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
在我看来,最好是创建一个新类,其中构造函数是你需要的函数,例如:
public class pairReturn{
//name your parameters:
public int sth1;
public double sth2;
public pairReturn(int param){
//place the code of your function, e.g.:
sth1=param*5;
sth2=param*10;
}
}
然后像使用函数一样使用构造函数:
pairReturn pR = new pairReturn(15);
你可以使用pR.sth1 pR.sth2作为函数的2个结果
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);
});
}
}