When passing a final object (A String in the code below) it shows up as null when printed from the anonymous inner class. However, when a final value type or straight final String is passed in, its value is correctly displayed. What does final really mean in context of the anonymous inner class and why is the object passed null?
public class WeirdInners
{
public class InnerThing
{
public InnerThing()
{
print();
}
public void print(){
}
}
public WeirdInners()
{
final String aString = "argh!".toString();
final String bString = "argh!";
System.out.println(aString);
System.out.println(bString);
InnerThing inner =new InnerThing(){
public void print()
{
System.out.println("inner"+aString); // When called from constructor, this value is null.
System.out.println("inner"+bString); // This value is correctly printed.
}
};
inner.print();
}
public static void main(String[] args)
{
WeirdInners test1 = new WeirdInners();
}
}
This is very strange behavior to me because the expectation is that the String is an object, why does calling toString() change things?
Other info: this behavior is only observed using Java 1.4, not in Java 5. Any suggestions on a workaround? Not calling toString() on an existing String is fair enough, but since this is only an example, it has real world implications if I perform it on a non-String object.
If you check the section on compile-time constants in the JLS, you'll see that calling .toString() does make a difference. As does nonsense like prefixing with false?null+"":.
What is important here is the relative ordering of setting the fields that are closed over and the constructor. If you use -target 1.4 or later (which is not the default in 1.4!) then the fields will be copied before calling the super. With the spec before 1.3 this was illegal bytecode.
As is often the case in these cases, javap -c is useful to see what the javac compiler is doing. The spec is useful for understanding why (should you have sufficient patience).
My guess would be that you trigger undefined behaviour when the constructor of InnerThing() passes (implicitly) its this to the print method of an anonymous subclass of InnerThing while the object is not fully constructed. This this in turn relies on an implicit reference to the this of WierdInners.
Calling .toString() moves initialisation of aString from compile time to runtime. Why the undefined behaviour differs between Java 1.4 and 1.5 is a JVM implementation detail probably.
It is dangerous to invoke overridden methods from a superclass constructor, as they will be called before the subclass-part of the object is initialized.
Moreover, an inner class accessing final variables of an enclosing scope will actually access copies of those variables (that's why they need to be final), and these copied fields reside in the anonymous subclass.
I suspect the reason bString is treated differently is that its value is known at compile time, which allows the compiler to inline the field access in the subclass, making the time that field is initialized irrelevant.
Related
I would appreciate help in understanding the following from 'Java Concurrency in Practice':
Calling an overrideable instance method(one that is neither
private nor final) from the constructor can also allow the
this reference to escape.
Does 'escape' here simply mean that we may probably be calling an instance method,before the instance is fully constructed?
I do not see 'this' escaping the scope of the instance in any other way.
How does 'final' prevent this from happening?Is there some aspect of 'final' in instance creation that I am missing?
It means calling code outside the class, and passing this.
That code will assume that the instance is fully initialized, and may break if it isn't.
Similarly, your class might assume that some methods will only be called after the instance is fully initialized, but the external code is likely to break those assumptions.
final methods cannot be overridden, so you can trust them to not pass this around.
If you call any non-final method in the constructor for a non-final class, a derived class might override that method and pass this anywhere.
Even when you call final methods, you still need to make sure that they are safely written – that they do not pass this anywhere, and that themselves don't call any non-final methods.
"Escape" means that a reference to the partially-constructed this object might be passed to some other object in the system. Consider this scenario:
public Foo {
public Foo() {
setup();
}
protected void setup() {
// do stuff
}
}
public Bar extends Foo implements SomeListener {
#Override protected void setup() {
otherObject.addListener(this);
}
}
The problem is that the new Bar object is being registered with otherObject before its construction is completed. Now if otherObject starts calling methods on barObject, fields might not have been initialized, or barObject might otherwise be in an inconsistent state. A reference to the barObject (this to itself) has "escaped" into the rest of the system before it's ready.
Instead, if the setup() method is final on Foo, the Bar class can't put code in there that will make the object visible before the Foo constructor finishes.
I believe the example is something like
public class Foo {
public Foo() {
doSomething();
}
public void doSomething() {
System.out.println("do something acceptable");
}
}
public class Bar extends Foo {
public void doSomething() {
System.out.println("yolo");
Zoom zoom = new Zoom(this); // at this point 'this' might not be fully initialized
}
}
Because the super constructor is always called first (either implicitly or explicitly), the doSomething will always get called for a child class. Because the above method is neither final nor private, you can override it in a child class and do whatever you want, which may conflict with what Foo#doSomething() was meant to do.
