[FIXED] Called method of inner generic class doesn't recognize the generic type

Issue

I struggle with the following inner class "inheriting" the generic parameter T from the outer class. The following lines are erroneous:

String string = "string";
new OuterClass<>()
    .inner(string)
    .call(s -> s.length());       // Cannot resolve method 'length' in 'Object'

Minimal and reproducible example of these classes:

public class OuterClass<T> {

    public InnerClass<T> inner(T object) {
        return new InnerClass<>(object);
    }

    @AllArgsConstructor(access = AccessLevel.PRIVATE)
    public static final class InnerClass<U> {

        U object;

        public final void call(Consumer<U> consumer) {
            consumer.accept(object);
        }
    }
}

Why does this happen? How am I supposed to specify the generic parameter through the creation of the inner class?

Solution

Essentially, this is because inference of type in the <> does not take into account the subsequent methods that you chain.

For example, this creates an ArrayList<Object>:

new ArrayList<>().add("");

If I insert the word String inside the <>, IntelliJ will not complain that the word String is redundant.

Similarly, this produces an error:

OuterClass<String>.InnerClass i = new OuterClass<>().inner(string);

Because the right hand side produces a OuterClass<Object>.Inner, while the left hand side is a OuterClass<String>.InnerClass.

To fix this, you need to specify the generic parameter:

new OuterClass<String>()
    .inner(string)
    .call(s -> s.length()); 

Regarding your edit, the edited inner method still depends on the generic type of Outer, which cannot be inferred correctly.

You can make the Inner object created not depend on the generic type of Outer by adding another generic parameter:

public <U> InnerClass<U> inner(U object) {
    return new InnerClass<>(object);
}

Answered By - Sweeper
Answer Checked By - David Goodson (JavaFixing Volunteer)