I'm using Java, and wondering if it's possible to put methods directly into arrays as elements. Something like:
...[] arrayName = {herp(), derp(), jinkies()};
When I looked this up prior, some people mentioned "reflection," but I don't know what this is (I'm very new to programming). Is there a simple way to put methods into arrays (my goal is to spit them out randomly with a Random, and eventually terminate with a String parameter in each method named "quit" or something)? If not, how does reflection work for this circumstance (if at all)?
I am NOT only asking what reflection is. My exact (main) question is "Is it possible to put methods into arrays," and if it is, how is that done? What is the syntax for it?
Thank you for your time
It is not possible in java to assign function as object.
But still you can use use lambda expression or similar structure for this. Which will not actually assign functions but you can use it like one.
Create a functional interface
#FunctionalInterface
interface MyFunction {
public void fun();
}
Use lambda expressions to initialize array of that functional interfaces
MyFunction[] functions = new MyFunction[]{
() -> System.out.println("I am herp"),
() -> {
int a = 2;
int b = 3;
System.out.println(a + b);
}
};
Use the array like array of functions.
functions[0].fun();
you can make a class for every method and make an array of classes.
also for it to be easier do it with inner classes and it will be more readable.
(of course every class will share the same interface)
Related
This question already has answers here:
Non-class functions in Java
(4 answers)
Closed 2 years ago.
When declaring methods in Java, do they need to be a part of a class? I am familiar with the idea of a Utility Class:
"Utility Class, also known as Helper class, is a class, which contains just static methods, it is stateless and cannot be instantiated. It contains a bunch of related methods, so they can be reused across the application."
However, can one just create a method separate from any class altogether? (I'd assume scope becomes public by default and declaring anything else for scope might result in an error).
If this is not possible, perhaps that would explain the need for Utility Classes, but I wasn't sure as I hadn't thought about this before - I assumed naturally you could make functions separate from any specific class, but I had been looking through various code samples and couldn't find a specific example where this was occurring.
Part of the reason I am asking this is I was reviewing this article (and mentioned in point 2):
https://www.geeksforgeeks.org/lambda-expressions-java-8/
In it, it states: Lambda expressions are added in Java 8 and provide below functionalities.
1) Enable to treat functionality as a method argument, or code as data.
2) A function that can be created without belonging to any class.
3) A lambda expression can be passed around as if it was an object and executed on demand.
Java is a sort of purely class-based programming language. So, Yes, it and everything needs to be a part of a class.
You are right, you can make a Utility class making methods public static in this way methods can be called without instantiating the class.
Answer to question in the comment:
Why would someone write Object.method() instead of just method()?
Object class is a standard class in java.lang package. You should not create your class named Object otherwise you will need to specify java.lang.Object everywhere you use java.lang.Object.
Now you probably meant
Why would someone write MyUtilClass.method() instead of just method()?
Suppose you have a class MyUtilClass as follows
public class MyUtilClass {
public static int utilMethodA() {
return 1;
}
public static boolean utilMethodB() {
int value = utilMethodA();
if(value == 1)
return true;
else
return false;
}
}
And suppose you have another class MyClass as
public class MyClass {
public void classMethod() {
int value = MyUtilClass.utilMethodA();
}
}
Here if you see in MyUtilClass, utilMethodB() uses utilMethodA() without writing MyUtilClass.utilMethodA() (however, we could write it that way also). Here we did not need to write it as MyUtilClass.utilMethodA() because compiler can find the utilMethodA() without fully specifying it's class because it is present inside it's own class.
Now, In Myclass's myMethod(), we must specify MyUtilClass.utilMethodA() (without it, it won't work), because the compiler has no way of figuring out that you meant to call utilMethodA() of MyUtilClass. There could be hundreds of classes with a method named utilMethodA(), the compiler has no way of finding out which one of the hundred methods you want to call.
Note:-
Also, you can do static import of MyUtilClass.myMethod() like
import static my.package.name.MyUtilClass.myMethodA()
and then use utilMethodA() inside MyClass without prefixing MyUtilClass (but you already informed compile by static import that you will be using utilMethodA() of MyUtilClass right?)
Looks cool to you? No!
This is rather a bad way because
It makes code looks unobvious. In a large class, it may seem that
method utilMethodA() is a local method defined somewhere in
MyClass.
