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Home/ Questions/Q 8689487
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T23:39:36+00:00 2026-06-12T23:39:36+00:00

I’m developing an Android 3.1 tablet application and I have a Fragment ( android.support.v4.app.Fragment

  • 0

I’m developing an Android 3.1 tablet application and I have a Fragment (android.support.v4.app.Fragment) with two galleries.

I use this code as for gallery (I use the same adapter for both Galleries):

public class ImageGalleryAdapter extends BaseAdapter
{
    private ArrayList<String> mImagesPath;
    private Context mContext;
    private ImageView.ScaleType mScaleType;
    private int mWidth;
    private int mHeight;

    public ArrayList<String> getmImagesPath()
    {
        return mImagesPath;
    }

    public void setmImagesPath(ArrayList<String> mImagesPath)
    {
        this.mImagesPath = mImagesPath;
    }

    public void addImage(String imagePath)
    {
        mImagesPath.add(imagePath);
    }

    public ImageGalleryAdapter(Context context, ImageView.ScaleType scaleType, int width, int height)
    {
        mContext = context;
        mWidth = width;
        mHeight = height;
        mScaleType = scaleType;

        mImagesPath = new ArrayList<String>();
    }

    @Override
    public int getCount()
    {
        return mImagesPath.size();
    }

    @Override
    public Object getItem(int position)
    {
        return position;
    }

    @Override
    public long getItemId(int position)
    {
        return position;
    }

    @Override
    public View getView(int position, View convertView, ViewGroup parent)
    {
        // Get a View to display image data                     
        ImageView imageView;
        if (convertView == null) { // if it's not recycled, initialize some attributes
            imageView = new ImageView(mContext);
            imageView.setScaleType(ImageView.ScaleType.CENTER_CROP);
            imageView.setLayoutParams(new GridView.LayoutParams(
                    LayoutParams.MATCH_PARENT, LayoutParams.MATCH_PARENT));
        } else {
            imageView = (ImageView) convertView;
        }

        File imageFile = new  File(mImagesPath.get(position));
        if(imageFile.exists())
        {
            BitmapFactory.Options options = new Options();
            options.inSampleSize = 32;
            imageView = new ImageView(this.mContext);
            Bitmap myBitmap = BitmapFactory.decodeFile(imageFile.getAbsolutePath(), options);
            imageView.setImageBitmap(myBitmap);

            imageView.setScaleType(mScaleType);

            // Set the Width & Height of the individual images
            imageView.setLayoutParams(new Gallery.LayoutParams(mWidth, mHeight));
        }

        return imageView;
    }
}

Somtimes, at the line Bitmap myBitmap = BitmapFactory.decodeFile(imageFile.getAbsolutePath());, I get an Out Of Memory error

One gallery has one image and the other one has three images.

The Out Of Memory occurs when I go to the fragment, tap on back button and then I come back to the fragment again.

I think I need to call myBitmap.recycle() but where…

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T23:39:38+00:00Added an answer on June 12, 2026 at 11:39 pm

    Try that in your else case when your convertview is not null:

    } else {
        imageView = (ImageView) convertView;
        Bitmap bm = ((BitmapDrawable) imageView.getDrawable()).getBitmap();
        if (bm != null) {
            bm.recycle();
        }
    }
    

    About how many images and at which size are we talking? Have you tried to set the adapter null in your onStop()? This could remove all references and free memory, too.

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