Recently I started exploring Java 8 and I can't quite understand the concept of "functional interface" that is essential to Java's implementation of lambda expressions. There is a pretty comprehensive guide to lambda functions in Java, but I got stuck on the chapter that gives definition to the concept of functional interfaces. The definition reads:
More precisely, a functional interface is defined as any interface that has exactly one abstract method.
An then he proceeds to examples, one of which is Comparator interface:
public interface Comparator<T> {
int compare(T o1, T o2);
boolean equals(Object obj);
}
I was able to test that I can use a lambda function in place of Comparator argument and it works(i.e. Collections.sort(list, (a, b) -> a-b)).
But in the Comparator interface both compare and equals methods are abstract, which means it has two abstract methods. So how can this be working, if the definition requires an interface to have exactly one abstract method? What am I missing here?
From the same page you linked to:
The interface Comparator is functional because although it declares two abstract methods, one of these—equals— has a signature corresponding to a public method in Object. Interfaces always declare abstract methods corresponding to the public methods of Object, but they usually do so implicitly. Whether implicitly or explicitly declared, such methods are excluded from the count.
I can't really say it better.
Another explanation is given in the #FunctionalInterface page:
Conceptually, a functional interface has exactly one abstract method. Since default methods have an implementation, they are not abstract. If an interface declares an abstract method overriding one of the public methods of java.lang.Object, that also does not count toward the interface's abstract method count since any implementation of the interface will have an implementation from java.lang.Object or elsewhere.
You can test which interface is a correct functional interface using #FunctionalInterface.
E.g.:
this works
#FunctionalInterface
public interface FunctionalInterf {
void m();
boolean equals(Object o);
}
this generates an error:
#FunctionalInterface
public interface FunctionalInterf {
void m();
boolean equals();
}
Multiple non-overriding abstract methods found in interface FunctionalInterf
Q. But in the Comparator interface both compare() and equals() methods are abstract, which means it has two abstract methods. So how can this be working, if the definition requires an interface to have exactly one abstract method? What am I missing here?
A.
A functional interface may specify any public method defined by Object, such as equals( ),
without affecting its “functional interface” status. The public Object methods are considered implicit
members of a functional interface because they are automatically implemented by an instance of a
functional interface.
An interface cannot extend Object class, because Interface has to have public and abstract methods.
For every public method in the Object class, there is an implicit public and abstract method in an interface.
This is the standard Java Language Specification which states like this,
“If an interface has no direct super interfaces, then the interface implicitly declares a public abstract member method m with signature s, return type r, and throws clause t corresponding to each public instance method m with signature s, return type r, and throws clause t declared in Object, unless a method with the same signature, same return type, and a compatible throws clause is explicitly declared by the interface.”
That's how Object class' methods are declared in an interface. And according to JLS, this does not count as interface' abstract method. Hence, Comparator interface is a functional interface.
A functional interface has only one abstract method but it can have multiple default and static methods.
Since default methods are not abstract you’re free to add default methods to your functional interface as many as you like.
#FunctionalInterface
public interface MyFuctionalInterface
{
public void perform();
default void perform1(){
//Method body
}
default void perform2(){
//Method body
}
}
If an interface declares an abstract method overriding one of the public methods of java.lang.Object, that also does not count toward the interface’s abstract method count since any implementation of the interface will have an implementation from java.lang.Object or elsewhere.
Comparator is a functional interface even though it declared two abstract methods. Because one of these abstract methods “equals()” which has signature equal to public method in Object class.
e.g. Below interface is a valid functional interface.
#FunctionalInterface
public interface MyFuctionalInterface
{
public void perform();
#Override
public String toString(); //Overridden from Object class
#Override
public boolean equals(Object obj); //Overridden from Object class
}
Here is a "show me the code" approach to understanding the definition:
we shall look into OpenJDK javac for how it checks validity of classes annotated with #FunctionalInterface.
The most recent (as of July, 2022) implementation lies here:
com/sun/tools/javac/code/Types.java#L735-L791:
/**
* Compute the function descriptor associated with a given functional interface
*/
public FunctionDescriptor findDescriptorInternal(TypeSymbol origin,
CompoundScope membersCache) throws FunctionDescriptorLookupError {
// ...
}
Class Restriction
if (!origin.isInterface() || (origin.flags() & ANNOTATION) != 0 || origin.isSealed()) {
//t must be an interface
throw failure("not.a.functional.intf", origin);
}
Pretty simple: the class must be an interface and must not be a sealed one.
