Develop Reference

Can a class have two no-args methods with same name?

Reading documentation:
https://docs.oracle.com/javase/9/docs/api/java/util/ServiceLoader.html
I was confused by this sentence:
The service provider class file has more than one public static no-args method named "provider".
Assuming JavaDoc is correct, and assuming that static members are not inherited, how this might be possible in Java? Or it's an error in JavaDoc?

how this might be possible in Java?
It is not, since method signatures must be unique, and the signature is the method name and parameter types.
Except, that is not how the JVM works, only how Java works.
The JVM includes the return type as part of the signature, so technically a class can have multiple methods with the same name and parameters, but different return types.
So it can definitely happen for classes written in other JVM languages, but can it happen for a Java class?
Yes it can, when you have covariant return types for overridden methods. See Can overridden methods differ in return type?
Given the example in that answer, what really happens is this fake code:
class ShapeBuilder {
...
public Shape build() {
....
}
class CircleBuilder extends ShapeBuilder{
...
#Override
public bridge Shape build() { // override matches full JVM signature
return <Circle>build(); // call other method with different return type
}
public Circle build() {
....
}
The "bridge" method is a hidden method generated by the compiler to make the difference between Java and the JVM work correctly.
FYI: In that aspect, "bridge" methods are similar to "synthetic" methods, which are generated by the compiler to allow outer classes access to private members of inner classes, or vice versa.

Why does Java not allow multiple inheritance but does allow conforming to multiple interfaces with default implementations

I am not asking this -> Why is there no multiple inheritance in Java, but implementing multiple interfaces is allowed?
In Java, multiple inheritance isn't allowed, but, after Java 8, Interfaces can have default methods (can implement methods itself), just like abstract classes. Within this context, it multiple inheritance should also be allowed.
interface TestInterface
{
// abstract method
public void square(int a);
// default method
default void show()
{
System.out.println("Default Method Executed");
}
}

Things are not so simple.
If a class implements multiple interfaces that defines default methods with the same signature the compiler will force you to override this method for the class.
For example with these two interfaces :
public interface Foo {
default void doThat() {
// ...
}
}
public interface Bar {
default void doThat() {
// ...
}
}
It will not compile :
public class FooBar implements Foo, Bar{
}
You should define/override the method to remove the ambiguity.
You could for example delegate to the Bar implementation such as :
public class FooBar implements Foo, Bar{
#Override
public void doThat() {
Bar.super.doThat();
}
}
or delegate to the Foo implementation such as : :
public class FooBar implements Foo, Bar {
#Override
public void doThat() {
Foo.super.doThat();
}
}
or still define another behavior :
public class FooBar implements Foo, Bar {
#Override
public void doThat() {
// ...
}
}
That constraint shows that Java doesn't allow multiple inheritancy even for interface default methods.
I think that we cannot apply the same logic for multiple inheritances because multiples issues could occur which the main are :
overriding/removing the ambiguity for a method in both inherited classes could introduce side effects and change the overall behavior of the inherited classes if they rely on this method internally. With default interfaces this risk is also around but it should be much less rare since default methods are not designed to introduce complex processings such as multiple internal invocations inside the class or to be stateful (indeed interfaces cannot host instance field).
how to inherit multiple fields ? And even if the language allowed it you would have exactly the same issue as this previously quoted : side effect in the behavior of the inherited class : a int foo field defined in a A and B class that you want to subclass doesn't have the same meaning and intention.

The language designers already thought about that, so these things are enforced by the compiler. So if you define:
interface First {
default void go() {
}
}
interface Second {
default void go() {
}
}
And you implement a class for both interfaces:
static class Impl implements First, Second {
}
you will get a compilation error; and you would need to override go to not create the ambiguity around it.
But you could be thinking that you can trick the compiler here, by doing:
interface First {
public default void go() {
}
}
static abstract class Second {
abstract void go();
}
static class Impl extends Second implements First {
}
You could think that First::go already provides an implementation for Second::go and it should be fine. This is too taken care of, thus this does not compile either.
JLS 9.4.1.3 : Similarly, when an abstract and a default method with matching signatures are inherited, we produce an error. In this case, it would be possible to give priority to one or the other - perhaps we would assume that the default method provides a reasonable implementation for the abstract method, too. But this is risky, since other than the coincidental name and signature, we have no reason to believe that the default method behaves consistently with the abstract method's contract - the default method may not have even existed when the subinterface was originally developed. It is safer in this situation to ask the user to actively assert that the default implementation is appropriate (via an overriding declaration).
The last point I would bring in, to solidify that multiple inheritance is not allowed even with new additions in java, is that static methods from interfaces are not inherited. static methods are inherited by default:
static class Bug {
static void printIt() {
System.out.println("Bug...");
}
}
static class Spectre extends Bug {
static void test() {
printIt(); // this will work just fine
}
}
But if we change that for an interface (and you can implement multiple interfaces, unlike classes):
interface Bug {
static void printIt() {
System.out.println("Bug...");
}
}
static class Spectre implements Bug {
static void test() {
printIt(); // this will not compile
}
}
Now, this is prohibited by the compiler and JLS too:
JLS 8.4.8 : A class does not inherit static methods from its superinterfaces.

