kotlin - 我可以将Kotlin中具有默认参数的函数的函数引用作为无参数函数得到吗？

Question

是否有可能获得对具有默认参数(指定为无参数调用)的函数的函数引用？
InputStream.buffered()是一种扩展方法，可将InputStream转换为缓冲区大小为8192字节的BufferedInputStream。

public inline fun InputStream.buffered(bufferSize: Int = DEFAULT_BUFFER_SIZE): BufferedInputStream =
    if (this is BufferedInputStream) this else BufferedInputStream(this, bufferSize)

我想使用默认参数有效地引用扩展方法，并将其传递给另一个函数。

fun mvce() {
    val working: (InputStream) -> InputStream = { it.buffered() }

    val doesNotCompile: (InputStream) -> BufferedInputStream = InputStream::buffered
    val alsoDoesNotCompile: (InputStream) -> InputStream = InputStream::buffered
}

doesNotCompile和 alsoDoesNotCompile产生以下错误

Type mismatch: inferred type is KFunction2 but (InputStream) -> BufferedInputStream was expected

Type mismatch: inferred type is KFunction2 but (InputStream) -> InputStream was expected

我知道错误是因为 InputStream.buffered()实际上不是 (InputStream) -> BufferedInputStream，而是 (InputStream, Int) -> BufferedInputStream的快捷方式，将缓冲区大小作为参数传递给BufferedInputStream构造函数。

动机主要是样式方面的原因，我宁愿使用已经存在的引用，也不愿在最后一刻创建引用。

val ideal: (InputStream) -> BufferedInputStream = InputStream::buffered// reference extension method with default parameter
val working: (InputStream) -> BufferedInputStream = { it.buffered() }// create new (InputStream) -> BufferedInputStream, which calls extension method

Answer 1

正如注释中提到的gpunto和Pawel一样，使用-XXLanguage:+NewInference编译器参数允许使用默认值引用函数。

已跟踪该问题here，并且针对Kotlin 1.4.0。