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#serialization
    g

    ghedeon

    2 years ago
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    Might be a strange question but how do you write a serializer for 
    kotlin.Result
    class? Let say I want to write a custom call adapter for Retrofit that wraps all my requests in 
    Result
    class. Right now serialization converter is failing with 
    Can't locate argument-less serializer for class kotlin.Result
    which sounds reasonable. How would one workaround this situation?
    s

    spand

    2 years ago
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    Seems like asking for trouble serializing its 
    Throwable
    no ?
    g

    ghedeon

    2 years ago
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    I don't expect serialization to happen at all, I can even leave it not implemented. The only reason it's an issue is that my requests return 
    Result<T>
    now, where 
    T
    is an actual object to deserialize from json and 
    Result
    is something that kotlin serialization is not aware of. As you can imagine, it's one way.
    Nikky

    Nikky

    2 years ago
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    so you want a Result.Success<T> to serialize like a T ?
    g

    ghedeon

    2 years ago
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    you can say that, yes. Although, this path should never be called, I can even throw 
    NotImplemented
    exception or something.
    Nikky

    Nikky

    2 years ago
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    val <T> T.result: KSerializer<Result<T>> = ResultSerializer<T>(this)
    for the shorthand
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    that way the 
    ResultSerializer
    has the child serializer and can call it
    g

    ghedeon

    2 years ago
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    Not sure where I'm supposed to put it. As you can imagine, I don't controll deserialization manually, it happes inside of Retrofit. So, by the time the call is performed, the Result Serializer is expected to be registered somewhere...
    Nikky

    Nikky

    2 years ago
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    you pass retrofit some sort of Json object right ?
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    anyways this is the Serializer i would use.. not sure if it will work..
    class ResultSerializer<T: Any>(val childSerializer: KSerializer<Result<T>>) : KSerializer<Result<T>> {
    override val descriptor: SerialDescriptor = SerialClassDescImpl(Result::class.simpleName!!)
    override fun deserialize(decoder: Decoder): Result<T> {
        TODO()
    }
    override fun serialize(encoder: Encoder, obj: Result<T>) {
        encoder.encodeSerializableValue(childSerializer,obj)
    }
}
    g

    ghedeon

    2 years ago
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    well, somewhere internally that's what is happening, yes. But I don't pass anything, it's already abstracted for the user. All you have to do is 
    fun foo(): Bar
    where 
    Bar
    must be serializable. And now it's 
    fun foo(): Result<Bar>
    .
    Nikky

    Nikky

    2 years ago
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    you register it to use kotlinx-serialization somehow though ?
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    otherwise does theu ser have no control over the Json options ?
    g

    ghedeon

    2 years ago
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    yep
    .addConverterFactory(Json.asConverterFactory(MediaType.get("application/json")))
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    i think you can use Json() too there
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    and install default modules or whatever they are called in there
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    otherwise ping the author i guess
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    that is what kotlinx-serialization should use to look up the serializer for Result .. not sure how to register it in a way that the retrofit adapter supports
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