Kotlin上限:`:Any`对Kotlin的泛型类型推断有何不同? [英] Kotlin Upper Bound: What difference does `:Any` make to Kotlin's generic type inference?

问题描述

在Kotlin for Android Developers书之后，我们遇到了扩展功能

Following the Kotlin for Android Developers book, we come across extension function

fun <T:Any> SelectQueryBuilder.parseList(parser: (Map<String,Any?>) -> T):List<T> = parseList(object:MapRowParser<T>{
    override fun parseRow(columns: Map<String, Any?>): T = parser(columns)
})

我不确定为什么需要:Any.

如果我将其写为fun <T> SelectQueryBuilder.parseList(...)，则Android Studio会抱怨

If I write it as fun <T> SelectQueryBuilder.parseList(...), Android Studio complains that

当您重新添加:Any时，该错误消失了.

whereas that error goes away when you add the :Any back.

现在，就我而言，T应该暗示T:Any，尽管显然不是这样.这是为什么?它有什么区别?

Now, as far as I'm concerned, T should imply T:Any, though that is clearly not the case. Why is that? And what difference does it make?

推荐答案

现在，就我而言，T应该暗示T:Any

T表示T:Any?，其中Any?与Java的Object最接近.使用T:Any您指定了不可为空的类型.

T implies T:Any?, where Any? is the closest equivalent to Java's Object. With T:Any you specified a non-nullable type.

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