Kotlin 序列化 : How to serialize MutableStateFlow?

Question

我有以下代码

@Serializeable
class Car {
   val speed = MutableStateFlow<Int>(0)
   var isMoving: Boolean = false
}

问题是如何序列化MutableStateFlow？

Answer 1

您需要为 MutableStateFlow<Int> 创建专门的序列化程序.

要么基于@Tenfour04 建议的通用序列化器

object MutableStateFlowOfIntsSerializer :
    KSerializer<MutableStateFlow<Int>> by MutableStateFlowSerializer(Int.serializer())

class MutableStateFlowSerializer<T>(private val dataSerializer: KSerializer<T>) : KSerializer<MutableStateFlow<T>> {
    override val descriptor: SerialDescriptor = dataSerializer.descriptor
    override fun serialize(encoder: Encoder, value: MutableStateFlow<T>) = dataSerializer.serialize(encoder, value.value)
    override fun deserialize(decoder: Decoder) = MutableStateFlow(dataSerializer.deserialize(decoder))
}

或从头开始:

object MutableStateFlowOfIntsSerializer : KSerializer<MutableStateFlow<Int>> {
    override val descriptor: SerialDescriptor = Int.serializer().descriptor
    override fun serialize(encoder: Encoder, value: MutableStateFlow<Int>) =
        Int.serializer().serialize(encoder, value.value)

    override fun deserialize(decoder: Decoder) = MutableStateFlow(Int.serializer().deserialize(decoder))
}

并将其连接到 speed领域:

@Serializable
class Car {
    @Serializable(with = MutableStateFlowOfIntsSerializer::class)
    val speed = MutableStateFlow(0)
    var isMoving: Boolean = false
}

请注意，此序列化方法将序列化 speed字段不管 encodeDefaults = false设置