Warning shows when i use Hash Map In android(Use new SparseArray<String>)

android hashmap
Naveen Kumar picture Naveen Kumar · Feb 9, 2013 · Viewed 20k times · Source

I am new to developing in android. In my android app I'm using HashMap, but I'm getting a warning: 

**"Use new SparseArray<String>(...) instead for better performance"**

What does this mean, and how can I use SparseArray<String> instead?

Answer

Bhavesh Patadiya picture Bhavesh Patadiya · Feb 9, 2013

Use new SparseArray<String>(...) instead for better performance

You are getting this warning because of reason described here.

SparseArrays map integers to Objects. Unlike a normal array of Objects, there can be gaps in the indices. It is intended to be more efficient than using a HashMap to map Integers to Objects.

Now

how i use SparseArray ?

You can do it by below ways:

  1. HashMap way:

    Map<Integer, Bitmap> _bitmapCache = new HashMap<Integer, Bitmap>();
private void fillBitmapCache() {
     _bitmapCache.put(R.drawable.icon, BitmapFactory.decodeResource(getResources(), R.drawable.icon));
     _bitmapCache.put(R.drawable.abstrakt, BitmapFactory.decodeResource(getResources(), R.drawable.abstrakt));
     _bitmapCache.put(R.drawable.wallpaper, BitmapFactory.decodeResource(getResources(), R.drawable.wallpaper));
     _bitmapCache.put(R.drawable.scissors, BitmapFactory.decodeResource(getResources(), 
 }

Bitmap bm = _bitmapCache.get(R.drawable.icon);

  2. SparseArray way:

    SparseArray<Bitmap> _bitmapCache = new SparseArray<Bitmap>();
private void fillBitmapCache() {
     _bitmapCache.put(R.drawable.icon, BitmapFactory.decodeResource(getResources(), R.drawable.icon));
     _bitmapCache.put(R.drawable.abstrakt, BitmapFactory.decodeResource(getResources(), R.drawable.abstrakt));
     _bitmapCache.put(R.drawable.wallpaper, BitmapFactory.decodeResource(getResources(), R.drawable.wallpaper));
     _bitmapCache.put(R.drawable.scissors, BitmapFactory.decodeResource(getResources(), 
 }

Bitmap bm = _bitmapCache.get(R.drawable.icon);

Hope it Will Help.

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