How memory allocation takes place in scala

Question

As we know that unlike java, scala has everything as object.

For example we have-

object A{
    val arg1=1
    def myFun(arg2:Int)=arg1
}

class A{  
    val arg1=1
    def myFun(arg2:Int)=arg1
}

trait A{
    val arg1=1
    def myFun(arg2:Int)=arg1
}

• Now as everything in scala is an object so how the memory allocation will take place?
• Is everything going to get memory in heap other than the reference variables ?
• And in java when the class instance is created then the methods and variables in that class gets the memory in heap. How does it happens for a Singleton Objects here?
• If everything is in Heap, will it not affect the performance?
• As in Java, Memory is divided into 5 sections i.e. Heap, Stack, MethodArea etc. What about in scala, how does the memory allocation takes place?

Answer 1

Scala runs on the JVM and it's base on the Java library.

scala file(*.scala) will be compiled to the java class bytecode and run these on JVM. for your example:

object A{
    val arg1=1
    def myFun(arg2:Int)=arg1
}

will be translated to(decompile bytecode by javap):

public class A$ extends java.lang.Object{
    public static final A$ MODULE$;
    private final int arg1;
    public static {};
    public int arg1();
    public int myFun(int);
    public A$();
}

---

class A{  
    val arg1=1
    def myFun(arg2:Int)=arg1
}

will be translated to(decompile bytecode by javap):

public class A extends java.lang.Object{
    private final int arg1;
    public int arg1();
    public int myFun(int);
    public A();
}

---

trait A{
    val arg1=1
    def myFun(arg2:Int)=arg1
}

will be translated to(decompile bytecode by javap):

public interface A{
public abstract void $line5$$read$A$_setter_$arg1_$eq(int);

public abstract int arg1();

public abstract int myFun(int);

}

so for your other memory questions, I think it's same as Java.