"没有足够的信息来推断参数T"与Kotlin和Android

Question

我正在尝试使用Kotlin在我的Android应用程序中复制以下ListView:https://github.com/bidrohi/KotlinListView.

不幸的是我收到了一个错误,我无法解决自己.这是我的代码:

MainActivity.kt:

override fun onCreate(savedInstanceState: Bundle?) {
    super.onCreate(savedInstanceState)
    setContentView(R.layout.activity_main)

    val listView = findViewById(R.id.list) as ListView
    listView.adapter = ListExampleAdapter(this)
}

private class ListExampleAdapter(context: Context) : BaseAdapter() {
    internal var sList = arrayOf("Eins", "Zwei", "Drei")
    private  val mInflator: LayoutInflater

    init {
        this.mInflator = LayoutInflater.from(context)
    }

    override fun getCount(): Int {
        return sList.size
    }

    override fun getItem(position: Int): Any {
        return sList[position]
    }

    override fun getItemId(position: Int): Long {
        return position.toLong()
    }

    override fun getView(position: Int, convertView: View?, parent: ViewGroup): View? {
        val view: View?
        val vh: ListRowHolder

        if(convertView == null) {
            view = this.mInflator.inflate(R.layout.list_row, parent, false)
            vh = ListRowHolder(view)
            view.tag = vh
        } else {
            view = convertView
            vh = view.tag as ListRowHolder
        }

        vh.label.text = sList[position]
        return view
    }
}

private class ListRowHolder(row: View?) {
    public val label: TextView

    init {
        this.label = row?.findViewById(R.id.label) as TextView
    }
}
}

布局完全如下:https://github.com/bidrohi/KotlinListView/tree/master/app/src/main/res/layout

我得到的完整错误消息是这样的: 错误:(92,31)类型推断失败:没有足够的信息来推断有趣的参数T findViewById(p0:Int):T!请明确说明.

我很感激我能得到任何帮助.

Answer 1

您必须使用API​​级别26(或更高级别).此版本更改了签名View.findViewById()- 请参阅此处https://developer.android.com/preview/api-overview.html#fvbi-signature

所以在你的情况下,结果findViewById是不明确的,你需要提供类型:

1 /改变

val listView = findViewById(R.id.list) as ListView 至

val listView = findViewById<ListView>(R.id.list)

2 /改变

this.label = row?.findViewById(R.id.label) as TextView 至

this.label = row?.findViewById<TextView>(R.id.label) as TextView

请注意,在2 /中只需要row强制转换因为可以为空.如果label 也可以为空,或者如果你row不可为空,那就不需要了.

Answer 2

Andoid O改变了findViewById api

public View findViewById(int id);

至

public final T findViewById(int id)

所以,如果你是API 26的目标,你可以改变

val listView = findViewById(R.id.list)作为ListView

至

val listView = findViewById(R.id.list)

要么

val listView:ListView = findViewById(R.id.list)

Answer 3

它的工作

API 级别 25 或以下使用此

    var et_user_name = findViewById(R.id.et_user_name) as EditText

API 级别 26 或更高级别使用此

    val et_user_name: EditText = findViewById(R.id.et_user_name)

快乐编码！