How does Kotlin's Json.decodeFromString work without a first DeserializationStrategy argument?

user_2800519 0
Viewed 5211

I'm very new to Kotlin and am curious what magic allows this code to work:

From https://github.com/Kotlin/kotlinx.serialization/blob/e2e764a132c8eebd31208120774baf9a71ec23c7/formats/json/commonTest/src/kotlinx/serialization/SerializerForNullableTypeTest.kt

@Serializable
data class Box(val s: StringHolder?)

val deserialized = Json.decodeFromString<Box>(string)

When the function definition seems to require an initial argument before the encoded JSON string.

https://github.com/Kotlin/kotlinx.serialization/blob/e2e764a132c8eebd31208120774baf9a71ec23c7/formats/json/commonMain/src/kotlinx/serialization/json/Json.kt

public final override fun <T> decodeFromString(deserializer: DeserializationStrategy<T>, string: String): T {
kotlin json
3 Answers

That method is currently part of an experimental API in SerialFormat.kt:

It is an extension function with the following interface:

@OptIn(ExperimentalSerializationApi::class)
public inline fun <reified T> StringFormat.encodeToString(value: T): String

You need to enable experimental APIs to be able to use it.

user_38486660
user_77662880

To use this you need to import import kotlinx.serialization.decodeFromString. But because this is an experimental API, you also need to mark your class or method with @ExperimentalSerializationApi. But to enable experimental switches you also need to add an compiler by languageSettings.optIn("kotlin.RequiresOptIn").

build.gradle.kts

plugins {
    kotlin("jvm") version "1.5.31"
    kotlin("plugin.serialization") version "1.5.31"
}

kotlin.sourceSets.all {
    languageSettings.optIn("kotlin.RequiresOptIn")
}

test.kts

import kotlinx.serialization.ExperimentalSerializationApi
import kotlinx.serialization.SerialFormat
import kotlinx.serialization.json.Json
import kotlinx.serialization.decodeFromString

@Serializable
data class Box(val s: String = "")

@OptIn(ExperimentalSerializationApi::class)
fun readJson(){
    val deserialized = Json.decodeFromString<Box>("""{"s" : "Hallo"}""")
}
user_41981700
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