3

有什么问题?

val is = IntegerSerializer.get
mutator.addInsertion(deviceId, COLUMN_FAMILY_CARSTATUS, createColumn("mileage", 111, ss, is))}


ModelOperation.scala:96: error: type mismatch;
[INFO]  found   : me.prettyprint.cassandra.serializers.IntegerSerializer
[INFO]  required: me.prettyprint.hector.api.Serializer[Any]
[INFO] Note: java.lang.Integer <: Any (and me.prettyprint.cassandra.serializers.IntegerSerializer <: me.prettyprint.cassandra.serializers.AbstractSerializer[java.lang.Integer]), but Java-defined trait Serializer is invariant in type T.
[INFO] You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
[INFO]      mutator.addInsertion(deviceId, COLUMN_FAMILY_CARSTATUS, createColumn("mileage", 111, ss, is))}
4

1 回答 1

6

错误是说createColumn需要 type 的序列化程序Serializer[Any],但您传递的是 type 之一Serializer[Integer]。这只有Serializer在其类型参数是协变的(即定义为Serializer[+T])时才有效。但相反,Serializer它来自 Java,其中协方差的工作方式不同。

该类型Serializer[Integer]可以安全地转换为Serializer[_ <: Any],因此 Scala 编译器建议可能createColumn应该编写为期望不太具体的通配符类型。

如果你不能修改createColumn,那么最后的手段是使用“类型系统逃生舱口”asInstanceOf来转换为预期的类型:

val is = IntegerSerializer.get.asInstanceOf[Serializer[Any]] // defeats type system
mutator.addInsertion(... is ...)
于 2011-08-31T14:20:15.557 回答