Android 作为函数中的参数扩展指定类的任何对象
我正在构建基本活动,在这里我有一个初始化Android 作为函数中的参数扩展指定类的任何对象,android,kotlin,android-recyclerview,Android,Kotlin,Android Recyclerview,我正在构建基本活动,在这里我有一个初始化RecyclerView的方法。 我希望这个方法有一个参数,它是一个对象/类,扩展了PagedListAdapter您的参数是class,它是类java.lang.class的实例,但您希望PagedListAdapter的实例如下: protected fun initRecyclerView(mAdapter: PagedListAdapter<out Any, out RecyclerView.ViewHolder>) { if
RecyclerView
的方法。
我希望这个方法有一个参数,它是一个对象/类,扩展了
PagedListAdapter您的参数是class
,它是类java.lang.class
的实例,但您希望PagedListAdapter
的实例如下:
protected fun initRecyclerView(mAdapter: PagedListAdapter<out Any, out RecyclerView.ViewHolder>) {
if (mRecyclerView?.layoutManager == null)
mRecyclerView?.layoutManager = component.linearLayoutManager
if (mRecyclerView?.adapter == null)
mRecyclerView?.adapter = mAdapter
}
protectedfun initRecyclerView(mAdapter:PagedListAdapter){
if(mRecyclerView?.layoutManager==null)
mRecyclerView?.layoutManager=component.linearLayoutManager
if(mRecyclerView?.adapter==null)
mRecyclerView?.adapter=mAdapter
}
您的参数是Class
,它是Classjava.lang.Class
的实例,但您希望PagedListAdapter
的实例如下:
protected fun initRecyclerView(mAdapter: PagedListAdapter<out Any, out RecyclerView.ViewHolder>) {
if (mRecyclerView?.layoutManager == null)
mRecyclerView?.layoutManager = component.linearLayoutManager
if (mRecyclerView?.adapter == null)
mRecyclerView?.adapter = mAdapter
}
protectedfun initRecyclerView(mAdapter:PagedListAdapter){
if(mRecyclerView?.layoutManager==null)
mRecyclerView?.layoutManager=component.linearLayoutManager
if(mRecyclerView?.adapter==null)
mRecyclerView?.adapter=mAdapter
}
Have look Have look
protected fun initRecyclerView(mAdapter: PagedListAdapter<out Any, out RecyclerView.ViewHolder>) {
if (mRecyclerView?.layoutManager == null)
mRecyclerView?.layoutManager = component.linearLayoutManager
if (mRecyclerView?.adapter == null)
mRecyclerView?.adapter = mAdapter
}