private ExecutorService exec = Executors.newSingleThreadExecutor(r -> {
Thread t = new Thread(r);
t.setDaemon(true); // allows app to exit if tasks are running
return t ;
});
I understand the idea behind an executor but, the paramater r
is confusing me. I used:
final ExecutorService exec = Executors.newSingleThreadExecutor(r -> {
Thread t = new Thread(r);
System.out.println("Class of r: " + r.getClass()+ ". r to string: " + r.toString());
System.out.println("Class of t: " + t.getClass() +". Name of t: "+ t.getName());
t.setDaemon(true);
return t;
});
to dig deeper and the result is:
Class of r: class java.util.concurrent.ThreadPoolExecutor$Worker. r to string: java.util.concurrent.ThreadPoolExecutor$Worker@1dc3963[State = -1, empty queue]
Class of t: class java.lang.Thread. Name of t: Thread-3
r
is being passed as a parameter to the Thread
object constructor.
r
indicating that the object being passed is a ThreadPoolExecutor
?ThreadPoolExecutor
passable as a parameter if it does not implement Runnable
as required by the by Thread's
constructor?If someone could provide me with a non-lambda version of the code as well, it would be of great benefit to my understanding.
Introduction. Lambda expressions are a new and important feature included in Java SE 8. They provide a clear and concise way to represent one method interface using an expression. Lambda expressions also improve the Collection libraries making it easier to iterate through, filter, and extract data from a Collection .
A lambda expression is an anonymous or unnamed method in Java. It doesn't execute on its own and used to implement methods that are declared in a functional interface. If we want to pass a lambda expression as a method parameter in java, the type of method parameter that receives must be of functional interface type.
Q 5 - Which of the following is correct about Java 8 lambda expression? A - Lambda expressions are used primarily to define inline implementation of a functional interface.
Lambda Expressions were added in Java 8. A lambda expression is a short block of code which takes in parameters and returns a value. Lambda expressions are similar to methods, but they do not need a name and they can be implemented right in the body of a method.
newSingleThreadExecutor
takes a ThreadFactory as an argument. ThreadFactory
defines a single method newThread
that takes a Runnable as an argument and returns a Thread.
The lambda may make more sense to you if we specify the type of r
explicitly:
(Runnable r) -> {
Thread t = new Thread(r);
return t;
}
Now it is more obvious this is a definition for the body of newThread
.
Except that since the lambda is immediately passed as an argument to a method that accepts a ThreadFactory, the compiler is able to infer that the type of r
must be Runnable. Therefore it can be omitted.
Without the lambda, this code translates to the following anonymous class definition and instantiation:
private ExecutorService exec = Executors.newSingleThreadExecutor(
new ThreadFactory() {
@Override
public Thread newThread(Runnable r) {
Thread t = new Thread(r);
t.setDaemon(true);
return t;
}
}
);
How is the simple letter
r
indicating that the object being passed is aThreadPoolExecutor
?
The type of r
is Runnable because the target type of the lambda expression defines its single method that way.
The object you are seeing passed is actually a ThreadPoolExecutor$Worker
which is a private inner class of ThreadPoolExecutor that implements Runnable.
How is a
ThreadPoolExecutor
passable as a parameter if it does not implementRunnable
as required by the byThread
's constructor?
See above (r
is a Runnable).
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