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
你也可以将可变对象作为参数发送，如果你使用方法来修改它们，那么当你从函数返回时，它们将被修改。它不能用于Float之类的东西，因为它是不可变的。
public class HelloWorld{
public static void main(String []args){
HelloWorld world = new HelloWorld();
world.run();
}
private class Dog
{
private String name;
public void setName(String s)
{
name = s;
}
public String getName() { return name;}
public Dog(String name)
{
setName(name);
}
}
public void run()
{
Dog newDog = new Dog("John");
nameThatDog(newDog);
System.out.println(newDog.getName());
}
public void nameThatDog(Dog dog)
{
dog.setName("Rutger");
}
}
结果是: 念完
其他回答
在Java中只能返回一个值，所以最简洁的方法是这样的:
return new Pair<Integer>(number1, number2);
这是你的代码的更新版本:
public class Scratch
{
// Function code
public static Pair<Integer> something() {
int number1 = 1;
int number2 = 2;
return new Pair<Integer>(number1, number2);
}
// Main class code
public static void main(String[] args) {
Pair<Integer> pair = something();
System.out.println(pair.first() + pair.second());
}
}
class Pair<T> {
private final T m_first;
private final T m_second;
public Pair(T first, T second) {
m_first = first;
m_second = second;
}
public T first() {
return m_first;
}
public T second() {
return m_second;
}
}
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);
});
}
}
您不需要创建自己的类来返回两个不同的值。就像这样使用HashMap:
private HashMap<Toy, GameLevel> getToyAndLevelOfSpatial(Spatial spatial)
{
Toy toyWithSpatial = firstValue;
GameLevel levelToyFound = secondValue;
HashMap<Toy,GameLevel> hm=new HashMap<>();
hm.put(toyWithSpatial, levelToyFound);
return hm;
}
private void findStuff()
{
HashMap<Toy, GameLevel> hm = getToyAndLevelOfSpatial(spatial);
Toy firstValue = hm.keySet().iterator().next();
GameLevel secondValue = hm.get(firstValue);
}
甚至还有类型安全的好处。
你也可以将可变对象作为参数发送，如果你使用方法来修改它们，那么当你从函数返回时，它们将被修改。它不能用于Float之类的东西，因为它是不可变的。
public class HelloWorld{
public static void main(String []args){
HelloWorld world = new HelloWorld();
world.run();
}
private class Dog
{
private String name;
public void setName(String s)
{
name = s;
}
public String getName() { return name;}
public Dog(String name)
{
setName(name);
}
}
public void run()
{
Dog newDog = new Dog("John");
nameThatDog(newDog);
System.out.println(newDog.getName());
}
public void nameThatDog(Dog dog)
{
dog.setName("Rutger");
}
}
结果是: 念完
