Question

我有两节课。

在第一堂课中，需要整数low和high。

在第二节课中，我想宣布并更改low和high。

public class DisplayMessageActivity extends ActionBarActivity { 
    int low = 0;
    int high = 101;

这很好 - 但是在以下函数运行之后，整数没有改变。

public int close (View v) {

    EditText low2 = (EditText) findViewById(R.id.editText2);
    EditText high2 = (EditText) findViewById(R.id.editText3);

    int low = Integer.parseInt(low2.getText().toString());
    int high = Integer.parseInt(high2.getText().toString());
    high ++;

    Toast.makeText(getApplicationContext(), "Low: " + low +" High: "+high, Toast.LENGTH_SHORT).show();

   if (low > high) { //you can ignore it
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        builder.setTitle("False Numbers");
        builder.setMessage("You set the lower number higher than the higher one.");
        builder.show();
    } else {
        this.finish();
    }

    return low; //I want to return low AND high - but this command will not change low or high!
}

编辑代码：

public class HighAndLow{
    int high= 0;
    int low = 101;
}



public int[] close(View v){


    EditText low2 = (EditText) findViewById(R.id.editText2);
    EditText high2 = (EditText) findViewById(R.id.editText3);

        low = Integer.parseInt(low2.getText().toString()); //cant find low now
        high = Integer.parseInt(high2.getText().toString()); //cant find high now
        high ++;

    Toast.makeText(getApplicationContext(), "Low: " + low +" High: "+high,
            Toast.LENGTH_SHORT).show();





   if (low > high) {
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        builder.setTitle("False Numbers");
        builder.setMessage("You set the lower number higher than the higher one.");
        builder.show();
    }
        else{

            this.finish();
        }




    return HighAndLow; //DONT WORK


}

Answer 1

在Java（和Android）中，您无法从函数返回多个值。您只能返回单个值。但是，该单个值可以是包含多个字段的对象。

您可以考虑创建一个包含两个值的新类。像这样：

public class HighAndLow{
   int high;
   int low;
}

然后你可以从你的方法返回这个类的实例，调用代码可以通过那种方式访问高值和低值。

事实上，Android已经有一个类来执行此操作：Range

Answer 2

为什么不将高低整数创建为类的属性？然后实现你通常的getter / setter。通过close方法设置它们，并在需要时从第一个实例化的类中获取它们。

Answer 3

这不是推荐和面向对象的方式，但如果以前的答案不满足，你可以这样做：

public int[] close (View v) {

    EditText low2 = (EditText) findViewById(R.id.editText2);
    EditText high2 = (EditText) findViewById(R.id.editText3);

    int low = Integer.parseInt(low2.getText().toString());
    int high = Integer.parseInt(high2.getText().toString());
    high ++;

    Toast.makeText(getApplicationContext(), "Low: " + low +" High: "+high, Toast.LENGTH_SHORT).show();

   if (low > high) { //you can ignore it
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        builder.setTitle("False Numbers");
        builder.setMessage("You set the lower number higher than the higher one.");
        builder.show();
    } else {
        this.finish();
    }
    int[] values = {high,low};
    return values; //I want to return low AND high - but this command will not change low or high!
}

之后

int[] myHighAndLow = close(someView);

//you need to remember what came first(First In First Out)
int high = myHighAndLow[0];
int low = myHighAndLow[1];

在int函数中返回2个整数

3 个答案: