[1.7.10] [Scala] Adding Recipes?

[JasonIEC]

Hello, I am JasonIEC, and I am wondering how to make recipes in Scala, or how to have mixed Scala/Java mods.

My specific problem is the fact that Scala doesn't seem to support java.lang.Object arrays.

Any help is appreciated

 

Thanks!

[[email protected]]

what kind of recipes and why in scala and not java.I have never worked with scala so I am Curious?

 

 

[sham1]

Hello, I am JasonIEC, and I am wondering how to make recipes in Scala, or how to have mixed Scala/Java mods.

My specific problem is the fact that Scala doesn't seem to support java.lang.Object arrays.

Any help is appreciated

 

Thanks!

 

Scala's AnyRef is a type alias for java.lang.Object. So to define your recipe, you go:

Array[AnyRef](/* your stuff comes here.*/)

 

Also @hsgill97, choosing scala is mostly about personal preference, as scala has some really convenient stuff about it. Also you can mix java and scala code so that will not be  a problem.

[shadowfacts]

Unless var args methods don't translate to Scala properly, you don't need an array to create a recipe.

[Choonster]

You can call vararg methods in Scala just like you would in Java, see this example.

[JasonIEC]

Thank you all! I will definitely try these things. Also, I chose Scala because I wanted to try it 

 

Edit: The Array[AnyRef]() hasn't worked for me

 

Also, the varargs aren't working because it won't let me cast java.lang.Character to java.lang.Object

[JasonIEC]

Okay, so I got it.

Basically, when you define the character, you have to put 'X'.asInstanceOf[java.lang.Character] instead of just 'X'.

Then it works!