Java doesn't allow multiple inheritance for fields. This would be difficult to support in the JVM as you can only have references to the start of an object where the header is, not arbitrary memory locations.
In Oracle/Openjdk, objects have a header followed by the fields of the most super class, then the next most super class, etc. It would be a significant change to allow the fields of a class to appear at different offsets relative to the header of an object for different subclasses. Most likely object references would have to become a reference to the object header and a reference to the fields to support this.

default methods in interfaces pose a problem that :
If both of the implemented interfaces define a default method with
same method signature, then the implementation class does not know
which default method to use.
The implementation class should define explicitly specify which default method to use or define it's own one.
Thus default methods in Java-8 do not facilitate multiple inheritance. The main motivation behind default methods is that if at some point we need to add a method to an existing interface, we can add a method without changing the existing implementation classes. In this way, the interface is still compatible with older versions. However, we should remember the motivation of using Default Methods and should keep the separation of interface and implementation.

The main issues with multiple inheritance are ordering (for overriding and calls to super), fields and constructors; interfaces don't have fields or constructors, so they don't cause problems.
If you look at other languages they usually fall in two broad categories:
Languages with multiple inheritance plus a few features to disambiguate special cases: virtual inheritance [C++], direct calls to all superconstructors in the most-derived class [C++], linearization of superclasses [Python], complex rules for super [Python], etc.
Languages with a differente concept, usually called interfaces, traits, mixins, modules, etc. that impose some limitations such as: no constructors [Java] or no constructors with parameters [Scala until very recently], no mutable fields [Java], specific rules for overriding (e.g. mixins take precedence over base classes [Ruby] so you can include them when you need a bunch of utility methods), etc. Java has become a language like these.
Why just by disallowing fields and constructors you solve many issues related to multiple inheritance?
You can't have duplicated fields in duplicated base classes.
The main class hierarchy is still linear.
You can't construct your base objects the wrong way.
Imagine if Object had public/protected fields and all subclasses had constructors setting those fields. When you inherit from more than one class (all of them derived from Object), which one gets to set the fields? The last class? They become siblings in the hierarchy, so they know nothing about each other. Should you have multiple copies of Object to avoid this? Would all classes interoperate correctly?
Remember that fields in Java are not virtual (overridable), they are simply data storage.
You could make a language where fields behave like methods and could be overridden (the actual storage would be always private), but that would be a much bigger change and problably wouldn't be called Java anymore.
Interfaces can't be instantiated by themselves.
You should always combine them with a concrete class. That eliminates the need for constructors and makes the programmer's intent clearer too (that is, what is meant to be a concrete class and what's an accessory interface/mixin). This also provides a well-defined place to solve all ambiguities: the concrete class.

That is mostly related to "diamonds problem" i think. Right now if you implement multiple interfaces with the same method, compiler forces you to override method the one you want to implement, because it don't know which on to use. I guess Java creators wanted to remove this problem back when interfaces couldn't use default methods. Now they came up with idea, that is good to be able to have methods with implementation in interfaces, as you can still use those as functional interfaces in streams / lambda expressions and utilize their default methods in processing. You cannot do that with classes but diamond problem still exist there. That is my guess :)

class A{
void m1(){
System.out.println("m1-A");
}
}
class B{
void m1(){
System.out.println("m1-B");
}
}
class C extends A, B{ // this will give an error
// inheritance means making all variables and/or methods available to the child class, here child class will get confused as which m1() method to inherit, hence an error
}

JAVA DOES SUPPORT MULTIPLE INHERITANCE.
If you make a OVERALL COMPARISON OF THE PROGRAMMING LANGUAGE,JAVA,THEN YOU COME TO KNOW THAT I AM TRUE.
Java's topclass or the root class in the Ancestor Hierarchy is the Object class.
This class is a Superclass of all other classes. Hence, each class in Java that we declare or is predefined in the API itself inherits this Object class.
Moreover, Java provides us to inherit one more class of our choice.
Hence, we can say that we are performing INTERLOCKED BUT MULTIPLE INHERITANCE.
2ND Way
Java supports Multiple Inheritance of Interfaces. So you can use as many interface implementations you want. But note, implementing an interface does not define IS A relationship as in case of Inheritance of Classes is possible.

Why in both the code snippets '#Override' annotation work properly?

Although the class B implements the interface A, it does not provide the implementation of the method show(). Again, D extends C but in class D the implementation of displayNothing() is the same as the implementation of displayNothing() method in its superclass C.
But in both the cases #Override annotation work properly, why?
//code snippet 1
interface A
{
void show();
}
abstract class B implements A
{
#Override
abstract public void show();
}
//code snippet 2
class C
{
void displayNothing()
{
}
}
class D extends C
{
#Override
void displayNothing()
{
}
}

#Override is only bothered with whether your function name along with the data type is just the same as the method in your super class (with no Alphabetical error). In your case it is the same.Try changing void as int in your class D. It shows a compilation error. As for your abstract class method, it needn't define anything coz the name says it all, it is abstract.

