I have edited this to a simpler form of the question to which @Zhi Yuan Wang responded :
object ContBound {
def f2[A: Seq, B: Seq]: Unit = {
val a1: Seq[A] = evidence$1
val b2: Seq[B] = evidence$2
}
def f3[A: Seq, B: Seq, C: Seq]: Unit = {
val a1: Seq[A] = evidence$1
val b2: Seq[B] = evidence$2
val a3: Seq[C] = evidence$3
}
}
I get the following errors:
not found value evidence$1
not found value evidence$2
type mismatch; found :Seq[A] required: Seq[C]
despite getting the following in the REPL:
def f3[A: Seq, B: Seq, C: Seq]: Unit =
| {
| val a1: Seq[A] = evidence$1
| val b2: Seq[B] = evidence$2
| val a3: Seq[C] = evidence$3
| }
f3: [A, B, C](implicit evidence$1: Seq[A], implicit evidence$2: Seq[B], implicit evidence$3: Seq[C])Unit
Zhi's awnser is correct. The following compiles:
object ContBound {
def f2[A: Seq, B: Seq]: Unit = {
val a1: Seq[A] = evidence$1
val b2: Seq[B] = evidence$2
}
def f3[A: Seq, B: Seq, C: Seq]: Unit = {
val a1: Seq[A] = evidence$3
val b2: Seq[B] = evidence$4
val a3: Seq[C] = evidence$5
}
}
However I still don't see this as correct behaviour, as these are parameters for two different methods and methods are normally allowed to reuse parameter names.
Have you tried
def comma3[A: RParse, B: RParse, C: RParse, D](f: (A, B, C) => D): D =
expr match {
case CommaExpr(Seq(e1, e2, e3)) =>
f(evidence$3.get(e1), evidence$4.get(e2), evidence$5.get(e3))
}
since the the evidence$1 already used by
def comma3[]??