我试图从Scala的准引用中调用一个无形宏,我没有得到我想要得到的。
我的宏没有返回任何错误,但它没有将证人(fieldName)展开为证人。Lt(字符串)
val implicits = schema.fields.map { field =>
val fieldName:String = field.name
val fieldType = TypeName(field.valueType.fullName)
val in = TermName("implicitField"+fieldName)
val tn = TermName(fieldName)
val cc = TermName("cc")
q"""implicit val $in = Field.apply[$className,$fieldType](Witness($fieldName), ($cc: $className) => $cc.$tn)"""
}
下面是我的Field定义:
sealed abstract class Field[CC, FieldName] {
val fieldName: String
type fieldType
// How to extract this field
def get(cc : CC) : fieldType
}
object Field {
// fieldType is existencial in Field but parametric in Fied.Aux
// used to explict constraints on fieldType
type Aux[CC, FieldName, fieldType_] = Field[CC, FieldName] {
type fieldType = fieldType_
}
def apply[CC, fieldType_](fieldWitness : Witness.Lt[String], ext : CC => fieldType_) : Field.Aux[CC, fieldWitness.T, fieldType_] =
new Field[CC, fieldWitness.T] {
val fieldName : String = fieldWitness.value
type fieldType = fieldType_
def get(cc : CC) : fieldType = ext(cc)
}
}
在这种情况下,我生成的隐式看起来像:
implicit val implicitFieldname : Field[MyCaseClass, fieldWitness.`type`#T]{
override type fieldType = java.lang.String
}
如果它被定义在准引用之外,它将生成如下内容:
implicit val implicitFieldname : Field.Aux[MyCaseClass, Witness.Lt[String]#T, String] = ...
有什么可以做的吗?