JavaScript closures vs. anonymous functions

Question

A friend of mine and I are currently discussing what is a closure in JS and what isn't. We just want to make sure we really understand it correctly.

Let's take this example. We have a counting loop and want to print the counter variable on the console delayed. Therefore we use setTimeout and closures to capture the value of the counter variable to make sure that it will not print N times the value N.

The wrong solution without closures or anything near to closures would be:

for(var i = 0; i < 10; i++) {
    setTimeout(function() {
        console.log(i);
    }, 1000);
}

which will of course print 10 times the value of i after the loop, namely 10.

So his attempt was:

for(var i = 0; i < 10; i++) {
    (function(){
        var i2 = i;
        setTimeout(function(){
            console.log(i2);
        }, 1000)
    })();
}

printing 0 to 9 as expected.

I told him that he isn't using a closure to capture i, but he insists that he is. I proved that he doesn't use closures by putting the for loop body within another setTimeout (passing his anonymous function to setTimeout), printing 10 times 10 again. The same applies if I store his function in a var and execute it after the loop, also printing 10 times 10. So my argument is that he doesn't really capture the value of i, making his version not a closure.

My attempt was:

for(var i = 0; i < 10; i++) {
    setTimeout((function(i2){
        return function() {
            console.log(i2);
        }
    })(i), 1000);
}

So I capture i (named i2 within the closure), but now I return another function and pass this around. In my case, the function passed to setTimeout really captures i.

Now who is using closures and who isn't?

Note that both solutions print 0 to 9 on the console delayed, so they solve the original problem, but we want to understand which of those two solutions uses closures to accomplish this.

Solution 1

Editor's Note: All functions in JavaScript are closures as explained in this post. However we are only interested in identifying a subset of these functions which are interesting from a theoretical point of view. Henceforth any reference to the word closure will refer to this subset of functions unless otherwise stated.

A simple explanation for closures:

1. Take a function. Let's call it F.
2. List all the variables of F.
3. The variables may be of two types:
   1. Local variables (bound variables)
   2. Non-local variables (free variables)
4. If F has no free variables then it cannot be a closure.
5. If F has any free variables (which are defined in a parent scope of F) then:
   1. There must be only one parent scope of F to which a free variable is bound.
   2. If F is referenced from outside that parent scope, then it becomes a closure for that free variable.
   3. That free variable is called an upvalue of the closure F.

Now let's use this to figure out who uses closures and who doesn't (for the sake of explanation I have named the functions):

Case 1: Your Friend's Program

for (var i = 0; i < 10; i++) {
    (function f() {
        var i2 = i;
        setTimeout(function g() {
            console.log(i2);
        }, 1000);
    })();
}

In the above program there are two functions: f and g. Let's see if they are closures:

For f:

1. List the variables:
   1. i2 is a local variable.
   2. i is a free variable.
   3. setTimeout is a free variable.
   4. g is a local variable.
   5. console is a free variable.
2. Find the parent scope to which each free variable is bound:
   1. i is bound to the global scope.
   2. setTimeout is bound to the global scope.
   3. console is bound to the global scope.
3. In which scope is the function referenced? The global scope.
   1. Hence i is not closed over by f.
   2. Hence setTimeout is not closed over by f.
   3. Hence console is not closed over by f.

Thus the function f is not a closure.

For g:

1. List the variables:
   1. console is a free variable.
   2. i2 is a free variable.
2. Find the parent scope to which each free variable is bound:
   1. console is bound to the global scope.
   2. i2 is bound to the scope of f.
3. In which scope is the function referenced? The scope of setTimeout.
   1. Hence console is not closed over by g.
   2. Hence i2 is closed over by g.

Thus the function g is a closure for the free variable i2 (which is an upvalue for g) when it's referenced from within setTimeout.

Bad for you: Your friend is using a closure. The inner function is a closure.

Case 2: Your Program

for (var i = 0; i < 10; i++) {
    setTimeout((function f(i2) {
        return function g() {
            console.log(i2);
        };
    })(i), 1000);
}

In the above program there are two functions: f and g. Let's see if they are closures:

For f:

1. List the variables:
   1. i2 is a local variable.
   2. g is a local variable.
   3. console is a free variable.
2. Find the parent scope to which each free variable is bound:
   1. console is bound to the global scope.
3. In which scope is the function referenced? The global scope.
   1. Hence console is not closed over by f.

Thus the function f is not a closure.

For g:

1. List the variables:
   1. console is a free variable.
   2. i2 is a free variable.
2. Find the parent scope to which each free variable is bound:
   1. console is bound to the global scope.
   2. i2 is bound to the scope of f.
3. In which scope is the function referenced? The scope of setTimeout.
   1. Hence console is not closed over by g.
   2. Hence i2 is closed over by g.

Thus the function g is a closure for the free variable i2 (which is an upvalue for g) when it's referenced from within setTimeout.

Good for you: You are using a closure. The inner function is a closure.

So both you and your friend are using closures. Stop arguing. I hope I cleared the concept of closures and how to identify them for the both of you.