The purpose of the #Override annotation is to prevent bugs. It's presence informs the compiler that there must be a inherited method with that same exact signature that this class is overriding. It does matter if there is an implementation of a method or not.
It is perfectly legal to not include an #Override annotation. The potential problem you might encounter, though, is if you accidentally had a typo or inadvertently changed the method signature, you would be overloading the method, not overriding it. The compiler would not be able to catch this bug. The program might crash or it might even run, but run incorrectly.
Because this might be a difficult bug to spot, especially with methods containing several parameters, the #Override annotation was created. The #Override annotation informs the compiler that this method is an override, not an overload and if the compiler fails to find a method to override, return a compilation error.
All these methods would compile and could be run but none would override the displayNothing() method.
void displayNothng();
void displaynothing();
void displayNothing(String value);

The #Override annotation "works properly" in both presented cases, because java reuses this annotation for interfaces and classes, even if you feel that the first and the second snippet of code do something different it's just not as important to introduce separated keywords/annotations to differentiate between overriding from an interface and from a class. Even an abstract class doesn't change things here.
The #Override indicates that a given method has a compatible representation in the super class/interface, it's used to ensure we don't have any syntax issue in the definition.

B is abstract. Abstract classes don’t need to provide implementations. If an abstract class doesn’t implement a required method it doesn’t matter. If a class isn’t abstract, then implementations of all abstract methods need to be there.
D provides an override of the displayNothing method in C. The compiler doesn’t check whether the overridden version is the same as the overriding version, and it doesn't care if the overriding implementation is empty. All the Override annotation does is confirm that there is a method with the same signature above the annotated one in the class hierarchy.
The purpose of the annotation is to confirm that you’re actually overriding an existing method and haven’t introduced a bug by misspelling the method name or otherwise not matching the signature of the method you mean to override. See the Java Language Specification, 9.6.4.4, where it says:
Programmers occasionally overload a method declaration when they mean to override it, leading to subtle problems. The annotation type Override supports early detection of such problems.
The classic example concerns the equals method. Programmers write the following in class Foo:
public boolean equals(Foo that) { ... }
when they mean to write:
public boolean equals(Object that) { ... }
This is perfectly legal, but class Foo inherits the equals implementation from Object, which can cause some very subtle bugs.
(Be aware when I describe what #Override does I'm skipping over override-equivalence because you didn't ask about it and there are already perfectly good answers that cover that.

How can Predicate be a Functional Interface if it has more than one abstract method? [duplicate]

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/

is abstract a must or not?

As you know, in a java interface, all methods have to be defined as abstract. But when I define a method as not typing abstract, the compiler says it is okay. I know that an abstract method must not have a body. Does a method somewhere in an interface necessarily have a name abstract or not? : What i mean is, what is the difference between:
public interface blabla {
public void aMethod();
//or
public abstract void aMethod();
}

No, marking an interface method as abstract has no meaning and is never required.
All interface methods are implicitly abstract (and public too btw).
From the JLS:
Every method declaration in the body of an interface is implicitly abstract, so its body is always represented by a semicolon, not a block.
Every method declaration in the body of an interface is implicitly public.
For compatibility with older versions of the Java platform, it is permitted but discouraged, as a matter of style, to redundantly specify the abstract modifier for methods declared in interfaces.
It is permitted, but strongly discouraged as a matter of style, to redundantly specify the public modifier for interface methods.
Related question (+ answer with a historical reference to a statement saying that abstract was once required for interface methods):
Java abstract interface

See the sample example below
interface xyz
{
void methodA();
}
Save this to xyz.java
Now compile this using javac tool
and then use the command given belo
javap xyz
the output would be
Compiled from "xyz.java"
interface xyz {
public abstract void methodA();
}
That means when you compile an interface, compiler makes its signature to public and abstract by default.
So it is not necessary to use abstract keyword for any method of interface.

I don't know that they have to be defined as abstract. Probably because they don't. See Oracle's tutorial.

you don't need to specify abstract (default) because within an interface it does not make sense as all the method of the interface needs to be implemented

All methods in an interface are abstract by definition.
You can't create an object out of an interface (e.g., using Interface i = new Interface();) so there's no difference between marking a method as abstract or not.
Any class that implements the interface needs to decide whether to implement it or to let a subclass do it. So as far as the interface is concerned, all methods are abstract by default.

An abstract method provides no implementation. A class which has an abstract method is necessarily abstract, which means that you cannot create instances of this class. To create an instance of that class, you need to subclass and provide non-abstract overwrites for the abstract methods.
An interface never provides an implementation of its methods and it cannot be instantiated. Therefore every method of an interface is per definition abstract. You do not need to provide the keyword abstract when declaring a method in an interface. And by convention the keyword abstract is not used within an interface.

The methods of an interface don't have to be explicitly defined as abstract because they are implicitly abstract and public as defined in the Java Language Specification §9.4. A redundant declaration is perfectly legal though.

If you forgot to put abstract keyword before interface method, Java will implicitly put public abstract keyword before it. Because all interface methods must be abstract.