Scala 类型不匹配错误尝试将参数化或抽象类型构造函数视为其上限
精简示例：
trait WithUpperBounds[AA[_] <: Set[_], Node, Edge] {
val nodes: AA[Node]
val edges: AA[Edge]
}
class WithoutUpperBounds[AA[_] <: Set[_], Node, Edge](
val nodes: AA[Node],
val edges: AA[Edge]
) extends WithUpperBounds[AA, Node, Edge] {
val nodes2Set: Set[Node] = nodes
val edges2Set: Set[Edge] = edges
}
REPL 输出：
scala> :l …/Sandbox.scala
Loading …/Sandbox.scala...
defined trait WithUpperBounds
<console>:10: error: type mismatch;
found : AA[Node]
required: Set[Node]
val nodes2Set: Set[Node] = nodes
^
<console>:11: error: type mismatch;
found : AA[Edge]
required: Set[Edge]
val edges2Set: Set[Edge] = edges
^
高阶函数，特别是对于理解，使我的问题更加复杂。
Stripped down example:
trait WithUpperBounds[AA[_] <: Set[_], Node, Edge] {
val nodes: AA[Node]
val edges: AA[Edge]
}
class WithoutUpperBounds[AA[_] <: Set[_], Node, Edge](
val nodes: AA[Node],
val edges: AA[Edge]
) extends WithUpperBounds[AA, Node, Edge] {
val nodes2Set: Set[Node] = nodes
val edges2Set: Set[Edge] = edges
}
REPL output:
scala> :l …/Sandbox.scala
Loading …/Sandbox.scala...
defined trait WithUpperBounds
<console>:10: error: type mismatch;
found : AA[Node]
required: Set[Node]
val nodes2Set: Set[Node] = nodes
^
<console>:11: error: type mismatch;
found : AA[Edge]
required: Set[Edge]
val edges2Set: Set[Edge] = edges
^
Higher-order functions especially for comprehensions compound my issue.
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评论（2）
将 Seq[_] 描述中的通配符更改为：
Change the wildcard in the Seq[_] description to this:
@thoredge 为我指明了正确的方向。我在 Predef.scala 中找到了解决方案：
通配符会丢弃类型等效性，但您可以抽象任何任意参数而不声明它。我的特质现在看起来像：
@thoredge pointed me in the right direction. I found the solution in Predef.scala:
Wildcards throws away type equivalence but you can instead abstract over any arbitrary parameter without declaring it. My trait now looks like: