[英]“error: missing parameter type” during macro splicing
编辑:我发现了我的错误-我的递归案例的准引用中有一个错误,导致它返回格式错误的序列
我正在尝试创建一个将案例类T
转换为updateMap: Map[String, (play.api.libs.json.JsValue) => Try[(T) => T]]
的宏updateMap: Map[String, (play.api.libs.json.JsValue) => Try[(T) => T]]
( 如何使用scala宏以创建函数对象(以创建Map [String,(T)=> T]) ),其中映射的键是case类的字段名称-这个想法是给定JsObject("key" -> JsValue)
我们可以使用key
从updateMap
检索适当的更新方法,然后使用JsValue
应用更新。 我在非递归情况下进行此工作,即给定一个没有任何其他案例类作为字段的案例类。 但是,我想扩展此宏,以便它可以为包含其他案例类的案例类生成updateMap
。
case class Inner(innerStr: String)
case class Outer(outerStr: String, inner: Inner)
updateMap[Outer] = {
// updateMap[Inner]
val innerMap = Map("innerStr" -> (str: String) =>
Try { (i: Inner) => i.copy(innerStr = str) } )
// updateMap[Outer]
Map("outerStr" -> (str: String) =>
Try { (o: Outer) => o.copy(outerStr = str) },
"inner.innerStr" -> (str: String) =>
Try { (o: Outer) => innerMap.get("innerStr").get(str).flatMap(lens => o.copy(inner = lens(o.inner))) })}
换句话说,给定updateMap[Outer]
,我将能够直接更新对象的outerStr
字段,否则,我将能够更新对象的inner.innerStr
字段,无论哪种方式都返回Try[Outer]
。
该代码对于非递归情况( copyMapRec[Inner]()
)正常工作,但是递归情况( copyMapRec[Outer]()
)给我一个“错误:缺少参数类型”错误-我假设我要么需要在某个地方提供一个隐式参数,否则我对拼接有一个基本的误解。
下面的代码使用(String) => Try[(T) => T]
而不是(JsValue) => Try[(T) => T]
因此我不需要将play框架导入到我的REPL。 我使用隐式转换将JsValue
(或String
)转换为适当的类型(在val x: $fieldType = str
发生val x: $fieldType = str
行,在基本情况下为准引号;如果没有适当的隐式转换,则得到编译器错误)。
import scala.language.experimental.macros
def copyMapImplRec[T: c.WeakTypeTag](c: scala.reflect.macros.Context)(blacklist: c.Expr[String]*): c.Expr[Map[String, (String) => scala.util.Try[(T) => T]]] = {
import c.universe._
// Fields that should be omitted from the map
val blacklistList: Seq[String] = blacklist.map(e => c.eval(c.Expr[String](c.resetAllAttrs(e.tree))))
def rec(tpe: Type): c.Expr[Map[String, (String) => scala.util.Try[(T) => T]]] = {
val typeName = tpe.typeSymbol.name.decoded
// All fields in the case class's primary constructor, minus the blacklisted fields
val fields = tpe.declarations.collectFirst {
case m: MethodSymbol if m.isPrimaryConstructor => m
}.get.paramss.head.filterNot(field => blacklistList.contains(typeName + "." + field.name.decoded))
// Split the fields into case classes and non case classes
val recursive = fields.filter(f => f.typeSignature.typeSymbol.isClass && f.typeSignature.typeSymbol.asClass.isCaseClass)
val nonRecursive = fields.filterNot(f => f.typeSignature.typeSymbol.isClass && f.typeSignature.typeSymbol.asClass.isCaseClass)
val recursiveMethods = recursive.map {
field => {
val fieldName = field.name
val fieldNameDecoded = fieldName.decoded
// Get the c.Expr[Map] for this case class
val map = rec(field.typeSignature)
// Construct an "inner.innerStr -> " seq of tuples from the "innerStr -> " seq of tuples
q"""{
val innerMap = $map
innerMap.toSeq.map(tuple => ($fieldNameDecoded + "." + tuple._1) -> {
(str: String) => {
val innerUpdate = tuple._2(str)
innerUpdate.map(innerUpdate => (outer: $tpe) => outer.copy($fieldName = innerUpdate(outer.$fieldName)))
}
})}"""
}
}
val nonRecursiveMethods = nonRecursive.map {
field => {
val fieldName = field.name
val fieldNameDecoded = fieldName.decoded
val fieldType = field.typeSignature
val fieldTypeName = fieldType.toString
q"""{
$fieldNameDecoded -> {
(str: String) => scala.util.Try {
val x: $fieldType = str
(t: $tpe) => t.copy($fieldName = x)
}.recoverWith {
case e: Exception => scala.util.Failure(new IllegalArgumentException("Failed to parse " + str + " as " + $typeName + "." + $fieldNameDecoded + ": " + $fieldTypeName))
}
}}"""
}
}
// Splice in all of the sequences of tuples, flatten the sequence, and construct a map
c.Expr[Map[String, (String) => scala.util.Try[(T) => T]]] {
q"""{ Map((List(..$recursiveMethods).flatten ++ List(..$nonRecursiveMethods)):_*) }"""
}
}
rec(weakTypeOf[T])
}
def copyMapRec[T](blacklist: String*) = macro copyMapImplRec[T]
我解决了这个问题-最初在我的recursiveMethods
准引用具有innerMap.toSeq(...)
而不是innerMap.toSeq.map(...)
-我忽略了先在REPL中测试代码
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