I have something like this:
Bitmap.Config conf = Bitmap.Config.ARGB_8888; WeakReference<Bitmap> bm = new WeakReference<Bitmap>(Bitmap.createBitmap(3000 + 3000, 2000, conf)); Canvas canvas = new Canvas(bm.get()); canvas.drawBitmap(firstBitmap, 0, 0, null); canvas.drawBitmap(bm, firstBitmap.getWidth(), 0, null); imageView.setImageBitmap(bm);
And I apply this on more than 10 imageView's which are created one by one. Whenever I create new ImageView, I want to recycle the 'bm' object from the first one, cause this code up there, causes my heap to grow more and more and then throw OutOfMemoryError, so I do:
bm.recycle()
right after I set the Bitmap (bm) to the imageView object. This causes exception that the ImageView's canvas wants to draw recycled Bitmap.
What is the way to recycle a Bitmap that has already been put as image on ImageView?
Thanksb
In your onDestroy method you could try something like this:
ImageView imageView = (ImageView)findViewById(R.id.my_image); Drawable drawable = imageView.getDrawable(); if (drawable instanceof BitmapDrawable) { BitmapDrawable bitmapDrawable = (BitmapDrawable) drawable; Bitmap bitmap = bitmapDrawable.getBitmap(); bitmap.recycle(); }
The cast should work since setImageBitmap is implemented as
public void setImageBitmap(Bitmap bm) { setImageDrawable(new BitmapDrawable(mContext.getResources(), bm)); }
If you set the same bitmap object on all your ImageView
s, it shouldn't throw an OutOfMemoryError
. Basically, this should work:
WeakReference<Bitmap> bm = new WeakReference<Bitmap>(Bitmap.createBitmap(3000 + 3000, 2000, Bitmap.Config.ARGB_8888)); Canvas canvas = new Canvas(bm.get()); canvas.drawBitmap(firstBitmap, 0, 0, null); canvas.drawBitmap(bm, firstBitmap.getWidth(), 0, null); imageView1.setImageBitmap(bm.get()); imageView2.setImageBitmap(bm.get()); imageView3.setImageBitmap(bm.get()); imageView4.setImageBitmap(bm.get()); imageView5.setImageBitmap(bm.get()); // ...
If this doesn't work, it simply means your bitmap is too large (6000x2000 pixels is about 12 megabytes, if I calculated right). You can either:
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