Also, it can generate ambiguity to the compiler if more than one static import of utilMethodA() is done. As compiler has no way of figuring out which of the two you intend to use.
(Edit) Regarding Lambda Expression
Lambda expression is pretty cool stuff added in Java 8. They are basically a kind of function. They provide you the power to define a function right where they need to be used. For example in this link that you provided, see the example shown below syntax of lambda, there the statement
ArrayList<Integer> arrL = new ArrayList<Integer>();
arrL.add(1);
arrL.add(2);
arrL.add(3);
arrL.add(4);
arrL.forEach(n -> { if (n%2 == 0) System.out.println(n); });
Basically, what we are doing here is, we are defining a function, if n is multiple of 2, we print n. We are doing it forEach element of arrL. Did you see, we defined the function to be executed on each element right inside a function call forEach(). That's the beauty of lambda expression.
Now, coming to your question,
So the primary benefit of lambda (besides syntax) is to make it easier to implement functional interfaces (compared to what alternative)?
Yes, sort of. Easy in terms of not creating a separate class implementing the interface and then implementing the abstract method and then calling that implemented method.
This becomes lots of work, especially if you need to call that method only once for example,
Consider the Functional Interface FuncInterface defined as in the link in your question:
interface FuncInterface {
// An abstract function
void abstractFun(int x);
// A non-abstract (or default) function
default void normalFun() {
System.out.println("Hello");
}
}
Now, you want two kind of implementation to your functional interface:
One that provides twice of the passed int x.
Another one that provides square of passed int x.
So, you make two implementations of it:
First FuncInterfaceTwiceImpl
public class FuncInferFaceTwiceImpl implements FuncInterface {
#Override
public void abstractFun(int x) {
System.out.println(2 * x);
}
}
Second, FuncInterfaceSquareImpl as
public class FuncInterfaceSquareImpl implements FuncInterface {
#Override
public void abstractFun(int x) {
System.out.println(x * x);
}
}
Now, you call them as
public class MyClass {
public static void main(String[] args) {
FuncInterface interfaceTwiceObject = new FuncInferFaceTwiceImpl();
interfaceTwiceObject.abstractFun(5);
FuncInterface interfaceSquareObject = new FuncInterfaceSquareImpl();
interfaceSquareObject.abstractFun(5);
}
}
It prints
10
25
Now, what you had to do?
You had to create two separate Classes (in separate new files or
could have made private classes in the same file that of MyClass),
each implementing the abstract method.
Then you instantiated objects of each class and called them
respectively in the main function.
What if this is the only place where you had to call this twice and square thing? You had to make two classes just to use them only once. This effort is too much!!
What if you want to call it without creating new classes and implementing methods in a class?
What if I tell you only provide me the method body, I will do the work for you without you to bother about implementing interface and overriding methods?
Here comes the Lambda magic. Instead of making any impl classes just
head straight towards the main method
Instantiate two objects of FuncInterface providing only method body in Lambda expression.
Call abstract method from objects just like below
public class MyClass {
public static void main(String[] args) {
FuncInterface interfaceTwiceObject = (n) -> System.out.println(2*n);
interfaceTwiceObject.abstractFun(5);
FuncInterface interfaceSquareObject = (n) -> System.out.println(n*n);
interfaceSquareObject.abstractFun(5);
}
}
And boom, the output is
10
25
Just one more time where Lambda saved your day!!
Yes all methods in Java have to be part of a class. You cannot create a method (static or otherwise) which is not associated with a class.
EDIT
Before I answer your question, I will point out that lambda expressions were introduced in Java 8 through the concept of SAM types. In addition, a bit of syntactic sugar was also introduced to facilitate the creation of these types.
When you hear the term "Lambda expression" in Java, you should always remember that they are expressions. Your confusion stems from thinking that lambda expressions evaluate to a pure function not associated with a class or object; well this is simply not the case in Java and I will show you why.
Lambda expressions are not functions
I can now see where your confusion comes from because that article you are reading made a false claim when they say that lambda expression is:
A function that can be created without belonging to any class.
This is simply not true. A lambda expression in Java is not a function. Take the example they give for instance.
interface FuncInterface
{
// An abstract function
void abstractFun(int x);
// A non-abstract (or default) function
default void normalFun()
{
System.out.println("Hello");
}
}
class Test
{
public static void main(String args[])
{
// lambda expression to implement above
// functional interface. This interface
// by default implements abstractFun()
FuncInterface fobj = (int x)->System.out.println(2*x);
// This calls above lambda expression and prints 10.
fobj.abstractFun(5);
}
}
Proof
Now take the comment they have in the main method:
lambda expression to implement above functional interface
From the start they admit that the next line of code implements a functional interface. However functions in Java do not implement interfaces, only classes or other interfaces can do that!
