final keyword in method parameters

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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...

Answer 6

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.

Answer 7

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

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Answer 8

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

Answer 9

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.