使用类声明中的类型参数，例如函数中的类型化参数 [英] Using type parameter from class declaration like reified type parameter in function

问题描述

我想编写一个函数，该函数需要使用类声明的类表示形式，例如MyClass::class.java.

I want to write a function, which needs to use a the class representation of a class declaration like MyClass::class.java.

我有一个基类ActivityStarter，它用作活动中伴侣对象的基础.

I have a base class ActivityStarter which I use as a base for companion objects in activities.

// declarations
open class ActivityCompanion<T: AppCompatActivity>(val activityClass : Class<T>) {
    fun startActivity(context: Context) {
        context.startActivity(Intent(context, activityClass))
    }
}

class MyActivity : AppCompatActivity() {
    companion object : ActivityStarter<MyActivity>(MyActivity::class:java)
...
}

// call
MyActivity.startActivity(this)

在声明以及伴随对象的继承中，该类基本上传递了两次.一次作为类型参数，一次作为常规参数. (ActivityStarter<MyActivity>(MyActivity::class:java)).

In the declaration, and the inheritence of the companion object, the class is basically passed twice. Once as type parameter and once as normal parameter. (ActivityStarter<MyActivity>(MyActivity::class:java)).

参数activityClass是必需的，因为我不能使用T::class.java. 不能将T用作化类型参数.使用类代替".

The parameter activityClass is necessary, because I cant use T::class.java. "Cannot use T as reified type parameter. Use class instead".

我曾经只是在函数调用中传递参数:

I have used to just pass the parameter in the function call:

// declarations
open class ActivityStarter {
    inline fun <reified T : AppCompatActivity>startActivity(context: Context) {
        context.startActivity(Intent(context, T::class.java))
    }
}

class MyActivity : AppCompatActivity() {
    companion object : ActivityStarter()
...
}

// call
MyActivity.startActivity<MyActivity>(this)

这从同伴对象声明中删除了冗余，但将其放入了调用中，并基本上使类中的type参数变得无用.

This removes the redundancy from the companion object declaration but puts it into the call and basically makes the type parameter in the class useless.

是否有一种方法可以使类声明reified中使用的类型参数，以便可以实现这样的实现?:

Is there a way to make the type parameter used in the class declaration reified, so that I can have an implementation like this?:

// declarations
open class ActivityCompanion<reified T: AppCompatActivity>() {
    fun startActivity(context: Context) {
        context.startActivity(Intent(context, T::class.java))
    }
}

class MyActivity : AppCompatActivity() {
    companion object : ActivityStarter<MyActivity>()
...
}

// call
MyActivity.startActivity(this)

或者是另一种取消activityClass参数的方式，这样我就可以拥有这样的东西:

Or another way to resctict the activityClass parameter so that I can have something like this:

// declarations
open class ActivityStarter(private val activityClass : Class<T : AppCompatActivity>) {
    inline fun startActivity(context: Context) {
        context.startActivity(Intent(context, activityClass))
    }
}

class MyActivity : AppCompatActivity() {
    companion object : ActivityStarter(MyActivity::class.java)
...
}

// call
MyActivity.startActivity(this)

推荐答案

有没有一种方法可以使在类声明中使用的类型参数正确化

Is there a way to make the type parameter used in the class declaration reified

不.但是，如果您使startActivity不是ActivityCompanion的成员，而是一个扩展函数，以便它 可以具有化类型参数，该怎么办?

No. But what if you make startActivity not a member of ActivityCompanion, but an extension function so it can have reified type parameters?

class ActivityCompanion<T: AppCompatActivity>()

fun <reified T: AppCompatActivity> ActivityCompanion<T>.startActivity(context: Context) {
    context.startActivity(Intent(context, T::class.java))
}

class MyActivity : AppCompatActivity() {
    companion object : ActivityStarter<MyActivity>()
    ...
}

MyActivity.startActivity(this)

或者是另一种取消activityClass参数的方式，这样我就可以拥有这样的东西:

Or another way to resctict the activityClass parameter so that I can have something like this:

如果我理解正确，并且您希望ActivityStarter接受AppCompatActivity的任何子类而没有类型参数本身，则可以将其编写为

If I understood correctly and you want ActivityStarter to accept any subclass of AppCompatActivity without having a type parameter itself, you can write it as

open class ActivityStarter(protected val activityClass: Class<out AppCompatActivity>) { ... }

请参见官方文档或

See official documentation or What is out keyword in kotlin for the meaning of out.

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