Member Restriction
for (Symbol sym : membersCache.getSymbols(new DescriptorFilter(origin))) { /* ... */ }
In this loop, javac goes through the members of the origin class, using a DescriptorFilter to retrieve:
Method members (of course)
&& that are abstract but not default,
&& and do not overwrite methods from Object,
&& with their top level declaration not a default one.
If there is only one method matching all the above conditions, then surely it is a valid functional interface, and any lambda will overwrite that very method.
However, if there are multiple, javac tries to merge them:
If those methods all share the same name, related by override equivalence:
then we filter them into a abstracts collection:
if (!abstracts.stream().filter(msym -> msym.owner.isSubClass(sym.enclClass(), Types.this))
.map(msym -> memberType(origin.type, msym))
.anyMatch(abstractMType -> isSubSignature(abstractMType, mtype))) {
abstracts.append(sym);
}
Methods are filtered out if:
their enclosing class is super class of that of another previously matched method,
and the signature of that previously matched method is subsignature of that of this method.
Otherwise, the functional interface is not valid.
Having collected abstracts, javac then goes to mergeDescriptors, which uses mergeAbstracts, which I will just quote from its comments:
/**
* Merge multiple abstract methods. The preferred method is a method that is a subsignature
* of all the other signatures and whose return type is more specific {#see MostSpecificReturnCheck}.
* The resulting preferred method has a thrown clause that is the intersection of the merged
* methods' clauses.
*/
public Optional<Symbol> mergeAbstracts(List<Symbol> ambiguousInOrder, Type site, boolean sigCheck) {
// ...
}
Conclusion
Functional interfaces must be interfaces :P , and must not be sealed or annotations.
Methods are searched in the whole inheritance tree.
Methods overlapping with those from Object are ignored.
default methods are ignored, unless they are later overridden by sub-interfaces as non-default.
Matching methods must all share the same name, related by override equivalence.
Either there is only one method matching the above conditions, or matching methods can get "merged" by their class hierarchy, subsignature relations, return types and thrown clauses.
The Java docs say:
Note that it is always safe not to override Object.equals(Object).
However, overriding this method may, in some cases, improve
performance by allowing programs to determine that two distinct
comparators impose the same order.
Maybe Comparator is special? Maybe, even though it's an interface, there is somehow a default implementation of equals() that calls compare()? Algorithmically, it's trivial.
I thought all methods that were declared in interfaces were abstract (i. e. no default implementation). But maybe I'm missing something.
Definition:
If an interface contains only one abstract method, then such type of interface is called functional interface.
Usage:
Once we write Lambda expression "->" to invoke its functionality ,
then in this context we require Functional Interface.
We can use the Functional Interface reference to refer Lambda
expression.
Inside functional interface we can have one abstract method and n
number of default/static methods.
Functional interface with respect to inheritance:
If an interface extends Functional interface and the child interface does not contain any abstract method , then the child interface is also considered to be Functional Interface.
Functional interface is not new to java, its already used in following interface API's:
Runnable : contains run() method only.
Callable : contains call() method only.
Comparable : contains compareTo() method only.
Before Java 8, an interface could only declare one or more methods also known as Abstract Method (method with no implementation, just the signature). Starting with Java 8 an interface can also have implementation of one or more methods (knows as Interface Default Method) and static methods along with abstract methods. Interface Default Methods are marked default keyword.
So the question is, what is Functional Interface?
An interface with Single Abstract Method (SAM) is called Functional Interface.
Which means -
An interface with Single Abstract Method is a Functional Interface
An interface with Single Abstract Method and zero or more default
methods and zero or more static method is also a valid Functional
Interface.
More detail with example code https://readtorakesh.com/functional-interface-java8/
Learning Java 8 Lambdas and just wondering how the compiler knows which method in Comparator to use for the lambda expression?
It doesn't seem to be a SAM interface? It has 2 abstract methods:
#FunctionalInterface
public interface Comparator<T> {
int compare(T o1, T o2);
boolean equals(Object obj);
}
equals() is not an abstract method. This method overrides Object.equals(Object), and is there only for the Comparator interface to be able to have javadoc attached to the method, explaining how comparators should implement equals().
See the javadoc of FunctionalInterface:
If an interface declares an abstract method overriding one of the public methods of java.lang.Object, that also does not count toward the interface's abstract method count since any implementation of the interface will have an implementation from java.lang.Object or elsewhere.
equals() is inherited from Object, and inherited public methods are not counted when you’re determining whether an interface is a functional interface. So even though equals() is abstract in Comparator, because it’s inherited, it doesn’t count.
RULE:
A functional interface is an interface that has one abstract method. Default methods don’t count; static methods don’t count; and methods inherited from Object don’t count.
All classes descend from Object class and Object contains an equal method.
