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author | Paul Phillips <paulp@improving.org> | 2013-05-20 21:56:14 -0700 |
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committer | Paul Phillips <paulp@improving.org> | 2013-05-21 01:18:20 -0700 |
commit | a2ad291acd159ed299eb30305e42a0bace95f02a (patch) | |
tree | 63dc189dea7b1096232da0a4e0d0edfdf676578f /build.number | |
parent | 085b4d9bdb7ba9f9fe00c63e998e93278a34b161 (diff) | |
download | scala-a2ad291acd159ed299eb30305e42a0bace95f02a.tar.gz scala-a2ad291acd159ed299eb30305e42a0bace95f02a.tar.bz2 scala-a2ad291acd159ed299eb30305e42a0bace95f02a.zip |
SI-3425 erasure crash with refinement members.
Checking that a refinement class symbol does not override
any symbols does mean it will have to be invoke reflectively;
but the converse is not true. It can override other symbols
and still have to be called reflectively, because the
overridden symbols may also be defined in refinement classes.
scala> class Foo { type R1 <: { def x: Any } ; type R2 <: R1 { def x: Int } }
defined class Foo
scala> typeOf[Foo].member(TypeName("R2")).info.member("x": TermName).overrideChain
res1: List[$r.intp.global.Symbol] = List(method x, method x)
scala> res1 filterNot (_.owner.isRefinementClass)
res2: List[$r.intp.global.Symbol] = List()
And checking that "owner.info decl name == this" only works if
name is not overloaded.
So the logic is all in "isOnlyRefinementMember" now, and
let's hope that suffices for a while.
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