kotlin 泛型函数作为参数

Question

我遇到了一些关于 kotlin 中 generic 类型的问题。最初，我想用值和调用函数调用函数。所以我在下面写了这个，但是这个值只允许我传递一个像 string/Int 这样的类型。我想在调用函数中传递一个函数作为参数。

A类

ClassA<T>{
var canCancel: Boolean = true
var recallFunction: (() -> T)? = null
var recallFunctionWithValue: ((value: T) -> Unit)? = null
var context: Context? = null
var value: T? = null
var functionPara :( ()->T)? = null
var recallFunctionWithFunction:((()->T) -> Unit)? = null

constructor(context: Context?, canCancel: Boolean, value: T, recallFunctionWithValue: (value: T) -> Unit) {
this.context = context
this.canCancel = canCancel
this.value = value
this.recallFunctionWithValue = recallFunctionWithValue
}
}

B类

ClassA(ctx, true,viewModel!!::testRefInside,viewModel!!::testRefOutside).createNativeNetworkDialog()

视图模型

fun testRefOutside(test: Observable<Any>? ){

}
fun testRefInside():Observable<Any>?{
    return null
}

---

我尝试做类似的事情

var canCancel: Boolean = true
var recallFunction: (() -> T)? = null
var recallFunctionWithValue: ((value: T) -> Unit)? = null
var context: Context? = null
var value: T? = null
var functionPara :( ()->T)? = null
var recallFunctionWithFunction:((()->T) -> Unit)? = null

    constructor(context: Context?, canCancel: Boolean, value: () -> T, recallFunctionWithValue: (value: ()->T) -> Unit) {
    this.context = context
    this.canCancel = canCancel
    this.functionPara = value
    this.recallFunctionWithFunction = recallFunctionWithValue
}

但它显示语法错误。有人可以帮忙吗？

语法错误

Error

Error2

Error show in this place only RYEasyDialog means ClassA

---

有效！！现在

  var recallFunctionWithFunction:((value: T) -> Unit)? = null
    constructor(context: Context?, canCancel: Boolean, value: () -> T, recallFunctionWithValue: (value: T) -> Unit) {
    this.context = context
    this.canCancel = canCancel
    this.functionPara = value
    this.recallFunctionWithFunction = recallFunctionWithValue

}

8/6 更新可变参数问题

我正在做一些高级的事情。我只是将它发布在完整的代码中。 如果我不使用 vararg ，则

EasyDialog<Observable<Any>,LiveData<Boolean>>(ctx, true,viewModel!!::callGetTeammateListAPITEST,recallFunctionWithFunction = viewModel!!::callMultipleAPI).createNativeNetworkDialog2()

不投射as ((Observable<Any>)->LiveData<Boolean>) .但是 callMuipleApi 接收 vararg 参数，我必须将其转换为 ((Observable)->LiveData) 但我没有显示任何错误，但未调用函数 callmuipleapi 和 callgetteammateslist 。 谁能帮忙？

视图模型

fun callMultipleAPI( vararg  observable: Observable<Any>):LiveData<Boolean>{
    if(progressDialogData==null){
        progressDialogData= MutableLiveData()
    }
    progressDialogData?.setValue(true)
    repo.callMultipleAPI<Any>(*observable, observer = RYEasyObserver(object : RyObserverSingleStatusListener<List<Any>> {
        override fun onNext(g: List<Any>) {
            Log.d("test",g.toString())
            for(model in g){
                when(model){
                    is GetTeammateListModel ->{setCallGetTeammateListResponse(model)}
                    is GetTeamListModel ->{setCallGetTeamListResponse(model)}
                    is GetMessageSumModel->{setCallGetMessageSumResponse(model)}
                    is GetCallModel->{setCallGetResponse(model)}
                }
            }
            progressDialogData?.setValue(false)
        }


        override fun onError() {
            progressDialogData?.setValue(false)
        }

    },true))
    return progressDialogData!!
}

片段

EasyDialog<Observable<Any>,LiveData<Boolean>>(ctx, true,viewModel!!::callGetTeammateListAPITEST,recallFunctionWithFunction = viewModel!!::callMultipleAPI as ((Observable<Any>)->LiveData<Boolean>)).createNativeNetworkDialog2()

EasyDialog

class EasyDialog<T,V> {
var canCancel: Boolean = true
var recallFunction: (() -> T)? = null
var recallFunctionWithValue: ((value: T) -> Unit)? = null
var context: Context? = null
var value: T? = null
var functionPara :(()->T)? = null
var recallFunctionWithFunction:((T) -> V)? = null



    constructor(
    context: Context?, canCancel: Boolean, value: ()->T, recallFunctionWithFunction:  (T) -> V  ){
    this.context = context
    this.canCancel = canCancel
    this.functionPara = value
    this.recallFunctionWithFunction = recallFunctionWithFunction
}
    fun createNativeNetworkDialog2() {
    val builder = AlertDialog.Builder(context)
    if (!canCancel) {
        builder.setCancelable(false)
    } else {
        builder.setNegativeButton(context?.resources?.getString(R.string.gloda_cancel)) { dialogInterface, id -> }
    }
    builder.setMessage(context?.resources?.getString(R.string.global_network_error))
    builder.setPositiveButton(context?.resources?.getString(R.string.global_retry)) { dialog, id ->
        recallFunction?.invoke()
        value?.let { it1 -> recallFunctionWithValue?.invoke(it1) }
        //functionPara?.let { it2->recallFunctionWithFunction?.invoke(it2) }
        functionPara?.let{ functionPara->{ recallFunctionWithFunction?.let {
                recallFunctionWithFunction-> compose<T,V>(recallFunctionWithFunction  ,functionPara).invoke()
        }}}
    }

    val dialog = builder.create()
    dialog.show()
}
    fun <T, V> compose( f: (T) -> V,    g:() -> T ): () -> V {
            return {  f(g()) }
}

Answer 1

最后一个构造函数参数是

(() -> T) -> Unit

根据您为第三个构造函数参数传递的内容，T 被推断为Observable<Any>?。所以它期待一个

(() -> Observable<Any>?) -> Unit

但testRefOutside 是

 Observable<Any>? -> Unit

我猜你可能不是想通过将函数的输入设为另一个函数来定义三阶函数。