Compilation of lambda in an interface, do we have a plan ?

[Remi Forax]

On 05/22/2013 12:29 PM, Maurizio Cimadamore wrote:
> I'm not sure as to why this hasn't been fixed by Robert recent changes 
> - Robert?

It seems that Robert as done it's change on jdk8/tl/langtools  and not 
lambda/lambda/langtools,
so the version of javac from jdk8/tl is more recent that the one from 
lambda/lambda/langtools.

>
> I believe the plan is to go static for the first example, and default 
> _private_ for the second.

Ok,
I think that in that case lambda in class should be translated the same way.
Currently, a lambda that capture this but doesn't use super is not 
translated to a static method.

>
> Maurizio

Rémi

>
> On 22/05/13 08:45, Remi Forax wrote:
>> The compiler doesn't like code like this:
>>
>> public interface Foo {
>>   default void bar() {
>>     Runnable r = () -> {
>>       System.out.println(this);
>>     };
>>   }
>> }
>>
>> in that case the lambda in bar can be compiled to a static method.
>>
>> But what if the lambda calls a super method like this:
>> interface I {
>>   default void m() { }
>> }
>>
>> public interface Foo extends I{
>>   default void bar() {
>>     Runnable r = () -> {
>>       I.super.m();
>>     };
>>   }
>> }
>>
>> given that I.super.m() is translated to an invokespecial call and that
>> as far as I know invokespecial needs to be called in an instance method.
>>
>> One possible translation is to translate the lambda to a default method
>> but given that a default method is public, an interface that extends
>> the interface may be able to override the code of the lambda
>> with another lambda declared in the class with the same signature
>> even if the two lambdas are unrelated.
>>
>> does it means that the lambda in interfaces must be numbered knowing
>> the whole hierarchy ?
>>
>> cheers,
>> Rémi
>>
>