Per secure coding
Example BAD code:
final class Publisher {
public static volatile Publisher published;
int num;
Publisher(int number) {
published = this;
// Initialization
this.num = number;
// ...
}
}
If an object's initialization (and consequently, its construction) depends on a security check within the constructor, the security check can be bypassed when an untrusted caller obtains the partially initialized instance. See rule OBJ11-J. Be wary of letting constructors throw exceptions for more information.
final class Publisher {
public static Publisher published;
int num;
Publisher(int number) {
// Initialization
this.num = number;
// ...
published = this;
}
}
Because the field is nonvolatile and nonfinal, the statements within
the constructor can be reordered by the compiler in such a way that
the this reference is published before the initialization statements
have executed.
Correct code:
final class Publisher {
static volatile Publisher published;
int num;
Publisher(int number) {
// Initialization
this.num = number;
// ...
published = this;
}
}
The this reference is said to have escaped when it is made available
beyond its current scope. Following are common ways by which the this
reference can escape:
Returning this from a non-private, overridable method that is invoked from the constructor of a class whose object is being
constructed. (For more information, see rule MET05-J. Ensure that
constructors do not call overridable methods.)
Returning this from a nonprivate method of a mutable class, which allows the caller to manipulate the object's state indirectly. This
commonly occurs in method-chaining implementations; see rule VNA04-J.
Ensure that calls to chained methods are atomic for more information.
Passing this as an argument to an alien method invoked from the constructor of a class whose object is being constructed.
Using inner classes. An inner class implicitly holds a reference to the instance of its outer class unless the inner class is declared
static.
Publishing by assigning this to a public static variable from the constructor of a class whose object is being constructed.
Throwing an exception from a constructor. Doing so may cause code to be vulnerable to a finalizer attack; see rule OBJ11-J. Be wary of
letting constructors throw exceptions for more information.
Passing internal object state to an alien method. This enables the method to retrieve the this reference of the internal member object.
This rule describes the potential consequences of allowing the this
reference to escape during object construction, including race
conditions and improper initialization. For example, declaring a field
final ordinarily ensures that all threads see the field in a fully
initialized state; however, allowing the this reference to escape
during object construction can expose the field to other threads in an
uninitialized or partially initialized state. Rule TSM03-J. Do not
publish partially initialized objects, which describes the guarantees
provided by various mechanisms for safe publication, relies on
conformance to this rule. Consequently, programs must not allow the
this reference to escape during object construction.
In general, it is important to detect cases in which the this
reference can leak out beyond the scope of the current context. In
particular, public variables and methods should be carefully
scrutinized.
I have been programming java professionally for more than ten years. This is one of the weirdest bugs I've ever tried to track down. I have a private member, I initialize it and then it changes to null all by itself.
public class MyObject extends MyParent
{
private SomeOtherClass member = null;
public MyObject()
{
super();
}
public void callbackFromParentInit()
{
member = new SomeOtherClass();
System.out.println("proof member initialized: " + member);
}
public SomeOtherClass getMember()
{
System.out.println("in getMember: " + member);
return member;
}
}
Output:
proof member initialized: SomeOtherClass#2a05ad6d
in getMember: null
If you run this code, obviously it will work properly. In my actual code there are only these three occurrences (five if you count the printlns) in this exact pattern.
Have I come across some bug in the JVM? Unless I'm wrong, the parent class can't interfere with a private member, and no matter what I put between the lines of code I've shown you, I can't change the value of member without using the identifier "member".
This happens because of the order in which member variables are initialized and constructors are called.
You are calling callbackFromParentInit() from the constructor of the superclass MyParent.
When this method is called, it will set member. But after that, the subclass part of the object initialization is performed, and the initializer for member is executed, which sets member to null.
See, for example:
What's wrong with overridable method calls in constructors?
State of Derived class object when Base class constructor calls overridden method in Java
Using abstract init() function in abstract class's constructor
In what order constructors are called and fields are initialized is described in paragraph 12.5 of the Java Language Specification.
Assignment of null to field member happens after executing parent constructor.
The fix is to change:
private SomeOtherClass member = null;
to:
private SomeOtherClass member;
Never, never ever call a non final method from the superclass' constructor.
It's considered bad practice, precisely because it can lead to nasty, hard-to-debug errors like the one you're suffering.
