是否可以获取对具有默认参数的函数的引用,该默认参数指定为无参数调用?
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
类型不匹配:推断的类型为KFunction2,但预期为(InputStream)-> BufferedInputStream
类型不匹配:推断的类型为KFunction2,但预期为(InputStream)-> InputStream
我理解错误是因为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