Java 方法给出错误，使方法返回无效

它显示错误：

make方法CreateStep返回语句void

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为什么？

如果发生异常，您必须返回一些内容：

public Step createStep(int position) {
    try {
        final StepFragmentSample step = new StepFragmentSample();
        Bundle b = new Bundle();
        b.putInt(CURRENT_STEP_POSITION_KEY, position);
        step.setArguments(b);
        return step;
    } catch (Exception e) {
        e.printStackTrace();
        Log.e("/Test", "/Excp due to" + e.toString());
    }
}

或

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您必须始终返回步骤对象。如果代码中有异常，您将永远不会返回Step对象

试着这样做：

public Step createStep(int position) {
    try {
        final StepFragmentSample step = new StepFragmentSample();
        Bundle b = new Bundle();
        b.putInt(CURRENT_STEP_POSITION_KEY, position);
        step.setArguments(b);
        return step;
    } catch (Exception e) {
        e.printStackTrace();
        Log.e("/Test", "/Excp due to" + e.toString());
    }
    return null;
}

像这样做

public Step createStep(int position) {
    try {
        final StepFragmentSample step = new StepFragmentSample();
        Bundle b = new Bundle();
        b.putInt(CURRENT_STEP_POSITION_KEY, position);
        step.setArguments(b);
        return step;
    } catch (Exception e) {
        e.printStackTrace();
        Log.e("/Test", "/Excp due to" + e.toString());
    }
    return new Step(); // or return null;
}

---

一旦方法被执行，您需要使用void或返回一些东西。但是我返回步骤，然后它为什么会给出错误？应该在catch语句之后调用return谢谢sir@Jens@萨加尔：很高兴见到您，先生，我可以在这里以外的地方与您联系吗？邮件还是其他什么

public Step createStep(int position) {
    try {
        final StepFragmentSample step = new StepFragmentSample();
        Bundle b = new Bundle();
        b.putInt(CURRENT_STEP_POSITION_KEY, position);
        step.setArguments(b);
        return step;
    } catch (Exception e) {
        e.printStackTrace();
        Log.e("/Test", "/Excp due to" + e.toString());
    }
    return new Step(); // or return null;
}

public Step createStep(int position) {
try {
    final StepFragmentSample step = new StepFragmentSample();
    Bundle b = new Bundle();
    b.putInt(CURRENT_STEP_POSITION_KEY, position);
    step.setArguments(b);

} catch (Exception e) {
    e.printStackTrace();
    Log.e("/Test", "/Excp due to" + e.toString());
}
 return step;
}