Now, they even go ahead and "call" this function:
This calls above lambda expression and prints 10.
except instead of directly invoking the function (as anyone would if this was really a function), they use the property accessor notation (.) to access the actual method they wanted to call, which means what we have here is not a function, but actually an instance of an anonymous class.
Furthermore, since this object actually contains another method (normalFun), one might ask the question, which one do I use when I want to pass this "function" to another method? This is not a question that is commonly (if ever) asked in the context of lambda functions because there is only one thing to do with a lambda function and that is to call it.
In closing
Java has lambda expressions, not lambda functions.
What makes it a lambda expression is simply the syntactic sugar introduced in Java 8 that uses the () -> { } notation. Unfortunately, many fans of functional programming began associating the term "Lambda function" with objects created using this syntax, and this has led to the confusion you have expressed in your question.
To rehash what I answered previously, all functions in Java are part of a class, and you cannot have a function which is not associated with an object, nor can you create a function outside a class.
HTH
say I have a functional interface and Lambda and the standard use:
interface NumericMethod{
int operation(int number);
}
NumericMethod add1 = (x) -> x+1
System.out.println(add1.operation(7))//but not (...add1(7))
So my question is, why not simply add1(7)? There is only one method that can be called.
Because in your code add1 isn't a method, it's an object. If, when defining lambdas, the Java 8 teams decided that classes with one method can be called directly, it would have left a whole lot of mess to clean up regarding the ever narrowing line between objects and methods.
"Java 8 Lambdas: Pragmatic Functional Programming" has an example for using peek method in Stream API. This piece of code prints artist nationalities whose name starts with "The":
Set<Nationality> nationalities = album.getMusician()
.filter(artist -> artist.getName().startsWith("The"))
.map(artist -> artist.getNationality())
.peek(nation -> System.out.println(nation))
.collect(Collectors.toList());
I want to rewrite this code with method references:
Set<Nationality> nationalities = album.getMusician()
.filter(artist -> artist.getName().startsWith("The"))
.map(Artist::getNationality)
.peek(System.out::println)
.collect(Collectors.toList());
Is there any solution to rewrite filter(artist -> artist.getName().startsWith("The"))?
You need to create a separate method that takes an Artist and returns a boolean:
private boolean nameStartsWithThe(Artist a) {
return a.getName().startsWith("The");
}
Set<Nationality> nationalities = album.getMusician()
.filter(this::nameStartsWithThe)
or with a static method:
private static boolean nameStartsWithThe(Artist a) {
return a.getName().startsWith("The");
}
Set<Nationality> nationalities = album.getMusician()
.filter(MyClass::nameStartsWithThe)
You'd need something that composes the two methods. There are some methods for composing methods (IntUnaryOperator has compose and andThen methods that can compose two IntUnaryOperators into a new IntUnaryOperator). But the ones I've found all seem specialized for certain types of functional interfaces; defining compose methods for every possible pair of functional interface types would be too unwieldy.
I did get something to work that would compose a Function and a Predicate to get a new Predicate:
static <T,U> Predicate<T> functionPredicate(Function<T,U> func, Predicate<U> pred) {
return obj -> pred.test(func.apply(obj));
}
That is, it can compose a predicate that operates on T from a function that takes a T and returns U, and a predicate that operates on U. This would almost work on your example, except that startsWith needs another parameter. But this does work:
static boolean startsWithThe(String s) {
return s.startsWith("The");
}
Predicate<Artist> pred = functionPredicate(Artist::getName, ThisClass::startsWithThe);
where ThisClass is whatever class contains startsWithThe. This works. If you want to avoid writing a new method (like startsWithThe), you could probably write a "parameterized predicate" generic method so that you write something like
Predicate<Artist> pred = functionPredicate(Artist::getName, parameterizedPredicate(String::startsWith, "The"));
but I haven't tried it.
