Question

代码B中的参数videoList是List<File>类型，而变量listFile是MutableList<File>。

我认为代码A会导致错误，但是实际上，它运行良好，为什么？

代码A

private lateinit var listFile: MutableList<File>


override fun onViewCreated(view: View, savedInstanceState: Bundle?) {
        super.onViewCreated(view, savedInstanceState)     

        val outputDirectoryOfVideo = getVideoOutputDirectory(requireContext())
        listFile =outputDirectoryOfVideo.listFiles{file -> VIDEO_EXTENSION_WHITELIST.contains(file.extension.toUpperCase())}
                   .sorted().reversed().toMutableList()

        videoRecyclerView.layoutManager =  LinearLayoutManager(requireContext(), RecyclerView.VERTICAL, false)
        val aVideoListAdapter=VideoListAdapter(listFile,requireActivity())
        videoRecyclerView.adapter=aVideoListAdapter        
}

代码B

class VideoListAdapter (private val videoList: List<File>, private val mActivity: Activity) : RecyclerView.Adapter<VideoListAdapter.MyViewHolder>() {

}

Answer 1

子类型

MutableList<E>是<div class="container"> <div class="row"> <div class="col-sm-12 border"> <p class="text-center">Need text-center and align-middle</p> </div> </div> </div>的子类型（List<E>可以是任何类型，包括E）。

File

这意味着可变列表可以执行列表可以执行的任何操作（并可以在任何地方使用）。

奖金：差异

列表接口实际上被声明为interface MutableList<E> : List<E>, MutableCollection<E>。这意味着接口只返回（E）对象（产生），而从不接收（消耗）对象。由于通常我们可以返回子类型来代替其超类型，因此List<out E>让我们这样做：

out

...或更与您的问题相关：

// This works
// - List<E> is a producer (covariant in E)
// - List<String> can fulfill all members of List<CharSequence>
val listStrings: List<String> = mutableListOf<String>("Some", "example", "Strings")
val listCharSeqs : List<CharSequence> = listStrings // use List<Strings> as List<CharSequence>

// This does NOT work
// - MutableList<E> is a producer AND consumer (invariant in E)
// - MutableList<String> cannot fulfill add(element: CharSequence)
val mutableListStrings: MutableList<String> = mutableListOf<String>("Some", "example", "Strings")
val mutableListCharSeqs : MutableList<CharSequence> = mutableListStrings // does not compile

您可以在此处了解更多信息（包括class Video {...} class TvShow : Video {...} class Movie : Video {...} class Adapter(private val inputOnly: List<out Video>) {...} val justMovies: MutableList<Movie> = mutableListOf(...) val adapter = Adapter(justMovies) // this is ok!的逆差，星形投影和约束条件）：https://kotlinlang.org/docs/reference/generics.html

当我在Kotlin中将MutableList <File>的变量传递给List <File>时，Android Studio为什么不启动错误？

1 个答案:

子类型

奖金：差异