Perform initialization of a class X within X's constructor. Don't rely on java's initialization order for hierarchies. If you can't initialize the class property i.e. because it has dependencies, use either the builder or the factory pattern.
Here, the subclass is resetting the attribute member to null, due to superclass and subclass constructors and initializer block execution order, in which, as already mentioned, you shouldn't rely.
Please refer to this related question for concepts regarding constructors, hierarchies and implicit escaping of the this reference.
I can only think about sticking to a (maybe incomplete) set of rules/principles to avoid this problem and others alike:
Only call private methods from within the constructor
If you like adrenaline and want to call protected methods from within the constructor, do it, but declare these methods as final, so that they cannot be overriden by subclasses
Never create inner classes in the constructor, either anonymous, local, static or non-static
In the constructor, don't pass this directly as an argument to anything
Avoid any transitive combination of the rules above, i.e. don't create an anonymous inner class in a private or protected final method that is invoked from within the constructor
Use the constructor to just construct an instance of the class, and let it only initialize attributes of the class, either with default values or with provided arguments
This question already has answers here:
Why should I use the keyword "final" on a method parameter in Java?
(12 answers)
Closed 7 years ago.
I often encounter methods which look like the following:
public void foo(final String a, final int[] b, final Object1 c){
}
What happens if this method is called without passing it final parameters. i.e. an Object1 that is later changed (so is not declared as final) can be passed to this method just fine
Java always makes a copy of parameters before sending them to methods. This means the final doesn't mean any difference for the calling code. This only means that inside the method the variables can not be reassigned.
Note that if you have a final object, you can still change the attributes of the object. This is because objects in Java really are pointers to objects. And only the pointer is copied (and will be final in your method), not the actual object.
There is a circumstance where you're required to declare it final —otherwise it will result in compile error—, namely passing them through into an anonymous class or a lambda. Here's a basic example using an anonymous class:
public FileFilter createFileExtensionFilter(final String extension) {
FileFilter fileFilter = new FileFilter() {
public boolean accept(File file) {
return file.getName().endsWith(extension);
}
};
// Imagine what would happen when we're allowed to change extension here?
// extension = "foo";
return fileFilter;
}
And here's the exact same example in lambda flavor:
public FileFilter createFileExtensionFilter(final String extension) {
FileFilter fileFilter = file -> file.getName().endsWith(extension);
// Imagine what would happen when we're allowed to change extension here?
// extension = "foo";
return fileFilter;
}
Removing the final modifier would result in compile error, because it isn't guaranteed anymore that the value is a runtime constant. Changing the value after creation of the anonymous class or lambda would namely cause the instance of the anonymous class or lambda to behave different after the moment of creation.
Java is only pass-by-value. (or better - pass-reference-by-value)
So the passed argument and the argument within the method are two different handlers pointing to the same object (value).
Therefore if you change the state of the object, it is reflected to every other variable that's referencing it. But if you re-assign a new object (value) to the argument, then other variables pointing to this object (value) do not get re-assigned.
The final keyword on a method parameter means absolutely nothing to the caller. It also means absolutely nothing to the running program, since its presence or absence doesn't change the bytecode. It only ensures that the compiler will complain if the parameter variable is reassigned within the method. That's all. But that's enough.
Some programmers (like me) think that's a very good thing and use final on almost every parameter. It makes it easier to understand a long or complex method (though one could argue that long and complex methods should be refactored.) It also shines a spotlight on method parameters that aren't marked with final.
Consider this implementation of foo():
public void foo(final String a) {
SwingUtilities.invokeLater(new Runnable() {
public void run() {
System.out.print(a);
}
});
}
Because the Runnable instance would outlive the method, this wouldn't compile without the final keyword -- final tells the compiler that it's safe to take a copy of the reference (to refer to it later). Thus, it's the reference that's considered final, not the value. In other words: As a caller, you can't mess anything up...
final means you can't change the value of that variable once it was assigned.
Meanwhile, the use of final for the arguments in those methods means it won't allow the programmer to change their value during the execution of the method.
This only means that inside the method the final variables can not be reassigned.
final keyword in the method input parameter is not needed. Java creates a copy of the reference to the object, so putting final on it doesn't make the object final but just the reference, which doesn't make sense
If you declare any parameter as final, you cannot change the value of it.
class Bike11 {
int cube(final int n) {
n=n+2;//can't be changed as n is final
n*n*n;
}
public static void main(String args[]) {
Bike11 b=new Bike11();
b.cube(5);
}
}
Output: Compile Time Error
For more details, please visit my blog: http://javabyroopam.blogspot.com
Strings are immutable, so actully you can't change the String afterwards (you can only make the variable that held the String object point to a different String object).
