Parameter extends class

Question

I want to make a class that accepts everything ordered and prints more. (I was just studying, so I know it's a little useless)

class PrinterOfGreater[T extends Ordered](val a:T, val b:T){println(a > b)} 

I know that it cannot be written with this style in scala, but I don’t know how to write it ... Does anyone know?

and why doesn't it compile? Serum line wrapper ordered

 class PrinterOfGreater[T <: Ordered[T]](a:T, b:T){println(a > b)} object A extends Application{new PrinterOfGreater("abc","abd")} 

Answer 1

Regarding your second question: String (which in Scala is java.lang.String , at least when targeting the Java / JVM platform) does not define the relational operator > . However, you can easily place this by replacing <: with <% , which indicates the so-called view boundary, which means that in A <% B A there is either a subtype of B , or there is an implicit transformation in the area that will give B when specifying A

This works for String because the Scala standard libraries supply implicit conversion from String to RichString (in Scala 2.7) or to StringOps (in Scala 2.8), where relational operators are defined.

Answer 2

Do you want to

 class PrinterOfGreater[T <: Ordered[T]](val a:T, val b:T){println(a > b)} 

<: means "is a subclass" (just like extends does in Java), and the Ordered parameter itself, and you want to clearly indicate that you are trying to compare T , so you specify Ordered[T] .