parameter extends a class
scala, type-parameter
Solution
Regarding your second question: `String` (which in Scala is just `java.lang.String`, at least when targeting the Java / JVM platform) does not define the relational operator `>`. However, you can accommodate this easily by replacing the `<:` with `<%` which specifies a so-called view bound, meaning that in `A <% B` `A` is either a subtype of `B` or that there is an implicit conversion in scope that will yield a `B` when given an `A`.
This works for `String` because Scala's standard libraries supply an implicit conversion from String to `RichString` (in Scala 2.7) or to `StringOps` (in Scala 2.8) where the relational operators are defined.
Problem
I want to do a class thats accepts anything ordered and prints greater. (I'm just learning so I know it's a bit useless) ``` class PrinterOfGreater[T extends Ordered](val a:T, val b:T){println(a > b)} ``` I know that it can't be written by this style in scala, but I don't know how to write it properly... Do anybody know? and why this doesn't compile? Whey the string wrapper is Ordered ``` class PrinterOfGreater[T <: Ordered[T]](a:T, b:T){println(a > b)} object A extends Application{new PrinterOfGreater("abc","abd")} ```