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 

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)