However, that is not the reason why you can bind any variable to a final parameter. All the compiler checks is that the parameter is not reassigned within the method. This is good for documentation purposes, arguably good style, and may even help optimize the byte code for speed (although this seems not to do very much in practice).
But even if you do reassign a parameter within a method, the caller doesn't notice that, because java does all parameter passing by value. After the sequence
a = someObject();
process(a);
the fields of a may have changed, but a is still the same object it was before. In pass-by-reference languages this may not be true.
GIVEN:
class A
{
String s = "A";
}
class B extends A
{
String s = "B";
}
public class C
{
public static void main(String[] args){ new C().go();}
void go()
{
A a = new B();
System.out.println(a.s);
}
}
Question:
What are the mechanics behind JVM when this code is run? How come a.s prints back as "A".
Field references are not subject to polymorphism, so at compile time the compiler is referencing A's field because your local variable is of type A.
In other words, the field behavior is like the Java overloading behavior on methods, not the Java overriding behavior.
You probably expect fields to be overridden like method, with dynamic dispatch based on the runtime type of the object.
That's not how Java works. Fields are not overridden, they are hidden. That means an object of class B has two fields named "s", but which of them is accessed depends on the context.
As for why this is so: it wouldn't really make sense to override fields, since there is no useful way to make it work when the types are different, and simply no point when the type is the same (as you can just use the superclass field). Personally, I think it should simply be a compiler error.
This isn't polymorphism (as tagged).
Java has virtual methods, not virtual member variables - i.e. you don't override a property - you hide it.
Although member variables are inherited from a base class, they are not invoked polymorphically (i.e dynamic invocation does not apply to member variables).
So, a.s will refer to the member in the base class and not the derived class.
Having said that, the code is not following OO principles. The members of a class need to be private/protected (not public or default) depending on the business use case and you need to provide public methods to get and set the values of the member.
I can't understand where the final keyword is really handy when it is used on method parameters.
If we exclude the usage of anonymous classes, readability and intent declaration then it seems almost worthless to me.
Enforcing that some data remains constant is not as strong as it seems.
If the parameter is a primitive then it will have no effect since the parameter is passed to the method as a value and changing it will have no effect outside the scope.
If we are passing a parameter by reference, then the reference itself is a local variable and if the reference is changed from within the method, that would not have any effect from outside of the method scope.
Consider the simple test example below.
This test passes although the method changed the value of the reference given to it, it has no effect.
public void testNullify() {
Collection<Integer> c = new ArrayList<Integer>();
nullify(c);
assertNotNull(c);
final Collection<Integer> c1 = c;
assertTrue(c1.equals(c));
change(c);
assertTrue(c1.equals(c));
}
private void change(Collection<Integer> c) {
c = new ArrayList<Integer>();
}
public void nullify(Collection<?> t) {
t = null;
}
Stop a Variable’s Reassignment
While these answers are intellectually interesting, I've not read the short simple answer:
Use the keyword final when you want the compiler to prevent a
variable from being re-assigned to a different object.
Whether the variable is a static variable, member variable, local variable, or argument/parameter variable, the effect is entirely the same.
Example
Let’s see the effect in action.
Consider this simple method, where the two variables (arg and x) can both be re-assigned different objects.
// Example use of this method:
// this.doSomething( "tiger" );
void doSomething( String arg ) {
String x = arg; // Both variables now point to the same String object.
x = "elephant"; // This variable now points to a different String object.
arg = "giraffe"; // Ditto. Now neither variable points to the original passed String.
}
Mark the local variable as final. This results in a compiler error.
void doSomething( String arg ) {
final String x = arg; // Mark variable as 'final'.
x = "elephant"; // Compiler error: The final local variable x cannot be assigned.
arg = "giraffe";
}
Instead, let’s mark the parameter variable as final. This too results in a compiler error.
void doSomething( final String arg ) { // Mark argument as 'final'.
String x = arg;
x = "elephant";
arg = "giraffe"; // Compiler error: The passed argument variable arg cannot be re-assigned to another object.
}
Moral of the story:
If you want to ensure a variable always points to the same object,
mark the variable final.