So, this means that every instance that implements Comparator will already have an implementation of equal method.
Therefore Only one method is required to override by the Implanting class of Comparator Interface.
This makes only one abstract method in Comparator interface
This is why Comparator is a functional interface
When I'm shooting a glance at lambda expressions, the book touches on a functional interface that has only one abstract method. My issue addresses on that quiz question
/* Which of these interfaces are functional interfaces? */
public interface Adder{
int add(int a, int b);
}
public interface SmartAdder extends Adder{
int add(double a, double b);
}
public interface Nothing{
}
I know the last one is not, but I think the first and second ones should be functional interface. But the book says second one is not. Why? Doesn't it overrides add method? So even in second, isn't there only one abstract method?
An easy way to find out would be to try to define a class that implements SmartAdder. The compiler will tell you you need to implement both add(int, int) and add(double, double).
It's understandable that you thought add(double, double) would override add(int, int), but they are in fact separate methods, and could potentially have totally unrelated implementations.
If SmartAdder had defined a default implementation of add(int, int) it would be a functional interface still:
public interface SmartAdder extends Adder {
int add(double a, double b);
default int add(int a, int b) {
return add((double)a, (double)b); // this calls the double method instead
}
}
You may also have come across the #FunctionalInterface annotation - this can be placed on an interface to enforce at compile-time that the interface has exactly one abstract method. If SmartAdder was annotated with #FunctionalInterface the interface itself would not compile.
SmartAdder has two methods. The method signatures are different. Functional Interface can have only one method.
In Java, a method in a subtype overrides a method of a parent type when it has the same signature. Signature means both name and arguments of the method. In particular, arguments must be of the same exact type and must be declared in the same exact order in both methods, i.e. the types of the arguments of the method declared in the subtype can't be subtypes or types wider than the types of the arguments declared in the parent type's method.
So, in your SmartAdder interface, the method with signature add(double a, double b) does not override the method add(int a, int b) of your Adder interface, because double is wider than int. When a type has two or more methods with the same name but with different arguments, it is called method overloading, and it's totally different than method overriding.
This is why SmartAdder ends up having two abstract methods, hence it's not a functional interface (which requires the type to have only one abstract method).
interface SmartAdder overloads, not overrides interface Adder. That's because the name is the same name, but parameters types are different. Therefore, it has 2 functions. To be a functional interface it needs to have only one function.
==> Only interface Adder is a functional interface.
This question already has answers here:
Precise definition of "functional interface" in Java 8
(9 answers)
Closed 6 years ago.
As everybody knows, Java 8 have introduced functional programming to Java developers.
Such interfaces as Comparator, Runnable, Callable and so on are functional.
As follows from definition: functional interfaces are interfaces that have only a single abstract method.
But for example the same interface Comparator have more than one abstract methods:
int compare(T o1, T o2);
boolean equals(Object obj); // inherited from Object class
// and a lot of concrete methods more
Well, is there any strict rule of how to determine if the interface is functional, so that it can be used as the assignment target for a lambda expression or method reference?
From the JLS §9.8 (highlighting mine):
A functional interface is an interface that has just one abstract method (aside from the methods of Object)
The rationale is
to use the interface, you have to instantiate it. Every instantiation necessarily inherits from Object, and thus implements these abstract methods anyway.
The other method - boolean equals(Object) - is an explicit declaration of an abstract method that would otherwise be implicitly declared, and will be automatically implemented by every class that implements the interface.
As a functional interface, it is very unlikely that you want to call a method that is defined by Object. Thus, these methods do not count when searching for a method to call (because the method of functional interfaces can be called without naming that method).
This is to allow functional treatment of an interface like java.util.Comparator<T> that declares multiple abstract methods of which only one is really "new" - int compare(T,T).
I understand that a bunch of methods with body are added using default keyword of Java 8. My question is where is this annotation useful? I see that only one abstract method is allowed to be added with such annotation. But Comparator has two abstract methods :
int compare(T o1, T o2);
boolean equals(Object obj);
Also , Please explain the use of this annotation? I see that its a runtime annotation, so what are its uses ?
From the documentation of FunctionalInterface :
If an interface declares an abstract method overriding one of the public methods of java.lang.Object, that also does not count toward the interface's abstract method count since any implementation of the interface will have an implementation from java.lang.Object or elsewhere.
Since equals is from java.lang.Object, it's not counted. It's the compare method that relates to FunctionInterface.
It is useful since if you add it and you have more than one method in the interface you will get a compile error.
#FunctionalInterfaces are used to let Java know that this piece of code might be replaced with a lambda expression since every lambda is an implementation of some #FunctionalInterface. It helps the compiler know what is the method signature.