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Fix #3433: Fix determining typeParams for structural type members #3733

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Merged
merged 1 commit into from
Feb 2, 2018
+5 −0
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Fix #3433: Fix determining typeParams for structural type members #3733

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1 change: 1 addition & 0 deletions compiler/src/dotty/tools/dotc/core/TypeApplications.scala
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Original file line number Diff line number Diff line change
Expand Up @@ -172,6 +172,7 @@ class TypeApplications(val self: Type) extends AnyVal {
case self: TypeRef =>
val tsym = self.symbol
if (tsym.isClass) tsym.typeParams
else if (!tsym.exists) self.info.typeParams
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.How come we don't have the symbol for the refinement member here? That seems likely to be problematic in a lot of other cases, for example in the IDE how do we do "go to definition" if the tree has no symbol?

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Refinements of types that do not exist in the baseclass don't have a symbol. Which symbol could they have?

else if (!tsym.isCompleting) tsym.info.typeParams
else Nil
case self: AppliedType =>
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4 changes: 4 additions & 0 deletions tests/pos/i3433.scala
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@@ -0,0 +1,4 @@
object Test {
type Const[t] = { type L[x] = t }
type Bar[a] = Const[Int]#L[a]
}