Is it possible in scala to expose the functionality of a class, based on the condition?
Say
class c(a: Int, b: Int){ if(a == 1){ val variable: Int = a def function: Int = b } Need to expose variable and function only when a is 1 during the usage of class
No, members must be known at compile time so that access to members can be checked by the compiler.
One option is to use Option:
class c(a: Int, b: Int){ val variable = if (a==1) Some(a) else None def function() = if (a==1) Some(f(b)) else None } You can use variable.getOrElse(default) to extract variable from an instance of c.
Another option is to use a class hierarchy
trait C class B() extends C class A(a: Int, b: Int) extends C { val variable: Int = a def function: Int = b } object C { def apply(a: Int, b: Int): C = if (a == 1) { new A(a, b) } else { new B() } } You create the instance with C(...,...) and when you need to access variable you use match to check that you have an instance of A rather than B.
It is possible with dynamic loading, but there are very few use cases where this would be the right solution.
Many languages support dynamic loading actual modification of the byte code at run-time. The main use-case is to upgrade and modifying applications that cannot stop or cannot be redeployed.
c, inside that function how would I know that I can (or can't) use eithervariableorfunction? I expect that explains why this won't be useful. - Now, why do you want this? What is your use case? What is the bigger picture? Probably there are ways to model what you want to express.class Cis a compile-time property. The value ofa(1, -2, 37, whatever) is a run-time property. "...and never the twain shall meet" -Kipling.