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Suppose I have next traits:

trait A { val a: String = "a" } trait B { def a: String = "b" }

And I want to mix both of these traits into some class C

class C extends B with A

Compiler doesn't allow me to create such class because I have to override method a

I want to override it using for example only A's implementaion. How can I do this?

EDIT

scala> class C extends B with A { | override val a = super.a | } <console>:10: error: super may be not be used on value a override val a = super.a ^
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2 Answers 2

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The compiler can't possibly know which one you intend to use, therefore you must specify it like:

class C extends B with A { override def a = super[A].a }

This approach allows you to choose the parent directly, regardless the trait order.

However, the traits define a differently (val and def) thus you must choose only one. You should use either def or val in both traits (not mix them).

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2 Comments

this will not compile
Have you unified def and val in traits? What scala version do you use?
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Provided you make a a def in the A trait, you can do

class C extends B with A { override val a = super.a } val c = new C c.a // "a"

This will work because A is extended after B, so super will be its implementation.

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4 Comments

this will not compile
@maks, read my answer carefully. You need to make a a def in A.
in my question I provided just an example. In reality A and B is a library code
you should specify it in the question then. This makes a big difference and you cannot refer to a super value from a subclass.

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