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JavaScript closures work?


Well, I tried to explain JavaScript closures to a 27-year-old friend and completely failed.

How would you explain it to someone with a knowledge of the concepts which make up closures (for example, functions, variables and the like), but does not understand closures themselves?

asked Jun 3, 2015 in JAVASCRIPT by rajesh
edited Jun 7, 2015 by rajesh
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1 Answer

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Whenever you see the function keyword within another function, the inner function has access to variables in the outer function.

function foo(x) {
  var tmp = 3;

  function bar(y) {
    alert(x + y + (++tmp)); // will alert 16
  }

  bar(10);
}

foo(2);

This will always alert 16, because bar can access the x which was defined as an argument to foo, and it can also access tmp from foo.

That is a closure. A function doesn't have to return in order to be called a closure. Simply accessing variables outside of your immediate lexical scope creates a closure.

function foo(x) {
  var tmp = 3;

  return function (y) {
    alert(x + y + (++tmp)); // will also alert 16
  }
}

var bar = foo(2); // bar is now a closure.
bar(10);

The above function will also alert 16, because bar can still refer to x and tmp, even though it is no longer directly inside the scope.

However, since tmp is still hanging around inside bar's closure, it is also being incremented. It will be incremented each time you call bar.

The simplest example of a closure is this:

var a = 10;
function test() {
  console.log(a); // will output 10
  console.log(b); // will output 6
}
var b = 6;
test();
When a JavaScript function is invoked, a new execution context is created. Together with the function arguments and the parent object, this execution context also receives all the variables declared outside of it (in the above example, both 'a' and 'b').

It is possible to create more than one closure function, either by returning a list of them or by setting them to global variables. All of these will refer to the same x and the same tmp, they don't make their own copies.

Here the number x is a literal number. As with other literals in JavaScript, when foo is called, the number x is copied into foo as its argument x.

On the other hand, JavaScript always uses references when dealing with Objects. If say, you called foo with an Object, the closure it returns will reference that original Object!

function foo(x) {
  var tmp = 3;

  return function (y) {
    alert(x + y + tmp);
    x.memb = x.memb ? x.memb + 1 : 1;
    alert(x.memb);
  }
}

var age = new Number(2);
var bar = foo(age); // bar is now a closure referencing age.
bar(10);

As expected, each call to bar(10) will increment x.memb. What might not be expected, is that x is simply referring to the same object as the age variable! After a couple of calls to bar, age.memb will be 2! This referencing is the basis for memory leaks with HTML objects.
answered Jun 3, 2015 by rajesh

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