Java 什么'；存储位置对象的好方法是什么？

我想存储一个位置对象，并试图选择一个好的方法来做到这一点。我只有一个小对象，我需要它是私有的，所以共享引用或内部存储对我来说是最有意义的

我看到一个物体是可以移动的

这是合理的做法吗？有更好的办法吗

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谢谢您的建议。

在保存简单数据时，我更喜欢使用JSON对象，因为对于未来的维护人员来说，代码比操作字节数组更简单、更容易理解。由于对象很小，因此使用字符串而不是字节数组的大小代价并不显著

private static final String LATITUDE = "com.somepackage.name.LATITUDE";
private static final String LONGITUDE = "com.somepackage.name.LONGITUDE";

/**
 * Save a location/key pair.
 * 
 * @param key the key associated with the location
 * @param location the location for the key
 * @return true if saved successfully false otherwise
 */
public boolean saveLocation(String key, Location location) {
    LOG.info("Saving location");
    try {
        JSONObject locationJson = new JSONObject();

        locationJson.put(LATITUDE, location.getLatitude());
        locationJson.put(LONGITUDE, location.getLongitude());
        //other location data
        SharedPreferences.Editor edit = preferences.edit();
        edit.putString(key, locationJson.toString());
        edit.commit();
    } catch (JSONException e) {
        LOG.error("JSON Exception", e);
        return false;
    }

    LOG.info("Location {}  saved successfully at key: {}", preferences.getString(key, null),key);
    return true;
}

/**
 * Gets location data for a key.
 * 
 * @param key the key for the saved location
 * @return a {@link Location} object or null if there is no entry for the key
 */
public Location getLocation(String key) {
    LOG.info("Retrieving location at key {} ", key);
    try {
        String json = preferences.getString(key, null);

        if (json != null) {
            JSONObject locationJson = new JSONObject(json);
            Location location = new Location(STORAGE);
            location.setLatitude(locationJson.getInt(LATITUDE));
            location.setLongitude(locationJson.getInt(LONGITUDE));
            LOG.info("Returning location: {}" , location);
            return location;
        }
    } catch (JSONException e) {
        LOG.error("JSON Exception", e);
    }

    LOG.warn("No location found at key {}",  key);
    //or throw exception depending on your logic
    return null;
}

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这是一个有趣的问题，这里有一些聪明的答案