Higher order function and closure in java

Question

The following code mimics both higher ordered function and closure.

static Function<Integer, Integer> getDoubleFunc(int multi) {
    return (val) -> {
        return val * multi;
    };
}

But why it's not considered as one? Is it because Function here is an interface?

What do you mean by "it's not considered as one"? What is your concern? — Tunaki, Jan 24 '16 at 10:34
I saw a deck by Bruce Eckel in which he said about java 8 not supporting closure. — barunsthakur, Jan 24 '16 at 10:36
Refer to this question: http://stackoverflow.com/q/17204279/1743880 — Tunaki, Jan 24 '16 at 10:38

score 6 · Answer 1 · answered Jan 24 '16 at 19:09

The whole "debate" about whether Java lambda expressions are "real closures" or not is a pretty unproductive one, because most participants are implicitly defining what "real closure" means in terms of what their favorite other language calls closures, and then turning around and saying "Language Y doesn't have 'real' closures."

Java lambda expressions close over values in the enclosing lexical scope, but not over variables.

Wikipedia says: "a closure is a record storing a function together with an environment, which is a mapping associating each free variable of the function (variables that are used locally, but defined in an enclosing scope) with the value or storage location to which the name was bound when the closure was created." (emphasis added).

Higher order function and closure in java

1 Answers1