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java – How to initialize a static SparseArray

Posted by: admin June 15, 2020 Leave a comment

Questions:

How can I initialize a static, unmodifiable instance of android.util.SparseArray?

How to&Answers:

You cannot do what you are attempting to. At least, not how you are attempting to do it. There is no implementation of SparseArray that is unmodifiable.

However, you could create one. Here’s how:

  • Create a class, say CustomSparseArray<E>, and have it extend SparseArray.
  • Override all methods that change the elements in the array, and replace them with something like this:

    @Override
    public void append(int key, E value) {
        if (mLocked)
            return; // Maybe throw an exception
        super.append(key, value);
    }
    
  • Then, add in a member variable to the class, boolean mLocked = false;.
  • Next, you need a method like the following:

    public void lock() { mLocked = true; }
    
  • Lastly, implement your static variable using a method similar to in the other post:

    public class Test {
        private static final CustomSparseArray<Integer> myArray;
        static {
            myArray = new CustomSparseArray<Integer>();
            myArray.append(1, 1);
            myArray.append(2, 5);
            myArray.lock();
        }
    }
    

Then you have an unmodifiable SparseArray in your static variable myArray.

Answer:

Here is a better way using an anonymous class:

static final SparseIntArray myArray = new SparseIntArray() {
    {
        append(1, 2);
        append(10, 20);
    }
};

Answer:

This works for me:

static final SparseIntArray CMyArray = new SparseIntArray();
static {
    CMyArray.append(2, 4);
    CMyArray.append(8, 3);
    CMyArray.append(255, 1);
}

as per: https://docs.oracle.com/javase/tutorial/java/javaOO/initial.html