deduplicating lambda methods

[Maurizio Cimadamore]

On 05/03/18 07:51, Tagir Valeev wrote:
> Hello!
>
>>   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.
>
> 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.

> 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.
>>