Never Reassign Arguments
As good programming practice (in any language), you should never re-assign a parameter/argument variable to an object other than the object passed by the calling method. In the examples above, one should never write the line arg = . Since humans make mistakes, and programmers are human, let’s ask the compiler to assist us. Mark every parameter/argument variable as 'final' so that the compiler may find and flag any such re-assignments.
In Retrospect
As noted in other answers…
Given Java's original design goal of helping programmers to avoid dumb mistakes such as reading past the end of an array, Java should have been designed to automatically enforce all parameter/argument variables as 'final'. In other words, Arguments should not be variables. But hindsight is 20/20 vision, and the Java designers had their hands full at the time.
So, always add final to all arguments?
Should we add final to each and every method parameter being declared?
In theory, yes.
In practice, no.➥ Add final only when the method’s code is long or complicated, where the argument may be mistaken for a local or member variable and possibly re-assigned.
If you buy into the practice of never re-assigning an argument, you will be inclined to add a final to each. But this is tedious and makes the declaration a bit harder to read.
For short simple code where the argument is obviously an argument, and not a local variable nor a member variable, I do not bother adding the final. If the code is quite obvious, with no chance of me nor any other programmer doing maintenance or refactoring accidentally mistaking the argument variable as something other than an argument, then don’t bother. In my own work, I add final only in longer or more involved code where an argument might mistaken for a local or member variable.
#Another case added for the completeness
public class MyClass {
private int x;
//getters and setters
}
void doSomething( final MyClass arg ) { // Mark argument as 'final'.
arg = new MyClass(); // Compiler error: The passed argument variable arg cannot be re-assigned to another object.
arg.setX(20); // allowed
// We can re-assign properties of argument which is marked as final
}
record
Java 16 brings the new records feature. A record is a very brief way to define a class whose central purpose is to merely carry data, immutably and transparently.
You simply declare the class name along with the names and types of its member fields. The compiler implicitly provides the constructor, getters, equals & hashCode, and toString.
The fields are read-only, with no setters. So a record is one case where there is no need to mark the arguments final. They are already effectively final. Indeed, the compiler forbids using final when declaring the fields of a record.
public record Employee( String name , LocalDate whenHired ) // 🡄 Marking `final` here is *not* allowed.
{
}
If you provide an optional constructor, there you can mark final.
public record Employee(String name , LocalDate whenHired) // 🡄 Marking `final` here is *not* allowed.
{
public Employee ( final String name , final LocalDate whenHired ) // 🡄 Marking `final` here *is* allowed.
{
this.name = name;
whenHired = LocalDate.MIN; // 🡄 Compiler error, because of `final`.
this.whenHired = whenHired;
}
}
Sometimes it's nice to be explicit (for readability) that the variable doesn't change. Here's a simple example where using final can save some possible headaches:
public void setTest(String test) {
test = test;
}
If you forget the 'this' keyword on a setter, then the variable you want to set doesn't get set. However, if you used the final keyword on the parameter, then the bug would be caught at compile time.
Yes, excluding anonymous classes, readability and intent declaration it's almost worthless. Are those three things worthless though?
Personally I tend not to use final for local variables and parameters unless I'm using the variable in an anonymous inner class, but I can certainly see the point of those who want to make it clear that the parameter value itself won't change (even if the object it refers to changes its contents). For those who find that adds to readability, I think it's an entirely reasonable thing to do.
Your point would be more important if anyone were actually claiming that it did keep data constant in a way that it doesn't - but I can't remember seeing any such claims. Are you suggesting there's a significant body of developers suggesting that final has more effect than it really does?
EDIT: I should really have summed all of this up with a Monty Python reference; the question seems somewhat similar to asking "What have the Romans ever done for us?"
Let me explain a bit about the one case where you have to use final, which Jon already mentioned:
If you create an anonymous inner class in your method and use a local variable (such as a method parameter) inside that class, then the compiler forces you to make the parameter final:
public Iterator<Integer> createIntegerIterator(final int from, final int to)
{
return new Iterator<Integer>(){
int index = from;
public Integer next()
{
return index++;
}
public boolean hasNext()
{
return index <= to;
}
// remove method omitted
};
}
Here the from and to parameters need to be final so they can be used inside the anonymous class.
The reason for that requirement is this: Local variables live on the stack, therefore they exist only while the method is executed. However, the anonymous class instance is returned from the method, so it may live for much longer. You can't preserve the stack, because it is needed for subsequent method calls.
So what Java does instead is to put copies of those local variables as hidden instance variables into the anonymous class (you can see them if you examine the byte code). But if they were not final, one might expect the anonymous class and the method seeing changes the other one makes to the variable. In order to maintain the illusion that there is only one variable rather than two copies, it has to be final.
