Java Static Class Variable Initialisation Efficiency

Question

I have the following:

public class ExampleObject extends GridObject { 
  private static Context c; 
  private static final String name = "Example Object"; 
  private static Bitmap skin = BitmapFactory.decodeResource(c.getResources(), R.drawable.defaultObject );
  private static float x,y; 

  public ExampleObject(Context c, float x, float y) { 
    this.c = c; 
    this.x = x; 
    this.y = y; 
  }
}

The class has 3 static class members, The image is a decoded bitmap, I want it to be decoded once and once only for use on ALL instances of this object.

In it's current state is this achieved? or is it decoded every time an instance of this class is created?

How should it be done?

You will get null pointer exception as c wont be initialized when decoding. — SMA
– SMA, Commented Jan 24, 2015 at 16:07
See javaworld.com/article/2073352/core-java/simply-singleton.html — Computer says 'no'--SOooooo
– Computer says 'no'--SOooooo, Commented Jan 24, 2015 at 16:07

Thorn G · Accepted Answer · 2015-01-24 16:14:45Z

2

A static field will only be initialized once; this is guaranteed by the JLS.

However, the decodeResource method will be called when the class is initialized, at which point your Context is null, so it will fail. You'll need something more complex if you want a static field which is only initialized once; something a bit closer to a singleton.

public class ExampleObject extends GridObject { 
  private static final String name = "Example Object"; 
  private static Bitmap skin;
  // = BitmapFactory.decodeResource(c.getResources(), R.drawable.defaultObject );
  private static float x,y; 

  public ExampleObject(Context c, float x, float y) { 
    synchronized(ExampleObject.class) {
      if(skin == null) {
        skin = BitmapFactory.decodeResource(c.getResources(), R.drawable.defaultObject);
      }
    }
    this.x = x; 
    this.y = y; 
  }
}

edited Jan 24, 2015 at 16:14

answered Jan 24, 2015 at 16:06

Thorn G

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5 Comments

Adam Pitt Over a year ago

public class ExampleObject extends GridObject { private static Context c; private static final String name = "Example Object"; private static Bitmap skin = BitmapFactory.decodeResource( c.getResources(), R.drawable.defaultObject ); private static float x,y; public ExampleObject(Context c, float x, float y) { this.c = c; this.x = x; this.y = y; } It's passed by constructor

Adam Pitt Over a year ago

so I would need to decode it within the constructor, but this would call it every time I create an instance of the object

Adam Pitt Over a year ago

Or will the JLS see that's it's static and already defined therefore ignore it?

Thorn G Over a year ago

You would need to check if the field was already assigned yourself. I will expand my answer with an example.

Adam Pitt Over a year ago

I could just store the r.drawable.defaultObject as an int and then decode in the onDraw method, however this decodes the image multiple times, exactly what I'm trying to avoid

frogatto · Accepted Answer · 2015-01-24 16:13:07Z

1

You can achieve your intended behavior by the following class definition.

public class ExampleObject extends GridObject {
    private static Bitmap skin;

    public static Bitmap getSkin(Context c){
        if(skin == null){
            skin = BitmapFactory.decodeResource( c.getResources(), R.drawable.defaultObject );
        }
        return skin;
    }
}

answered Jan 24, 2015 at 16:13

frogatto

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2 Comments

Adam Pitt Over a year ago

perfect thanks, so trivial I'm such a fool sometimes! Thank-you very much much appreciated

Simon Over a year ago

A singleton is a better solution.

AADProgramming · Accepted Answer · 2015-01-24 16:14:42Z

1

Static variables are initialized only once and A single copy of it is to be shared by all instances of the class.

This single initialization procedure is run automatically, one time only, when the class is first loaded.

You may want to use "static block" to initialize your classes's static fields. For example:

// start of static block 
    static {
        //initialize your static fields
        System.out.println("static block called ");
    }
    // end of static block

answered Jan 24, 2015 at 16:14

AADProgramming

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