JavaScript: Why does closure only happen if I assign the return function to a variable?

Question

Even after reading You don't know JS and JavaScript: The Core I still couldn't understand the behavior of the following code.

Why, when I call counter()(), do I get no closure, but if I assign a variable to the result of counter(), like var getClosure = counter(), I then get a closure when calling getClosure()?

function counter() {

    var _counter = 0;

    function increase() { return _counter++ }

    return increase;
}

// Double ()() to call the returned function always return 0, so no closure. 
counter()() // returns 0
counter()() // returns 0
counter()() // returns 0
counter()() // returns 0

var createClosure = counter();

createClosure() // returns 0
createClosure() // returns 1
createClosure() // returns 2
createClosure() // returns 3

question from:https://stackoverflow.com/questions/47342760/javascript-why-does-closure-only-happen-if-i-assign-the-return-function-to-a-va

Answer 1

_counter is the local variable within function counter(). Every time when you call counter() a new _counter will be created.

But var createClosure = counter() only called the function 1 time, that's why _counter is not newly created every time and 'remembered' there (that's where closure happens)