Kotlin避免对null进行明智的强制转换 [英] Kotlin avoid smart cast for null check

问题描述

因此，我正在尝试减少此代码，并避免来自IDE的智能转换提示. 我的想法是，我有一个nullable类型的nullable变量，我想将其映射到R，或者我只是从供应商那里获得R，以防该变量为null.

So I'm trying to reduce this code and avoid the smart cast hint from IDE. The idea is I have a nullable variable of type T and I want to either map it to R or I just get R from a supplier in case the variable is null.

我尝试了不同的方法并提出了这个方法.仍然给了我聪明的演员提示.

I've tried different approaches and came up with this one. Still it gives me the smart cast hint.

fun <T, R> T?.func(mapper: (T) -> R, supplier: () -> R): R =
    when(this) {
        null -> supplier()
        else -> mapper(this) // smart cast
    }

但是我不喜欢将其中一个lambda括在括号中.例如.

But I don't like the need for wrapping one of the lambdas in parenthesis. For example.

fun foo(value: String?): Int =
    value.func({ it.length + 20}) { 30 }

这似乎很奇怪，但我的观点是将变量不是nullable传递给生成R的函数，或者调用生成R的函数.

This may seem odd but the ideia in my context was to pass the variable as not nullable to a function that produced a R or call a function that generated a R.

fun bar(value: T?): R =
    when(value) {
        null -> func1()
        else -> func2(value) // smart cast
    }

注意:我已经阅读了此，但它并不相同.

Note: I've read this but its not the same.

推荐答案

以下应避免使用智能转换提示

Following should avoid the smart cast hint

fun <T, R> T?.func(mapper: (T) -> R, supplier: () -> R): R {
    return this?.let { mapper(it) } ?: supplier()
}

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