Scala: Why does isInstanceOf[List[Any]] not work? -

i trying solve problem 7 of 99 scala problems , encountered difficulty in figuring out type of list can contain type. looked @ answer , saw list[any]. proceeded code out implementation follows:

def flatten(lst: list[any]): list[any] = lst match {     case nil => nil     case x::xs =>         (if (x.isinstanceof[list[any]]) { flatten(x) } else { list(x) }) :::         flatten(xs) } 

however, gives me following compile error:

[error] <filename omitted>:<line number omitted>: type mismatch; [error] found   : [error] required: list[any] [error]                (if (x.isinstanceof[any]) { flatten(x) } else {list(x) }) [error]                                                    ^ [error] 1 error found 

changing isinstanceof[list[any]] isinstanceof[list[_]] gives same compilation error.

after short google search , consulting this solution, implemented this:

def flatten(lst: list[any]): list[any] = lst match {     case nil => nil     case x::xs => x match {         case x: list[_] => flatten(x) ::: flatten(xs)         case _ => x :: flatten(xs)     } } 

which works fine. why scala compiler think x has type any when, in order inside block, has pass x.isinstanceof[any], makes of type list[any] ? compiler bug, or part of scala don't understand?

thank you!

in code, x head of list[any]: it's any, hence error message you're getting. need cast list[any], pattern matching lets quite elegantly:

def flatten(lst: list[any]): list[any] = lst match {   case nil               => nil   case (x:list[any])::xs => flatten(x) ::: flatten(xs)   case x::xs             => list(x) ::: flatten(xs) }