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Precise definition of "functional interface" in Java 8
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According to definition of functional interface - A functional interface is an interface that contains only one abstract method.
But Comparator<T> has two abstract methods:
int compare(T o1, T o2);
boolean equals(Object obj);
others are default and static.
JavaDocs mentions it as functional interface. How can it be?
You're reading the wrong definition, or at least, an (over)simplified one.
The proper definition of a FunctionalInterface is:
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. 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.
SOURCE: Java Language Specification section 9.8
If you look at the source code of Comparator<T> , it is like below:
#FunctionalInterface
public interface Comparator<T> {
// abstract method
int compare(T o1, T o2);
// abstract method, overriding public methods of `java.lang.Object`, so it does not count
boolean equals(Object obj);
}
The equals is an abstract method overriding one of the public methods of java.lang.Object, this doesn’t count as an abstract method.
So In fact The Comparator only has one abstract method i.e int compare(T o1, T o2), and it meet the definition of functional interface.
equals is only explicitly included in the Comparator interface so that they can add some additional JavaDocs to it, e.g. some comparator-specific requirements:
this method can return true only if the specified object is also a comparator and it imposes the same ordering as this comparator
Source: Comparator JavaDocs
The JLS says:
interfaces do not inherit from Object, but rather implicitly declare
many of the same methods as Object (§9.2)
So all the Comparator authors have done is declare something explicitly that is normally implicit.
If the methods of Object counted as abstract methods for the purposes of defining a functional interface then no functional interface would have just a single abstract method. As such, they are not considered:
A functional interface is an interface that has just one abstract
method (aside from the methods of Object)
JLS, emphasis mine
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
While reading JLS Specification for Interfaces I came across following phrase:
If an interface has no direct superinterfaces, 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.
So My question is when we implement an interface why we are not forced to override implicit methods which are declared in Object class even though they are implicitly defined as abstract in Interface.
Hope I put myself correctly.
Thanks.
All classes implicitly extend Object, either directly or through some chain of superclasses. Thus, you don't have to explicitly override the implicit methods declared in an interface because you inherited the implementations from Object.
The point of an interface is to force you to implement some method. The reason everything extends Object is that we want to have some way of dealing with all classes independent of implementation. There is no reason that each class should implement something like getClass(), because the behaviour is always going to be the same.
I have an interface and an abstract class:
interface MyInterface {int presents();};
public abstract class MyAbstractClass implements MyInterface {
abstract void presents();
abstract void chill();
}
The compiler complains that in
abstract void presents();
The return type is incompatible with
MyInterface.presents()
Offering the following options:
make it int type
make MyInterface.presents void type
Why is this happening?
You defined method presents() that returns int in Gift interface.
Then you try to redefine method presents() with return type void in class named myClass.
It is pretty obvious that int is not void. Thism means that return type is incompatible.
You are confusing with methods inheritance. The problem is that syntax of java (and all c-like languages) does not dictate usage of return value, so even if you method returns int you can still call it like following:
presents();
So, assume that compiler allows you to have 2 different methods named presents(): one that returns int, other returns void. How can compiler understand which one of the methods to use in our example? The answer is no way. This is the reason that you can override methods using different parameters signature, e.g.
presents();
presents(String str);
presents(int n);
You cannot however have 2 methods with the same parameters signature and different return type.
In the interface, presents is defined as returning an int, but in your class that claims to implement the interface, it's void. The two need to match.
Because interface having the method presents() and you are trying to change the method signature here.
If you dont want that method write new one.
abstract void anotherPresents();
or
satisfy the original signature
abstract int presents();