Proper usage of Optional.ifPresent()

Question

I am trying to understand the ifPresent() method of the Optional API in Java 8.

I have simple logic:

Optional<User> user=...
user.ifPresent(doSomethingWithUser(user.get()));

But this results in a compilation error:

ifPresent(java.util.functionError:(186, 74) java: 'void' type not allowed here)

Of course I can do something like this:

if(user.isPresent())
{
  doSomethingWithUser(user.get());
}

But this is exactly like a cluttered null check.

If I change the code into this:

 user.ifPresent(new Consumer<User>() {
            @Override public void accept(User user) {
                doSomethingWithUser(user.get());
            }
        });

The code is getting dirtier, which makes me think of going back to the old null check.

Any ideas?

Answer 1

Optional<User>.ifPresent() takes a Consumer<? super User> as argument. You're passing it an expression whose type is void. So that doesn't compile.

A Consumer is intended to be implemented as a lambda expression:

Optional<User> user = ...
user.ifPresent(theUser -> doSomethingWithUser(theUser));

Or even simpler, using a method reference:

Optional<User> user = ...
user.ifPresent(this::doSomethingWithUser);

This is basically the same thing as

Optional<User> user = ...
user.ifPresent(new Consumer<User>() {
    @Override
    public void accept(User theUser) {
        doSomethingWithUser(theUser);
    }
});

The idea is that the doSomethingWithUser() method call will only be executed if the user is present. Your code executes the method call directly, and tries to pass its void result to ifPresent().