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kotlin - Kotlin 警告：在构造函数中访问非最终属性

问题描述

我有一些以下代码并收到producer变量警告accessing non-final property in constructor

class KafkaService {

  val producer: KafkaProducer<String, String>

  init {
    val props = Properties()
    props[ProducerConfig.BOOTSTRAP_SERVERS_CONFIG] = "127.0.0.1:9092"
    props[ProducerConfig.CLIENT_ID_CONFIG] = "DemoProducer"
    props[ProducerConfig.KEY_SERIALIZER_CLASS_CONFIG] = StringSerializer::class.java.name
    props[ProducerConfig.VALUE_SERIALIZER_CLASS_CONFIG] = StringSerializer::class.java.name

    producer = KafkaProducer(props)
  }

  fun sendToKafka(topic: String, message: String) {
    val producerRecord: ProducerRecord<String?, String> = ProducerRecord(topic, null, message)
    producer.send(producerRecord)
  }
}

解决此问题的最佳方法是什么？

标签： kotlin

解决方案

您应该将producer变量初始化为惰性：

class KafkaService {

  val producer: KafkaProducer<String, String> by lazy {
    val props = Properties()
    props[ProducerConfig.BOOTSTRAP_SERVERS_CONFIG] = "127.0.0.1:9092"
    props[ProducerConfig.CLIENT_ID_CONFIG] = "DemoProducer"
    props[ProducerConfig.KEY_SERIALIZER_CLASS_CONFIG] = StringSerializer::class.java.name
    props[ProducerConfig.VALUE_SERIALIZER_CLASS_CONFIG] = StringSerializer::class.java.name

    KafkaProducer(props)
  }

  fun sendToKafka(topic: String, message: String) {
    val producerRecord: ProducerRecord<String?, String> = ProducerRecord(topic, null, message)
    producer.send(producerRecord)
  }
}

这样，您的变量将在您第一次访问它时被初始化，并且您不会重新分配最终变量。阅读本文以更好地理解。

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