Lambda This Reference in Java

Lambda this reference in java

You can't reference to this in a lambda expression. The semantic of this has been changed to reference the instance of the surrounding class only, from within the lambda. There is no way to reference to the lambda expression's this from inside the lambda.

The problem is that you use this in the main() method. The main method is static and there is no reference to an object that represents this.

When you use this inside an instance of an inner class you are referencing to the instance of the inner class.
A lambda expression is not an inner class, this is not referencing to the instance of the lambda expression. It is referencing to the instance of the class you define the lambda expression in. In your case it would be a instance of Main. But since your are in a static method, there is no instance.

This is what your second compilation error is telling you. You hand over an instance of Main to your method. But your method signature requires an instance of Observer.

Update:

The Java Language Specification 15.27.2 says:

Unlike code appearing in anonymous class declarations, the meaning of names and the this and super keywords appearing in a lambda body, along with the accessibility of referenced declarations, are the same as in the surrounding context (except that lambda parameters introduce new names).

The transparency of this (both explicit and implicit) in the body of a lambda expression - that is, treating it the same as in the surrounding context - allows more flexibility for implementations, and prevents the meaning of unqualified names in the body from being dependent on overload resolution.

Practically speaking, it is unusual for a lambda expression to need to talk about itself (either to call itself recursively or to invoke its other methods), while it is more common to want to use names to refer to things in the enclosing class that would otherwise be shadowed (this, toString()). If it is necessary for a lambda expression to refer to itself (as if via this), a method reference or an anonymous inner class should be used instead.

this' reference escape in case of lambda in constructor

In the original question the code

public class ThisEscape {
public ThisEscape(EventSource source) {
source.registerListener(
new EventListener() {
public void onEvent(Event e) {
doSomething(e);
}
});
}
}

is problematic because the object have been registered in the constructor and then may be used by the event managing system while not being fully constructed. In fact, this is dangerous only if doSomething access something external to the ThisEscape instance.

And this is the same with your lambda "equivalent"

public class ThisEscape {
public ThisEscape(EventSource source) {
source.registerListener(e -> doSomething(e));
}
}

But don't be fooled, anonymous inner classes are not strictly equivalent to lambdas... this refers to the current instance in case of anonymous inner class but refers to the enclosing instance of the lambda (this is in the closure of the lambda) :

interface doable {
public void doIt();
}
public class Outer {
public Outer() {
doable d1 = new doable() {
public void doIt() {
System.out.println(this);
}
};
d1.doIt();

doable d2 = ()->System.out.println(this);
d2.doIt();
}
public static void main(String []argv) {
new Outer();
}
}

produces something like :

Outer$1@3764951d
Outer@3cd1a2f1

The second line clearly shows that a lambda is not an instance of any class, it is not an object.

Variable referencing from a lambda expression in Java

The first example works, because a = true is actually shorthand for this.a = true, and this is always final (so says the Java Specification).

Getting an interface reference inside a lambda function

Since default lambda conversion gives you lambda with signature of () -> Unit, that means underlying Runnable is completely hidden.

You have to either deal with manual object creation, or write a wrapper extension function that will consume lambda with another signature:

// custom extension function for handler
inline fun Handler.postDelayed(delayMilis: Long, crossinline runnable: (Runnable) -> Unit) = postDelayed(object : Runnable{
override fun run() {
runnable(this)
}
}, delayMilis)

Then at calling side you will be provided with Runnable object as lambda parameter (only parameter: it):

hwnd.postDelayed(5000){ 
// it : Runnable
hwnd.postDelayed(it, 5000)
}

Or if you want to get really fancy, change extension parameter to Handler.(Runnable) -> Unit, then you will be able to call:

hwnd.postDelayed(5000){ 
// this : Handler, it : Runnable
postDelayed(it, 5000)
}

How to make a self-referencing lambda function that also references locals in Java?

Although I do not understand why such a construct would be needed, you could do:

    Consumer<String> func = argument -> {}; // pre-declare func
List<Consumer<String>> one = new ArrayList<>(Collections.singletonList(func));
one.set(0, argument -> {
async_fn(one.get(0)); // Call an function with func as callback
});

Or use an array.

new Keyword In Java Lambda Method Reference

String::toUpperCase is a method reference that can be applied to any String instance.

new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).

Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).

On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.

It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:

Supplier<String> emptyStringToUpperCase = new String()::toUpperCase; 

This is similar to:

Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();

while this:

UnaryOperator<String> stringToUpperCase = String::toUpperCase;

is similar to:

UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();

What would be the problem in providing this keyword for Java's lambda body?

The Java Language Specification 15.27.2 says:

Unlike code appearing in anonymous class declarations, the meaning of names and the this and super keywords appearing in a lambda body, along with the accessibility of referenced declarations, are the same as in the surrounding context (except that lambda parameters introduce new names).

The transparency of this (both explicit and implicit) in the body of a lambda expression - that is, treating it the same as in the surrounding context - allows more flexibility for implementations, and prevents the meaning of unqualified names in the body from being dependent on overload resolution.

Practically speaking, it is unusual for a lambda expression to need to talk about itself (either to call itself recursively or to invoke its other methods), while it is more common to want to use names to refer to things in the enclosing class that would otherwise be shadowed (this, toString()). If it is necessary for a lambda expression to refer to itself (as if via this), a method reference or an anonymous inner class should be used instead.

Java method reference with return result using lambda expression

Using BooleanSupplier as proposed by @Holger is a possible solution.

However I would recommend using an IntPredicate, because this allows you to pass the testValue from handleSomethingto the predicate:

import java.util.function.IntPredicate;

public class Test {
public static void main(String[] args) {
int testValue = 23;
handleSomething(true, testValue, Test::checkIfZero);
handleSomething(false, testValue, Test::checkIfLargerThanZero);
}

private static boolean checkIfZero(int value) {
if (value == 0)
return true;
return false;
}

private static boolean checkIfLargerThanZero(int value) {
if (value > 0)
return true;
return false;
}

private static void handleSomething(boolean test, int value, IntPredicate function) {
if (test) {
System.out.println("Ignore");
return;
}

if (function.test(value))
System.out.println("Passed");
else
System.out.println("Failed");
}
}


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