The reason why the code would not initially compile is because defaultValue is an instance variable of the class Test, meaning that when an object of type Test is created, a unique instance of defaultValue is also created and attached to that particular object. Because of this, it is not possible to reference defaultValue in … Read more
Yes, they are initialized in the order in which they appear in the source. You can read all of the gory details in The Java Language Specification, §12.4.2. See step 9, which reads: … execute either the class variable initializers and static initializers of the class, or the field initializers of the interface, in textual … Read more
I do not really understand your question. That public class Test3 { private final String test = “test123”; public Test3() { System.out.println(“Test = “+test); } public static void main(String[] args) { Test3 t = new Test3(); } } executes as follows: $ javac Test3.java && java Test3 Test = test123
Because that’s what the language specification says. According to Guy Steele, the rationale behind this choice is that programmers would expect the declaration int x = 0 in a method to result in stack-allocated storage, but if you can return a new myRun() from the method (or otherwise let a myRun persist past the function’s … Read more
Vals are initialized in the order they are declared (well, precisely, non-lazy vals are), so properties is getting initialized before loadedProps. Or in other words, loadedProps is still null when propertiesis getting initialized. The simplest solution here is to define loadedProps before properties: class Config { private val loadedProps = { val p = new … Read more
Adding final to methods does not improve performance with Sun HotSpot. Where final could be added, HotSpot will notice that the method is never overridden and so treat it the same. In Java private methods are non-virtual. You can’t override them, even using nested classes where they may be accessible to subclasses. For instance methods … Read more
There are two main reasons you might want to mark an argument final. First, if you’re planning on using the argument in an anonymous inner class, then you must mark it final so that it can be referenced in that class. This is actually a pretty common use case for marking arguments final. The other … Read more
Effectively final means that it is never changed after getting the initial value. A simple example: public void myMethod() { int a = 1; System.out.println(“My effectively final variable has value: ” + a); } Here, a is not declared final, but it is considered effectively final since it is never changed. Starting with Java 8, … Read more
http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28 says that Simple names (§6.5.6.1) that refer to constant variables (§4.12.4) are constant expressions. http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28 also says: A constant expression is an expression denoting a value of primitive type or a String that does not complete abruptly and is composed using only the following: Literals of primitive type and literals of type String (§3.10.1, … Read more
You can’t, basically. It looks like the developer who designed Bar intended for it not to be extended – so you can’t extend it. You’ll need to look for a different design – possibly one which doesn’t use inheritance so much… assuming you can’t change the existing code, of course. (It’s hard to give more … Read more