
    
    • You’re creating a new map Map[Int,Double]; Int being the type of the index created by zipWithIndex.

    • If T were Int, then you can use the constructor (m:Map[T,Double].

    • However: T is not yet bound to a type since you’re defining the class. Nor will the type matching bind T to Int at this point.

    • Therefore the type matching fails.

    Solutions:

    How to fix it depends on what you’re trying to do.

    • If it were the case that T <: Int, then bounding the type-param with <: Int could resolve your problem; however it seems a bit unlikely that T is a subclass of Int…

    • If it is always true that T : Int, then drop the generic T.

    • If T is to remain generic and unbounded then that leaves you with making a special case for when T : Int; senia’s solution looks good for that.

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