I use final all the time on parameters.
Does it add that much? Not really.
Would I turn it off? No.
The reason: I found 3 bugs where people had written sloppy code and failed to set a member variable in accessors. All bugs proved difficult to find.
I'd like to see this made the default in a future version of Java. The pass by value/reference thing trips up an awful lot of junior programmers.
One more thing.. my methods tend to have a low number of parameters so the extra text on a method declaration isn't an issue.
Using final in a method parameter has nothing to do with what happens to the argument on the caller side. It is only meant to mark it as not changing inside that method. As I try to adopt a more functional programming style, I kind of see the value in that.
Personally I don't use final on method parameters, because it adds too much clutter to parameter lists.
I prefer to enforce that method parameters are not changed through something like Checkstyle.
For local variables I use final whenever possible, I even let Eclipse do that automatically in my setup for personal projects.
I would certainly like something stronger like C/C++ const.
Since Java passes copies of arguments I feel the relevance of final is rather limited. I guess the habit comes from the C++ era where you could prohibit reference content from being changed by doing a const char const *. I feel this kind of stuff makes you believe the developer is inherently stupid as f*** and needs to be protected against truly every character he types. In all humbleness may I say, I write very few bugs even though I omit final (unless I don't want someone to override my methods and classes). Maybe I'm just an old-school dev.
Short answer: final helps a tiny bit but... use defensive programming on the client side instead.
Indeed, the problem with final is that it only enforces the reference is unchanged, gleefully allowing the referenced object members to be mutated, unbeknownst to the caller. Hence the best practice in this regard is defensive programming on the caller side, creating deeply immutable instances or deep copies of objects that are in danger of being mugged by unscrupulous APIs.
I never use final in a parameter list, it just adds clutter like previous respondents have said. Also in Eclipse you can set parameter assignment to generate an error so using final in a parameter list seems pretty redundant to me.
Interestingly when I enabled the Eclipse setting for parameter assignment generating an error on it caught this code (this is just how I remember the flow, not the actual code. ) :-
private String getString(String A, int i, String B, String C)
{
if (i > 0)
A += B;
if (i > 100)
A += C;
return A;
}
Playing devil's advocate, what exactly is wrong with doing this?
One additional reason to add final to parameter declarations is that it helps to identify variables that need to be renamed as part of a "Extract Method" refactoring. I have found that adding final to each parameter prior to starting a large method refactoring quickly tells me if there are any issues I need to address before continuing.
However, I generally remove them as superfluous at the end of the refactoring.
Follow up by Michel's post. I made myself another example to explain it. I hope it could help.
public static void main(String[] args){
MyParam myParam = thisIsWhy(new MyObj());
myParam.setArgNewName();
System.out.println(myParam.showObjName());
}
public static MyParam thisIsWhy(final MyObj obj){
MyParam myParam = new MyParam() {
#Override
public void setArgNewName() {
obj.name = "afterSet";
}
#Override
public String showObjName(){
return obj.name;
}
};
return myParam;
}
public static class MyObj{
String name = "beforeSet";
public MyObj() {
}
}
public abstract static class MyParam{
public abstract void setArgNewName();
public abstract String showObjName();
}
From the code above, in the method thisIsWhy(), we actually didn't assign the [argument MyObj obj] to a real reference in MyParam. In instead, we just use the [argument MyObj obj] in the method inside MyParam.
But after we finish the method thisIsWhy(), should the argument(object) MyObj still exist?
Seems like it should, because we can see in main we still call the method showObjName() and it needs to reach obj. MyParam will still use/reaches the method argument even the method already returned!
How Java really achieve this is to generate a copy also is a hidden reference of the argument MyObj obj inside the MyParam object ( but it's not a formal field in MyParam so that we can't see it )
As we call "showObjName", it will use that reference to get the corresponding value.
But if we didn't put the argument final, which leads a situation we can reassign a new memory(object) to the argument MyObj obj.
Technically there's no clash at all! If we are allowed to do that, below will be the situation:
We now have a hidden [MyObj obj] point to a [Memory A in heap] now live in MyParam object.
We also have another [MyObj obj] which is the argument point to a [Memory B in heap] now live in thisIsWhy method.
No clash, but "CONFUSING!!" Because they are all using the same "reference name" which is "obj".
To avoid this, set it as "final" to avoid programmer do the "mistake-prone" code.