我试图从一个Java方法返回2个值,但我得到这些错误。这是我的代码:

// Method code
public static int something(){
    int number1 = 1;
    int number2 = 2;

    return number1, number2;
}

// Main method code
public static void main(String[] args) {
    something();
    System.out.println(number1 + number2);
}

错误:

Exception in thread "main" java.lang.RuntimeException: Uncompilable source code - missing return statement
    at assignment.Main.something(Main.java:86)
    at assignment.Main.main(Main.java:53)

Java结果:1


当前回答

返回一个对象数组

private static Object[] f () 
{ 
     double x =1.0;  
     int y= 2 ;
     return new Object[]{Double.valueOf(x),Integer.valueOf(y)};  
}

其他回答

使用Pair/Tuple类型的对象,如果你依赖于Apache common -lang,你甚至不需要创建一个。只需使用Pair类。

您可以创建一个记录(Java 14起可用),以返回具有类型安全、命名和简洁的值。

public record MyResult(int number1, int number2) {
}

public static MyResult something() {
    int number1 = 1;
    int number2 = 2;

    return new MyResult(number1, number2);
}

public static void main(String[] args) {
    MyResult result = something();
    System.out.println(result.number1() + result.number2());
}
public class Mulretun
{
    public String name;;
    public String location;
    public String[] getExample()
    {
        String ar[] = new String[2];
        ar[0]="siva";
        ar[1]="dallas";
        return ar; //returning two values at once
    }
    public static void main(String[] args)
    {
        Mulretun m=new Mulretun();
        String ar[] =m.getExample();
        int i;
        for(i=0;i<ar.length;i++)
        System.out.println("return values are: " + ar[i]);      

    }
}

o/p:
return values are: siva
return values are: dallas

您必须使用集合来返回多个返回值

在您的情况下,您将代码编写为

public static List something(){
        List<Integer> list = new ArrayList<Integer>();
        int number1 = 1;
        int number2 = 2;
        list.add(number1);
        list.add(number2);
        return list;
    }

    // Main class code
    public static void main(String[] args) {
      something();
      List<Integer> numList = something();
    }

我很好奇为什么没有人提出更优雅的回调解决方案。所以不是使用返回类型,而是使用传递给方法的处理程序作为参数。下面的例子有两种截然不同的方法。我知道这两件事对我来说哪一件更优雅。:-)

public class DiceExample {

    public interface Pair<T1, T2> {
        T1 getLeft();

        T2 getRight();
    }

    private Pair<Integer, Integer> rollDiceWithReturnType() {

        double dice1 = (Math.random() * 6);
        double dice2 = (Math.random() * 6);

        return new Pair<Integer, Integer>() {
            @Override
            public Integer getLeft() {
                return (int) Math.ceil(dice1);
            }

            @Override
            public Integer getRight() {
                return (int) Math.ceil(dice2);
            }
        };
    }

    @FunctionalInterface
    public interface ResultHandler {
        void handleDice(int ceil, int ceil2);
    }

    private void rollDiceWithResultHandler(ResultHandler resultHandler) {
        double dice1 = (Math.random() * 6);
        double dice2 = (Math.random() * 6);

        resultHandler.handleDice((int) Math.ceil(dice1), (int) Math.ceil(dice2));
    }

    public static void main(String[] args) {

        DiceExample object = new DiceExample();


        Pair<Integer, Integer> result = object.rollDiceWithReturnType();
        System.out.println("Dice 1: " + result.getLeft());
        System.out.println("Dice 2: " + result.getRight());

        object.rollDiceWithResultHandler((dice1, dice2) -> {
            System.out.println("Dice 1: " + dice1);
            System.out.println("Dice 2: " + dice2);
        });
    }
}