Treat S<dynamic> as S<Object>?

[vsmenon]

We've discussed mapping these to the same:

S<T>                =>   S == S<Object>
S<T extends Base>   =>   S == S<Base>

Should we?
@jacob314 @leafpetersen

[sigmundch]

(FYI you can use markdown code blocks to preserve the format, I just went ahead and edited the code above in place :))

[leafpetersen]

I think this is fine. One subtlety that should probably be dealt with is that the following:

class Foo<T extends num, S extends Comparable<T>>
var x = new Foo();

should probably be interpreted as

...
var x = new Foo<num, Comparable<num>>();

[jacob314]

SGTM

[jmesserly]

I think we decided the distinction of dynamic vs Object needs to be preserved at runtime, although it will rarely make a difference. (when T is substituted into function parameter type positions, where dynamic is treated as bottom instead of top).

@leafpetersen @vsmenon thoughts on this one?
S<T extends Base> => S == S<Base>

[vsmenon]

@leafpetersen - is this still an open issue?

[leafpetersen]

I believe this is still open.

[jmesserly]

moved to dart-lang/sdk#25877