So it seems it's possible to come up with something that will let you use method references instead of lambdas. I question whether it's worthwhile. To me, a method reference is just a shorthand for certain kinds of lambdas; and unless you can do what you want with a simple method reference, I'd think using a lambda is concise and clear enough, and you don't need to add all the extra rigmarole like my functionPredicate method. I've seen several questions that ask something like "How can I use a method reference for this instead of a lambda?", and I honestly don't understand why.
There is no way to replace that line with a method reference.
Method reference works by using the fact that there is only one object being used in entire lambda expression and the compiler can infer it (reference does not matter and type can be inferred) using target typing.
So,
artist -> artist.getNationality()
is replaced with
Artist::getNationality
Here Artist::getNationality method matches with the target type without requiring any further information.
In case of artist -> artist.getName().startsWith("The"), there are two method calls in the lambda expression. The order, parameters are important, and have to be specified.
It looks as if the artist reference should be inferred, but the compiler won't know what object should the startsWith("The") method be called on.
Hope this helps.
Am having some arguments say (String a, Treeset b, Set c)
and I try to get the class by arguments[i].getClass(); of the above arguments..
is Iit possible to get the class of the interface <Set>.
For example:
Class[] argumentTypes = new Class [arguments.length];
for (int i = 0 ; i < arguments.length ; i++)
{
argumentTypes[i] = arguments[i].getClass();
}
The code you've given will find the classes of the arguments (i.e. the values provided to the method) - but those can never be interfaces; they'll always be concrete implementations. (You can never pass "just a set" - always a reference to an object which is an instance of an implementation of the interface, or a null reference.)
It sounds like you want the types of the parameters - which you'd get via reflection if you absolutely had to, finding the Method and then getting the parameters from that with getParameterTypes. But given that you're within the method, you already know the parameter types, because they're at the top of the method... I'm not sure the best way of finding "the currently executing" method, if that's what you're after.
If you're just trying to get the class associated with Set, you can use Set.class of course. But again, it's not really clear what you're trying to do.
EDIT: Okay, judging from your comment, there are some logical problems with what you're trying to do. Going from the values of arguments to which method would be invoked is impossible in the general case, because you've lost information. Consider this, for example:
void foo(String x) {}
void foo(Object y) {}
foo("hello"); // Calls first method
foo((Object) "hello"); // Calls second method
Here the argument values are the same - but the expressions have a different type.
You can find all methods which would be valid for the argument values - modulo generic information lost by type erasure - using Class.isAssignableFrom. Does that help you enough?
Note that you'll also need to think carefully about how you handle null argument values, which would obviously be valid for any reference type parameter...
You can use http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Class.html#getInterfaces()
You will get the class what the caller provided.
I mean,in below class you will get HashSet.
Set set=new HashSet();
System.out.println(set.getClass());
You can do this in two ways given below
Set s = new //any class that implements it for example HashSet or TreeSet etc.;
s.getClass().getName(); //This will return the name of the subclass which is refered by s.
or if in other way can do it
Set s = null;
s.getClass();//This causes NullPointer Exception
Is there anyway to override a method at run time? Even if it requires dynamically creating a subclass from that instance?
With plain Java, no.
With ByteBuddy(preferred), asm, cglib or aspectj, yes.
In plain Java, the thing to do in a situation like that is to create an interface-based proxy that handles the method invocation and delegates to the original object (or not).
You could create an anonymous class that overrides the method and uses the strategy pattern to decide what to do.
If you are looking for dynamic compilation from code, you can follow these instructions
As others said, no, you can't override a method at runtime. However, starting with Java 8 you can take the functional approach. Function is a functional interface that allows you to treat functions as reference types. This means that you can create several ones and switch between them (dynamically) a-la strategy pattern.
Let's look at an example:
public class Example {
Function<Integer, Integer> calculateFuntion;
public Example() {
calculateFuntion = input -> input + 1;
System.out.println(calculate(10));
// all sorts of things happen
calculateFuntion = input -> input - 1;
System.out.println(calculate(10));
}
public int calculate(int input) {
return calculateFuntion.apply(input);
}
public static void main(String[] args) {
new Example();
}
}
Output:
11
9
I don't know under what circumstances and design you intend to override, but the point is that you replace the behavior of the method, which is what overriding does.
I think it not possible with simple Java.
With reflection and/or cglib probally you can do it.
Look at these links:
http://www.rgagnon.com/javadetails/java-0039.html
http://www.javaworld.com/javaworld/jw-06-2006/jw-0612-dynamic.html