Android - RxJava - 竞争条件

Question

我是多线程概念的新手，也是 RxJava 的新手。我想我在我的 Android 项目中遇到了竞争条件。 为了能够更好地演示它，我下载了 google codelabs 示例 RoomWithAView 并实现了与我自己的项目中相同的竞争条件。

WordDao

@Dao
public interface WordDao {
    @Insert()
    void insert(Word word);

    @Query("SELECT EXISTS(SELECT word FROM WORD_TABLE WHERE word = :word)")
    Single<Integer> wordExists(String word);
}

词库

class WordRepository {

    private WordDao mWordDao;
    private boolean mWordRedundant;
    private final CompositeDisposable mCompositeDisposable = new CompositeDisposable();

    WordRepository(Application application) {
        WordRoomDatabase db = WordRoomDatabase.getDatabase(application);
        mWordDao = db.wordDao();
    }

    void insert(Word word) {
        WordRoomDatabase.databaseWriteExecutor.execute(() -> {
            mWordDao.insert(word);
        });
    }

    public boolean isWordRedundant(String word) {
        mCompositeDisposable.add(
        mWordDao.wordExists(word)
                .subscribeOn(Schedulers.computation())
                .subscribe(integer -> mWordRedundant = integer == 1));
        return mWordRedundant;
    }

WordViewModel

public class WordViewModel extends AndroidViewModel {

    private WordRepository mRepository;

    public WordViewModel(Application application) {
        super(application);
        mRepository = new WordRepository(application);
    }

    public boolean isWordRedundant(String word) {
        return mRepository.isWordRedundant(word);
    }

    public void insert(Word word) {
        mRepository.insert(word);
    }
}

新词活动

public class NewWordActivity extends AppCompatActivity {

    public static final String EXTRA_REPLY = "com.example.android.wordlistsql.REPLY";

    private EditText mEditWordView;
    private WordViewModel mWordViewModel;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_new_word);
        mWordViewModel = new ViewModelProvider(this).get(WordViewModel.class);
        mEditWordView = findViewById(R.id.edit_word);

        final Button button = findViewById(R.id.button_save);
        button.setOnClickListener(view -> {
            Intent replyIntent = new Intent();

            if (TextUtils.isEmpty(mEditWordView.getText())) {
                System.out.println("Word empty");
                setResult(RESULT_CANCELED, replyIntent);
            }

            // possible race condition?
            else if (mWordViewModel.isWordRedundant(mEditWordView.getText().toString())) {
                System.out.println("Word redundant");
                setResult(RESULT_CANCELED, replyIntent);
            }

            else {
                String word = mEditWordView.getText().toString();
                System.out.println("Word acceptable");
                replyIntent.putExtra(EXTRA_REPLY, word);
                setResult(RESULT_OK, replyIntent);
            }
            finish();
        });
    }
}

在我的例子中，方法isPlayerNameRedundant()总是返回false。我在 App Inspection 中测试了 SQL 查询，它返回 1，所以 isPlayerNameRedundant() 方法应该返回 true。 我怀疑因为我在存储库中使用了Schedulers.computation()，所以查询是在后台线程上执行的，并且在该线程完成其任务之前，主线程返回mWordRedundant。 那是对的吗？

如果是这样，解决这个问题的方法是什么？

先感谢您。

Answer 1

你在滥用 Rx-java。

正确方法：

词库

public Single<Boolean> isWordRedundant(String word) {
    return mWordDao.wordExists(word)
            .subscribeOn(Schedulers.io()) //computation() is for CPU intensive operation. Not for DB read/write. This is a tiny error.
            .map(i -> i == 1 ? true : false)
}

WordViewModel

public Single<Boolean> isWordRedundant(String word) {
    return mRepository.isWordRedundant(word);
}

新词活动

    button.setOnClickListener(view -> {
        Intent replyIntent = new Intent();

        if (TextUtils.isEmpty(mEditWordView.getText())) {
            System.out.println("Word empty");
            setResult(RESULT_CANCELED, replyIntent);
            finish();
        } else {
            //Your CompositDisposable in activity
            compositDisposable.add(
                mWordViewModel.isWordRedundant(mEditWordView.getText().toString())
                    .observeOn(AndroidSchedulers.mainThread())
                    .subscribe(isRedundant -> {
                        if (isRedundant) {
                            System.out.println("Word redundant");
                            setResult(RESULT_CANCELED, replyIntent);
                        } else {
                            String word = mEditWordView.getText().toString();
                            System.out.println("Word acceptable");
                            replyIntent.putExtra(EXTRA_REPLY, word);
                            setResult(RESULT_OK, replyIntent);
                        }
                        finish();
                    }, error -> {/* handle error here. Maybe a Toast or something. */})
            );
        }
    });

那么，为什么您之前的代码不起作用？让我们看看代码词库之前（注意cmets）：

class WordRepository {

    private WordDao mWordDao;
    private boolean mWordRedundant; //field variable. Not initiate. So is the default value: **false**.
    private final CompositeDisposable mCompositeDisposable = new CompositeDisposable();

    public boolean isWordRedundant(String word) {
        //Main thread enter here.
        //Main thread step 1: Create an Observable i.e.(mWordDao.wordExists(word)). And subscribe on it.
        mCompositeDisposable.add(
            mWordDao.wordExists(word)
                .subscribeOn(Schedulers.computation())
                .subscribe(integer -> mWordRedundant = integer == 1)); //We process it and observe the result in computation() thread. Then change the filed variable "mWordRedundant" value.
        
        //Main thread step 2: return the field variable **immediately**! Not wait for the result from computation() thread. That is why you can not get the right result.
        return mWordRedundant;
    }
}