scala – 方法“|”中的预期类型的??解析器

import scala.util.parsing.combinator.{syntactical,PackratParsers}
import syntactical.StandardTokenParsers

object MyParser extends StandardTokenParsers with PackratParsers{
  lexical.reserved ++= Set("int","char","boolean")

  lazy val primitiveType:PackratParser[PrimitiveType[_]] = primitiveChar | primitiveInt | primitiveBool

  lazy val primitiveInt:PackratParser[PrimitiveType[Int]] = "int" ^^ { _ => PrimitiveType[Int]() }

  lazy val primitiveChar:PackratParser[PrimitiveType[Char]] = "char" ^^ { _ => PrimitiveType[Char]() }

  lazy val primitiveBool:PackratParser[PrimitiveType[Boolean]] = "boolean" ^^ { _ => PrimitiveType[Boolean]() }
}

object MyParser2 extends StandardTokenParsers with PackratParsers{
  lexical.reserved ++= Set("int","boolean")

  lazy val primitiveType:PackratParser[PrimitiveType[_]] =  primitiveChar | primitiveIntOrBool

  lazy val primitiveIntOrBool:PackratParser[PrimitiveType[_]] = "int" ^^ { _ => PrimitiveType[Int]() } | "boolean" ^^ {_ => PrimitiveType[Boolean]()}

  lazy val primitiveChar:PackratParser[PrimitiveType[Char]] = "char" ^^ { _ => PrimitiveType[Char]()} 
}

case class PrimitiveType[T]()

编译MyParser1给出：

error: inferred type arguments  [this.PrimitiveType[_ >: _1 with Boolean <: AnyVal]] do not conform to method |'s type parameter bounds [U >: this.PrimitiveType[_ >: Char with Int <: AnyVal]]

我相信它因为|而失败了方法类型签名,定义为：

def | [U >: T](q: => Parser[U]): Parser[U]

为什么U必须是T的超类型？什么应该是“primitiveType”的返回值？