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Incorrect compiler error message overriding java raw type with a recursive type bound #10260

jeantil opened this issue Apr 11, 2017 · 1 comment


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commented Apr 11, 2017

Given the following java code:

public class A<T extends A<T>> {}
interface IA {
    void foo(A a);

when trying to implement the interface from scala with

class IAImpl extends IA{
  override def foo(a: A[_]): Unit = ???

then the scala compiler (2.10.6, 2.11.8, 2.12.1) fails with the following message:

class IAImpl needs to be abstract, since method foo in trait IA of type (a: A)Unit is not defined
(Note that A does not match A[_]. To implement a raw type, use A[_])
class IAImpl extends IA{

Which is fairly confusing since the recommendation is the same as the actual error.

After discussing this on the scala user's board it appears the correct syntax to override such a method is

class IAImpl extends IA { 
  override def foo(a: A[X] forSome {type X <: A[X]}): Unit = ??? 

Maybe the compiler could detect that and print a better error message ?

Jasper-M added a commit to Jasper-M/scala that referenced this issue May 9, 2017
Display raw type as existential in error message
Before, when implementing a raw type in a method override in the wrong way, the error message always suggested a simple wildcard type.
This commit changes that error message to always suggest an existential type that the user can copy paste into his code.
To that end `typeParamsToExistentials` is changed to preserve the names of the type parameters instead of changing them to "?0".."?N".

fixes scala/bug#10260

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commented Sep 18, 2017

thank you !!

@adriaanm adriaanm added this to the 2.12.4 milestone Sep 18, 2017

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