Java: Anonymous inner class using a local variable

Question

How can I get the value of userId passed to this method in my anonymous inner subclass here?

public void doStuff(String userID) {
    doOtherStu         


        

Answer 1

declare the method

public void doStuff(final String userID)

The value needs to be final so that the compiler can be sure it doesn't change. This means the compiler can bind the value to the inner class at any time, without worrying about updates.

The value isn't changing in your code so this is a safe change.

Answer 2

As everyone else here has said, local variables have to be final to be accessed by an inner class.

Here is (basically) why that is... if you write the following code (long answer, but, at the bottom, you can get the short version :-):

class Main
{
    private static interface Foo
    {
        void bar();
    }

    public static void main(String[] args)
    {
        final int x;
        Foo foo;

        x = 42;
        foo = new Foo()
        {
            public void bar()
            {
                System.out.println(x);
            }
        };

        foo.bar();
    }
}

the compiler translates it roughly like this:

class Main
{
    private static interface Foo
    {
        void bar();
    }

    public static void main(String[] args)
    {
        final int x;
        Foo foo;

        x = 42;

        class $1
            implements Foo
        {
            public void bar()
            {
                System.out.println(x);
            }
        }

        foo = new $1();
        foo.bar();
    }
}

and then this:

class Main
{
    private static interface Foo
    {
        void bar();
    }

    public static void main(String[] args)
    {
        final int x;
        Foo foo;

        x = 42;
        foo = new $1(x);
        foo.bar();
    }

    private static class $1
        implements Foo
    {
        private final int x;

        $1(int val)
        {
           x = val;
        }

        public void bar()
        {
            System.out.println(x);
        }
    }
}

and finally to this:

class Main
{
    public static void main(String[] args) 
    {
        final int x;
        Main$Foo foo;

        x = 42;
        foo = new Main$1(x);
        foo.bar();
    }
}

interface Main$Foo
{
    void bar();
}

class Main$1
    implements Main$Foo
{
    private final int x;

    Main$1(int val)
    {
       x = val;
    }

    public void bar()
    {
        System.out.println(x);
    }
}

The important one is where it adds the constructor to $1. Imagine if you could do this:

class Main
{
    private static interface Foo
    {
        void bar();
    }

    public static void main(String[] args)
    {
        int x;
        Foo foo;

        x = 42;
        foo = new Foo()
        {
            public void bar()
            {
                System.out.println(x);
            }
        };

        x = 1;

        foo.bar();
    }
}

You would expect that foo.bar() would print out 1 but it would actually print out 42. By requiring local variables to be final this confusing situation cannot arise.

Answer 3

What's the problem with making it final as in

public void doStuff (final String userID)

Answer 4

In Java 8, this has changed a little bit. You can now access variables that are effectively final. Relevant snippet and example from the Oracle documentation (emphasis mine):

However, starting in Java SE 8, a local class can access local variables and parameters of the enclosing block that are final or effectively final. A variable or parameter whose value is never changed after it is initialized is effectively final. For example, suppose that the variable numberLength is not declared final, and you add the highlighted assignment statement in the PhoneNumber constructor:

PhoneNumber(String phoneNumber) {
    numberLength = 7; // From Kobit: this would be the highlighted line
    String currentNumber = phoneNumber.replaceAll(
        regularExpression, "");
    if (currentNumber.length() == numberLength)
        formattedPhoneNumber = currentNumber;
    else
        formattedPhoneNumber = null;
}

Because of this assignment statement, the variable numberLength is not effectively final anymore. As a result, the Java compiler generates an error message similar to "local variables referenced from an inner class must be final or effectively final" where the inner class PhoneNumber tries to access the numberLength variable:

if (currentNumber.length() == numberLength)

Starting in Java SE 8, if you declare the local class in a method, it can access the method's parameters. For example, you can define the following method in the PhoneNumber local class:

public void printOriginalNumbers() {
    System.out.println("Original numbers are " + phoneNumber1 +
        " and " + phoneNumber2);
}

The method printOriginalNumbers accesses the parameters phoneNumber1 and phoneNumber2 of the method validatePhoneNumber

Answer 5

Sure you can assign it as final - just put that keyword in the declaration of the parameter:

public void doStuff(final String userID) {
   ...

I'm not sure what you meant about it being a variable with an unknown value; all that final means is that once a value is assigned to the variable, it cannot be re-assigned. Since you're not changing the value of the userID within your method, there's no problem making it final in this case.