Question

以下代码是关于反思的。它试图做两件事：

case1()从SimpleStudent类创建一个实例，它可以正常工作。
case2()从Student类创建一个实例，而不是工作。

case2()无法解决问题的原因在于generateValue()：

我不知道怎么检查它是kotlin类型还是我自己的类型（我有一种肮脏的方式来检查param.type.toString() not contain "kotlin"但我想知道是否有更好的解决方案
当它是自定义类时，我不知道如何获取它的类引用。问题在于即使param.type.toString() == "Lesson"，当我尝试获取param.type::class时，它是class kotlin.reflect.jvm.internal.KTypeImpl

那么，如何解决呢？感谢

==============

import kotlin.reflect.KParameter
import kotlin.reflect.full.primaryConstructor
import kotlin.test.assertEquals

data class Lesson(val title:String, val length:Int)
data class Student(val name:String, val major:Lesson )
data class SimpleStudent(val name:String, val age:Int )

fun generateValue(param:KParameter, originalValue:Map<*,*>):Any? {
    var value = originalValue[param.name]

//    if (param.type is not Kotlin type){
//        // Get its ::class so that we could create the instance of it, here, I mean Lesson class?
//    }

    return value
}

fun case1(){
    val classDesc = SimpleStudent::class
    val constructor = classDesc.primaryConstructor!!

    val value = mapOf<Any,Any>(
            "name" to "Tom",
            "age" to 16
    )

    val params = constructor.parameters.associateBy (
        {it},
        {generateValue(it, value)}
    )

    val result:SimpleStudent = constructor.callBy(params)

    assertEquals("Tom", result.name)
    assertEquals(16, result.age)
}

fun case2(){
    val classDesc = Student::class
    val constructor = classDesc.primaryConstructor!!

    val value = mapOf<Any,Any>(
            "name" to "Tom",
            "major" to mapOf<Any,Any>(
                    "title" to "CS",
                    "length" to 16
            )
    )

    val params = constructor.parameters.associateBy (
        {it},
        {generateValue(it, value)}
    )

    val result:Student = constructor.callBy(params)

    assertEquals("Tom", result.name)
    assertEquals(Lesson::class, result.major::class)
    assertEquals("CS", result.major.title)
}

fun main(args : Array<String>) {
    case1()
    case2()
}

Answer 1

问题解决了：

您可以::class使用param.type.classifier as KClass<T>

来获取param is KParameter

如何在kotlin中从KParameter获取类引用？

1 个答案: