scala - 在 Scala 中尝试惰性 val 上的模式匹配？

Question

是否可以对声明为 Try 的 lazy val 进行模式匹配，如下所示？

lazy val kafkaProducer: Try[producer.KafkaProducer[Array[Byte], String]] = Try(kafkaProducerSettings.createKafkaProducer())
...
kafkaProducer.get match {
  case Success(_) => Source.single(producerRecord()).runWith(Producer.plainSink(kafkaProducerSettings, kafkaProducer.get))
  case Failure(x) => Future.failed(x)
}

我收到此错误:

constructor cannot be instantiated to expected type;
[error]  found   : akka.actor.Status.Success
[error]  required: org.apache.kafka.clients.producer.KafkaProducer[Array[Byte],String]
[error]         case Success(_) => Source.single(producerRecord()).runWith(Producer.plainSink(kafkaProducerSettings, kafkaProducer.get))

注意，这个替代代码可以工作，但我不确定它是“Scala 方式”:

lazy val kafkaProducer: producer.KafkaProducer[Array[Byte], String] = kafkaProducerSettings.createKafkaProducer()
...
val tryAccessLazyKafkaProducer = Try(kafkaProducer)
if (tryAccessLazyKafkaProducer.isSuccess) {
  Source.single(producerRecord()).runWith(Producer.plainSink(kafkaProducerSettings, kafkaProducer))
} else {
  Future.failed(tryAccessLazyKafkaProducer.failed.get)
}

Answer 1

这绝对有可能，您只是导入了错误的 Success 类型:

found   : akka.actor.Status.Success

您需要 scala.util.Success 而不是

有一件事你不能做的是调用Try.get，如果返回的类型是Failure，它就会爆炸。相反，请执行以下操作:

import scala.util.Success
import scala.util.Failure

kafkaProducer match {
    case Success(producer) => Source.single(producerRecord()).runWith(Producer.plainSink(kafkaProducerSettings, producer))
    case failure: Failure => failure
}

lazy 只是一种语言构造，可确保该值仅被评估一次。底层类型，无论是否惰性，仍然是一个 Try，您可以用它做您想做的事情。