方法"|“中需要分析器的类型

Question

我有以下针对scala 2.8.0进行编译的代码：

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","char","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可以实现以下功能：

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]]

我认为它失败是因为|方法类型签名，定义如下：

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

为什么你必须是T的超类型？primitiveType的返回值应该是什么？

Answer 1

您需要将最后一行更改为

case class PrimitiveType[+T]()

这允许使用PrimitiveTypeInt <：PrimitiveTypeAnyVal，当您想要通过|合并PrimitiveTypeBoolean和PrimitiveTypeInt解析器的结果时，需要使用它。

顺便说一句，我也建议写

PrimitiveType[AnyVal]

而不是

PrimitiveType[_]

因为这在你的情况下更精确。