deduplicating lambda methods

[Tagir Valeev]

Hello!

By the way how types are matched? I assume that you won't merge `s -> s`
> and `x -> x`
> if `s` is String and `x` is an int, right? However merging two lambdas like
> `s -> true` and `sb -> true`
> where `s` is a String and `sb` is a StringBuilder seems possible (creating
> a synthetic method which
> receives an Object).
>
Yes, if parameters are unused, then I'd expect that an AST diff will just
say that the body is the same for both.


What if unused parameter have different primitive type or primitive and
object type? Will we tolerate unnecessary boxing? Well, it's possible just
to drop an unused parameter from synthetic method declaration, though this
might require changes in LMF. Also in this case we will not see the
variable in debugger.

This reminds me about (postponed?) lambda leftovers JEP. Probably if that
JEP will be implemented, duplication should ignore underscore parameters
only. In this case absence in debugger is not a problem.

Tagir.



Also two lambdas like `list -> list.size()` can be
> merged if the parameter types
> differ in type parameters (e.g. List<String> and List<Integer>), but have
> the same erasure.
>
Yes, the diff should work on erased types, which is what really matters in
terms of bytecode generation. There are of course caveat - if you call
List::get instead of List::size, then you can get extra synthetic casts,
and, in that case, you need to avoid deduplication so that each lambda can
have its own synthetic cast.

Maurizio


> Tagir.
>
> On Mon, Mar 5, 2018 at 10:59 AM, Liam Miller-Cushon <cushon at google.com>
> wrote:
>
> On Fri, Mar 2, 2018 at 8:01 PM, Vicente Romero <vicente.romero at oracle.com>
>> wrote:
>>
>> 1) How to identifying duplicates: I have a prototype that runs during
>>>
>>>> lambda desugaring and identifies duplicates by diffing ASTs. Is that the
>>>> best place for deduplication, or it worth considering comparing
>>>>
>>> generated
>>
>>> code instead of ASTs?
>>>>
>>>> are you doing an exact diff? I assume that we want: s -> s to be equal
>>> to
>>> z -> z provided that the target is the same
>>>
>>
>> Lambda parameters are currently handled as a special case. The diff tests
>> that the syntax for the two trees is the same, and that the symbols
>> associated with identifier and select nodes are equivalent. It accepts
>> symbols that correspond to the same parameter in each of the associated
>> lambda methods.
>>
>> E.g. when diffing the lambda bodies for `s -> s` and `x -> x` it sees that
>> `s` and `x` are both the first parameter of their enclosing lambdas.
>>
>>