scala variables initialization - scala

Question

package Controls
object TestBreak extends App {
def values = {
x1 = x
y1 = y
}
val (x, y) = (1, 2)
values
var (x1, y1) = (2, 3)
println((x, y))
println((x1, y1))
}
I could see here, the program is Executing successfully without any error.
When i call method values ,even before the intialization of variables x1,y1,..
How does scala handles this case ?
How compilation of code is taking palce ?

Answer 1

App extends DelayedInit
As App extends DelayedInit and Main extends App. Code in the Main constructor executes as part of delayed init. Before jumping into how delayed init works, Lets see the differnce between constructor variable declaration and method variable declaration.
scala> class A {
| def a(): Unit = { println(b) }
| val b = 1
| }
defined class A
In the above example b is declared later and function a() is declared first. As it the constructor of the class Compiler accepts it.
scala> def x: Int = {
| def y: Int = z
| val z = 1
| y
| }
<console>:12: error: forward reference extends over definition of value z
def y: Int = z
^
In the above case compiler complains saying that its a forward reference.
Delayed Init is deprecated
Classes and objects (but note, not traits) inheriting the DelayedInit marker trait will have their initialization code rewritten as follows: code becomes delayedInit(code).
Initialization code comprises all statements and all value definitions that are executed during initialization.
Example:
trait Helper extends DelayedInit {
def delayedInit(body: => Unit) = {
println("dummy text, printed before initialization of C")
body // evaluates the initialization code of C
}
}
class C extends Helper {
println("this is the initialization code of C")
}
object Test extends App {
val c = new C
}
Should result in the following being printed:
dummy text, printed before initialization of C
this is the initialization code of C