You can return only one value in Java. If needed you can return multiple values using array or an object.
JavaScript doesn't support functions that return multiple values. However, you can wrap multiple values into an array or an object and return the array or the object. Use destructuring assignment syntax to unpack values from the array, or properties from objects.
The recommended option: create a specific class that "wraps around" those three values. And your method returns an instance of that class! But for the record: don't just create a class with three fields that everybody else can read and write to.
No, you can not have two returns in a function, the first return will exit the function you will need to create an object.
If you want to return two objects you usually want to return a single object that encapsulates the two objects instead.
You could return a List of NamedObject
objects like this:
public class NamedObject<T> {
public final String name;
public final T object;
public NamedObject(String name, T object) {
this.name = name;
this.object = object;
}
}
Then you can easily return a List<NamedObject<WhateverTypeYouWant>>
.
Also: Why would you want to return a comma-separated list of names instead of a List<String>
? Or better yet, return a Map<String,TheObjectType>
with the keys being the names and the values the objects (unless your objects have specified order, in which case a NavigableMap
might be what you want.
If you know you are going to return two objects, you can also use a generic pair:
public class Pair<A,B> {
public final A a;
public final B b;
public Pair(A a, B b) {
this.a = a;
this.b = b;
}
};
Edit A more fully formed implementation of the above:
package util;
public class Pair<A,B> {
public static <P, Q> Pair<P, Q> makePair(P p, Q q) {
return new Pair<P, Q>(p, q);
}
public final A a;
public final B b;
public Pair(A a, B b) {
this.a = a;
this.b = b;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((a == null) ? 0 : a.hashCode());
result = prime * result + ((b == null) ? 0 : b.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
@SuppressWarnings("rawtypes")
Pair other = (Pair) obj;
if (a == null) {
if (other.a != null) {
return false;
}
} else if (!a.equals(other.a)) {
return false;
}
if (b == null) {
if (other.b != null) {
return false;
}
} else if (!b.equals(other.b)) {
return false;
}
return true;
}
public boolean isInstance(Class<?> classA, Class<?> classB) {
return classA.isInstance(a) && classB.isInstance(b);
}
@SuppressWarnings("unchecked")
public static <P, Q> Pair<P, Q> cast(Pair<?, ?> pair, Class<P> pClass, Class<Q> qClass) {
if (pair.isInstance(pClass, qClass)) {
return (Pair<P, Q>) pair;
}
throw new ClassCastException();
}
}
Notes, mainly around rustiness with Java & generics:
a
and b
are immutable. makePair
static method helps you with boiler plate typing, which the diamond operator in Java 7 will make less annoying. There's some work to make this really nice re: generics, but it should be ok-ish now. (c.f. PECS)hashcode
and equals
are generated by eclipse.cast
method is ok, but doesn't seem quite right.isInstance
are necessary.In the event the method you're calling is private, or called from one location, try
return new Object[]{value1, value2};
The caller looks like:
Object[] temp=myMethod(parameters);
Type1 value1=(Type1)temp[0]; //For code clarity: temp[0] is not descriptive
Type2 value2=(Type2)temp[1];
The Pair example by David Hanak has no syntactic benefit, and is limited to two values.
return new Pair<Type1,Type2>(value1, value2);
And the caller looks like:
Pair<Type1, Type2> temp=myMethod(parameters);
Type1 value1=temp.a; //For code clarity: temp.a is not descriptive
Type2 value2=temp.b;
You may use any of following ways:
private static final int RETURN_COUNT = 2;
private static final int VALUE_A = 0;
private static final int VALUE_B = 1;
private static final String A = "a";
private static final String B = "b";
1) Using Array
private static String[] methodWithArrayResult() {
//...
return new String[]{"valueA", "valueB"};
}
private static void usingArrayResultTest() {
String[] result = methodWithArrayResult();
System.out.println();
System.out.println("A = " + result[VALUE_A]);
System.out.println("B = " + result[VALUE_B]);
}
2) Using ArrayList
private static List<String> methodWithListResult() {
//...
return Arrays.asList("valueA", "valueB");
}
private static void usingListResultTest() {
List<String> result = methodWithListResult();
System.out.println();
System.out.println("A = " + result.get(VALUE_A));
System.out.println("B = " + result.get(VALUE_B));
}
3) Using HashMap
private static Map<String, String> methodWithMapResult() {
Map<String, String> result = new HashMap<>(RETURN_COUNT);
result.put(A, "valueA");
result.put(B, "valueB");
//...
return result;
}
private static void usingMapResultTest() {
Map<String, String> result = methodWithMapResult();
System.out.println();
System.out.println("A = " + result.get(A));
System.out.println("B = " + result.get(B));
}
4) Using your custom container class
private static class MyContainer<M,N> {
private final M first;
private final N second;
public MyContainer(M first, N second) {
this.first = first;
this.second = second;
}
public M getFirst() {
return first;
}
public N getSecond() {
return second;
}
// + hashcode, equals, toString if need
}
private static MyContainer<String, String> methodWithContainerResult() {
//...
return new MyContainer("valueA", "valueB");
}
private static void usingContainerResultTest() {
MyContainer<String, String> result = methodWithContainerResult();
System.out.println();
System.out.println("A = " + result.getFirst());
System.out.println("B = " + result.getSecond());
}
5) Using AbstractMap.simpleEntry
private static AbstractMap.SimpleEntry<String, String> methodWithAbstractMapSimpleEntryResult() {
//...
return new AbstractMap.SimpleEntry<>("valueA", "valueB");
}
private static void usingAbstractMapSimpleResultTest() {
AbstractMap.SimpleEntry<String, String> result = methodWithAbstractMapSimpleEntryResult();
System.out.println();
System.out.println("A = " + result.getKey());
System.out.println("B = " + result.getValue());
}
6) Using Pair of Apache Commons
private static Pair<String, String> methodWithPairResult() {
//...
return new ImmutablePair<>("valueA", "valueB");
}
private static void usingPairResultTest() {
Pair<String, String> result = methodWithPairResult();
System.out.println();
System.out.println("A = " + result.getKey());
System.out.println("B = " + result.getValue());
}
I almost always end up defining n-Tuple classes when I code in Java. For instance:
public class Tuple2<T1,T2> {
private T1 f1;
private T2 f2;
public Tuple2(T1 f1, T2 f2) {
this.f1 = f1; this.f2 = f2;
}
public T1 getF1() {return f1;}
public T2 getF2() {return f2;}
}
I know it's a bit ugly, but it works, and you just have to define your tuple types once. Tuples are something Java really lacks.
EDIT: David Hanak's example is more elegant, as it avoids defining getters and still keeps the object immutable.
Before Java 5, I would kind of agree that the Map solution isn't ideal. It wouldn't give you compile time type checking so can cause issues at runtime. However, with Java 5, we have Generic Types.
So your method could look like this:
public Map<String, MyType> doStuff();
MyType of course being the type of object you are returning.
Basically I think that returning a Map is the right solution in this case because that's exactly what you want to return - a mapping of a string to an object.
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