Java 功能在android上无法正常工作

我在android应用程序上使用的一个函数有点问题。 这个函数不太难，它只是一个简单的计算，有两个随机整数。 问题是我有一个标签，我想把它的文本改成“问题”，这样用户就可以解决它了。我知道函数有问题，但我不太明白是什么。我最接近的猜测是返回部分是错误的（因为它是空的），但到目前为止我无法修复它

这是我的代码：

import java.util.Random;
import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.TextView;

public class Game extends Activity implements OnClickListener {

private static final String TAG = "Game";

int game[];

public static final String KEY_DIFFICULTY = "com.example.coursework1.dificulty";

public static final int DIFFICULTY_NOVICE = 0;
public static final int DIFFICULTY_EASY = 1;
public static final int DIFFICULTY_MEDIUM = 2;
public static final int DIFFICULTY_GURU = 3;

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.game);

    int diff = getIntent().getIntExtra(KEY_DIFFICULTY, DIFFICULTY_NOVICE);
    game = getGame(diff);


}

public int[] getGame(int diff) {
    // TODO Auto-generated method stub
    String operators [] =  {"+","-","*","/"};
    Random op1, op2, op3, op4, op5, op6;
    if(diff == getIntent().getIntExtra(KEY_DIFFICULTY, DIFFICULTY_NOVICE)){
        op1 = new Random();
        op2 = new Random();
        int operator = new Random().nextInt(operators.length);
        String random = (operators[operator]);
        int generate1 = op1.nextInt(10);
        int generate2 = op2.nextInt(10);
        int operation;
        if(random == "+"){
            operation = generate1 + generate2;
            TextView setProblemToView = (TextView) findViewById(R.id.problem);
            setProblemToView.setText("generate1" + random + "generate2");
        }
    }


    return null;
}

@Override
public void onClick(View arg0) {
    // TODO Auto-generated method stub
}
}

---

为什么要使用int-ruturn而必须始终返回null？这样使用：

public void getGame(int diff) {

    String operators [] =  {"+","-","*","/"};
    Random op1, op2, op3, op4, op5, op6;
    if(diff == getIntent().getIntExtra(KEY_DIFFICULTY, DIFFICULTY_NOVICE)){
        op1 = new Random();

        int operator = new Random().nextInt(operators.length);
        String random = (operators[operator]);
        int generate1 = op1.nextInt(10);
        int generate2 = op1.nextInt(10);
        int operation;
        if(  "+".equalsIgnoreCase(random) ){
            operation = generate1 + generate2;
            TextView setProblemToView = (TextView) findViewById(R.id.problem);
            setProblemToView.setText("generate1" + random + "generate2");
        }
    }

}

getGame(diff);

就这样称呼它：

public void getGame(int diff) {

    String operators [] =  {"+","-","*","/"};
    Random op1, op2, op3, op4, op5, op6;
    if(diff == getIntent().getIntExtra(KEY_DIFFICULTY, DIFFICULTY_NOVICE)){
        op1 = new Random();

        int operator = new Random().nextInt(operators.length);
        String random = (operators[operator]);
        int generate1 = op1.nextInt(10);
        int generate2 = op1.nextInt(10);
        int operation;
        if(  "+".equalsIgnoreCase(random) ){
            operation = generate1 + generate2;
            TextView setProblemToView = (TextView) findViewById(R.id.problem);
            setProblemToView.setText("generate1" + random + "generate2");
        }
    }

}

getGame(diff);

---

在我的编辑中解决了相同的随机问题，请查看更多信息。

注意：不要使用

==

比较字符串值。使用

equals（）

。这里到底出了什么问题？这里的问题是，当我在avd上测试应用程序时，问题字符串的值仍然是相同的（这就是问题）。可能重复@ZouZou感谢提示。我现在就改！当然，不要返回相同的值。它就在我面前，我没有看到它。谢谢：）