GSF ID KEY (google service framework id) as Android device unique identifier

Question

I need to uniquely identify an Android device. I read about ANDROID_ID but it seems to have problems with Android 2.2. Then there are other identifiers related to TelephonyManager, but I reckon they don't work on tablets.
So, looking for something working on every device I stumbled upon the GSF ID KEY (google service framework id). Do you guys think this is a reliable and always working solution? This is the code I found to retrieve the GSF ID KEY:

private static String getGsfAndroidId(Context context) 
{
    Uri URI = Uri.parse("content://com.google.android.gsf.gservices");
    String ID_KEY = "android_id";
    String params[] = {ID_KEY};
    Cursor c = context.getContentResolver().query(URI, null, null, params, null);
    if (!c.moveToFirst() || c.getColumnCount() < 2)
        return null;
    try 
    {
        return Long.toHexString(Long.parseLong(c.getString(1)));
    } 
    catch (NumberFormatException e) 
    {
        return null;
    }
}

Answer 1

In case someone is wondering if this method works the answer is yes, I tried it (and used it in an app I put on the Android market with thousands of downloads) and it works. Note: the GSF ID KEY changes every time the user does a factory reset or messes up with Google Services, but it was good enough for my purpose.

Answer 2

using Android Studio, I get auto-recommendations from lint. here is the code, after revised. it may solve exceptions reported by https://stackoverflow.com/users/423171/cprcrack

private static String getGsfAndroidId(Context context)
{
    Uri URI = Uri.parse("content://com.google.android.gsf.gservices");
    String ID_KEY = "android_id";
    String params[] = {ID_KEY};
    Cursor c = context.getContentResolver().query(URI, null, null, params, null);
    if (c != null && (!c.moveToFirst() || c.getColumnCount() < 2)){
        if(!c.isClosed())
            c.close();
        return null;
    }

    try {
        if (c != null) {
            String result = Long.toHexString(Long.parseLong(c.getString(1)));
            if(!c.isClosed())
                c.close();
            return result;
        }else {
            return null;
        }
    } catch (NumberFormatException e) {
        if(!c.isClosed())
            c.close();
        return null;
    }
}