I'm using an LRUCache
to cache bitmaps which are stored on the file system. I built the cache based on the examples here: http://developer.android.com/training/displaying-bitmaps/cache-bitmap.html
The problem is that I'm seeing OutOfMemory crashes frequently while using the app. I believe that when the LRUCache evicts an image to make room for another one, the memory is not being freed.
I added a call to Bitmap.recycle() when an image is evicted:
// use 1/8 of the available memory for this memory cache final int cacheSize = 1024 * 1024 * memClass / 8; mImageCache = new LruCache<String, Bitmap>(cacheSize) { @Override protected int sizeOf(String key, Bitmap bitmap) { return bitmap.getByteCount(); } @Override protected void entryRemoved(boolean evicted, String key, Bitmap oldBitmap, Bitmap newBitmap) { oldBitmap.recycle(); oldBitmap = null; } };
This fixes the crashes, however it also results in images sometimes not appearing in the app (just a black space where the image should be). Any time that occurs I see this message in my Logcat: Cannot generate texture from bitmap
.
A quick google search reveals that this is happening because the image which is displaying has been recycled.
So what is happening here? Why are recycled images still in the LRUCache if I'm only recycling them after they've been removed? What is the alternative for implementing a cache? The Android docs clearly state that LRUCache is the way to go, but they do not mention the need to recycle bitmaps or how to do so.
RESOLVED: In case its useful to anyone else, the solution to this problem as suggested by the accepted answer is to NOT do what I did in the code example above (don't recycle the bitmaps in the entryRemoved()
call).
Instead, when you're finished with an ImageView (such as onPause()
in an activity, or when a view is recycled in an adapter) check if the bitmap is still in the cache (I added a isImageInCache()
method to my cache class) and, if it's not, then recycle the bitmap. Otherwise, leave it alone. This fixed my OutOfMemory
exceptions and prevented recycling bitmaps which were still being used.
If you're displaying large amounts of bitmap data in your app, you're likely to run into OutOfMemoryError errors. The recycle() method allows an app to reclaim memory as soon as possible. Caution: You should use recycle() only when you are sure that the bitmap is no longer being used.
Choose the most appropriate decode method based on your image data source. These methods attempt to allocate memory for the constructed bitmap and therefore can easily result in an OutOfMemory exception. Each type of decode method has additional signatures that let you specify decoding options via the BitmapFactory.
I believe that when the LRUCache evicts an image to make room for another one, the memory is not being freed.
It won't be, until the Bitmap
is recycled or garbage-collected.
A quick google search reveals that this is happening because the image which is displaying has been recycled.
Which is why you should not be recycling there.
Why are recycled images still in the LRUCache if I'm only recycling them after they've been removed?
Presumably, they are not in the LRUCache
. They are in an ImageView
or something else that is still using the Bitmap
.
What is the alternative for implementing a cache?
For the sake of argument, let's assume you are using the Bitmap
objects in ImageView
widgets, such as in rows of a ListView
.
When you are done with a Bitmap
(e.g., row in a ListView
is recycled), you check to see if it is still in the cache. If it is, you leave it alone. If it is not, you recycle()
it.
The cache is simply letting you know which Bitmap
objects are worth holding onto. The cache has no way of knowing if the Bitmap
is still being used somewhere.
BTW, if you are on API Level 11+, consider using inBitmap
. OutOMemoryErrors
are triggered when an allocation cannot be fulfilled. Last I checked, Android does not have a compacting garbage collector, so you can get an OutOfMemoryError
due to fragmentation (want to allocate something bigger than the biggest single available block).
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