scala – 方法“|”中的预期类型的解析器

我有针对 scala 2.8.0编译的以下代码：

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","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？

解决方法

您需要将最后一行更改为

case class PrimitiveType[+T]()

这允许PrimitiveType [Int]&lt ;: PrimitiveType [AnyVal],当您想通过|合并PrimitiveType [Boolean]和PrimitiveType [Int]解析器的结果时需要它.

顺便说一句,我也建议写

PrimitiveType[AnyVal]

代替

PrimitiveType[_]

因为在你的情况下这更精确.

scala – 方法“|”中的预期类型的​​解析器

解决方法

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