Lambda and parameterized ref of an inline type doesn't work well

[Remi Forax]

[move to valhalla-dev]

Hi Srikanth,
This seems to fix the lambda bug but not the one with the enhanced for loop.
Here, the inline type is itself a generics so it should work as usual.

  @__inline__ static class Foo<V> implements Iterable<V> {
    int x = 3;
    
    public Iterator<V> iterator() {
      return null;
    }
  }


  public static void main(String[] args) {
    Foo<String> foo = new Foo<String>();
    for(var s: foo) {
    }
  }


$javac Hello.java 
Hello.java:15: error: for-each not applicable to expression type
    for(var s: foo) {
               ^
  required: array or java.lang.Iterable
  found:    Foo<String>
1 error

regards,
Rémi

----- Mail original -----
> De: "Srikanth" <srikanth.adayapalam at oracle.com>
> À: "amber-dev" <amber-dev at openjdk.java.net>
> Envoyé: Vendredi 8 Mai 2020 15:00:46
> Objet: Re: Lambda and parameterized ref of an inline type doesn't work well

> Hello Remi,
> 
> Thanks for the report - this should be fixed by the commit
> https://github.com/openjdk/valhalla/commit/e586cdbe made for
> https://bugs.openjdk.java.net/browse/JDK-8244559
> 
> With this all three problems reported are solved and should work now.
> 
> (The inline + record synthesized methods is open)
> 
> Srikanth
> 
> On 07/05/20 4:07 am, Remi Forax wrote:
>>    @__inline__ static class Foo<V> {
>>      int x;
>>      Foo(int x) { this.x = x; }
>>    }
>>
>>    static void m(Foo foo) {
>>      System.out.println("inline");
>>    }
>>    static void m(Foo.ref foo) {
>>      System.out.println("ref");
>>    }
>>
>>    public static void main(String[] args) {
>>      List<Foo.ref<Integer>> list = List.of(new Foo<Integer>(3));
>>      list.stream().forEach(e -> m(e));
>>    }
>>
>> I don't even understand the error message when playing with parameterized ref of
>> an inline type.
>>
>> error: cannot select a static class from a parameterized type
>> Hello.java:19: error: incompatible types: incompatible parameter types in lambda
>> expression
>>      list.stream().forEach(e -> m(e));
>>                            ^
>>
>> regards,
> > Rémi