I did this:
public class LambdaConflict
{
public static void main(String args[]){
//*
System.out.println(LambdaConflict.get(
Just for reference and to enrich the answers already given:
As per JSR-335: Lambda Expressions for the Java Programming Language, in section Lambda Specification, Part A: Functional Interfaces it says:
A functional interface is an interface that has just one abstract method (aside from the methods of Object), and thus represents a single function contract. (In some cases, this "single" method may take the form of multiple abstract methods with override-equivalent signatures inherited from superinterfaces; in this case, the inherited methods logically represent a single method.)
So, what you need is to either provide a default implementation for one of your methods or put one of your methods in a different interface.
As stated by @Thomas-Uhrig, Functional Interfaces can only have one method.
A way to fix this, primarily because you never use public T get2(T arg1);
, is to change the Intf<T>
interface to:
@FunctionalInterface
interface Intf<T>
{
public T get1(T arg1);
}
No. There is no way to "overcome" this. A functional interface must have only one abstract method. Your interface has two:
interface Intf<T> {
public T get1(T arg1);
public T get2(T arg1);
}
Note: You don't need to annotate your interface as mentioned in comments. But you can use the @FunctionalInterface
annotation to get compile time errors if your interface is not a valid functional interface. So it brings you a little bit more security in your code.
For more see e.g. http://java.dzone.com/articles/introduction-functional-1
Think about it:
How should the compiler know if you want to override get1
or get2
?
If you only override get1
, what will be get2
's implementation? Even the code you commented out won't work because you don't implement get2
...
There are reasons for this limitation...