Edit: A simple explanation as to why are all functions closures (credits @Peter):

First let's consider the following program (it's the control):

lexicalScope();

function lexicalScope() {
    var message = "This is the control. You should be able to see this message being alerted.";

    regularFunction();

    function regularFunction() {
        alert(eval("message"));
    }
}

1. We know that both lexicalScope and regularFunction aren't closures from the above definition.
2. When we execute the program we expect message to be alerted because regularFunction is not a closure (i.e. it has access to all the variables in its parent scope - including message).
3. When we execute the program we observe that message is indeed alerted.

Next let's consider the following program (it's the alternative):

var closureFunction = lexicalScope();

closureFunction();

function lexicalScope() {
    var message = "This is the alternative. If you see this message being alerted then in means that every function in JavaScript is a closure.";

    return function closureFunction() {
        alert(eval("message"));
    };
}

1. We know that only closureFunction is a closure from the above definition.
2. When we execute the program we expect message not to be alerted because closureFunction is a closure (i.e. it only has access to all its non-local variables at the time the function is created (see this answer) - this does not include message).
3. When we execute the program we observe that message is actually being alerted.

What do we infer from this?

1. JavaScript interpreters do not treat closures differently from the way they treat other functions.
2. Every function carries its scope chain along with it. Closures don't have a separate referencing environment.
3. A closure is just like every other function. We just call them closures when they are referenced in a scope outside the scope to which they belong because this is an interesting case.

Solution 2

According to the closure definition:

A "closure" is an expression (typically a function) that can have free variables together with an environment that binds those variables (that "closes" the expression).

You are using closure if you define a function which use a variable which is defined outside of the function. (we call the variable a free variable).
They all use closure(even in the 1st example).

Solution 3

In a nutshell Javascript Closures allow a function to access a variable that is declared in a lexical-parent function.

Let's see a more detailed explanation. To understand closures it is important to understand how JavaScript scopes variables.

Scopes

In JavaScript scopes are defined with functions. Every function defines a new scope.

Consider the following example;

function f()
{//begin of scope f
  var foo='hello'; //foo is declared in scope f
  for(var i=0;i<2;i++){//i is declared in scope f
     //the for loop is not a function, therefore we are still in scope f
     var bar = 'Am I accessible?';//bar is declared in scope f
     console.log(foo);
  }
  console.log(i);
  console.log(bar);
}//end of scope f

calling f prints

hello
hello
2
Am I Accessible?

Let's now consider the case we have a function g defined within another function f.

function f()
{//begin of scope f
  function g()
  {//being of scope g
    /*...*/
  }//end of scope g
  /*...*/
}//end of scope f

We will call f the lexical parent of g. As explained before we now have 2 scopes; the scope f and the scope g.

But one scope is "within" the other scope, so is the scope of the child function part of the scope of the parent function? What happens with the variables declared in the scope of the parent function; will I be able to access them from the scope of the child function? That's exactly where closures step in.

Closures

In JavaScript the function g can not only access any variables declared in scope g but also access any variables declared in the scope of the parent function f.

Consider following;

function f()//lexical parent function
{//begin of scope f
  var foo='hello'; //foo declared in scope f
  function g()
  {//being of scope g
    var bar='bla'; //bar declared in scope g
    console.log(foo);
  }//end of scope g
  g();
  console.log(bar);
}//end of scope f

calling f prints

hello
undefined

Let's look at the line console.log(foo);. At this point we are in scope g and we try to access the variable foo that is declared in scope f. But as stated before we can access any variable declared in a lexical parent function which is the case here; g is the lexical parent of f. Therefore hello is printed.
Let's now look at the line console.log(bar);. At this point we are in scope f and we try to access the variable bar that is declared in scope g. bar is not declared in the current scope and the function g is not the parent of f, therefore bar is undefined

Actually we can also access the variables declared in the scope of a lexical "grand parent" function. Therefore if there would be a function h defined within the function g

function f()
{//begin of scope f
  function g()
  {//being of scope g
    function h()
    {//being of scope h
      /*...*/
    }//end of scope h
    /*...*/
  }//end of scope g
  /*...*/
}//end of scope f

then h would be able to access all the variables declared in the scope of function h, g, and f. This is done with closures. In JavaScript closures allows us to access any variable declared in the lexical parent function, in the lexical grand parent function, in the lexical grand-grand parent function, etc. This can be seen as a scope chain; scope of current function -> scope of lexical parent function -> scope of lexical grand parent function -> ... until the last parent function that has no lexical parent.

The window object

Actually the chain doesn't stop at the last parent function. There is one more special scope; the global scope. Every variable not declared in a function is considered to be declared in the global scope. The global scope has two specialities;

• every variable declared in the global scope is accessible everywhere
• the variables declared in the global scope correspond to the properties of the window object.

Therefore there are exactly two ways of declaring a variable foo in the global scope; either by not declaring it in a function or by setting the property foo of the window object.

Both attempts uses closures

Now that you have read a more detailed explanation it may now be apparent that both solutions uses closures. But to be sure, let's make a proof.

Let's create a new Programming Language; JavaScript-No-Closure. As the name suggests, JavaScript-No-Closure is identical to JavaScript except it doesn't support Closures.

In other words;

var foo = 'hello';
function f(){console.log(foo)};
f();
//JavaScript-No-Closure prints undefined
//JavaSript prints hello

Alright, let's see what happens with the first solution with JavaScript-No-Closure;

for(var i = 0; i < 10; i++) {
  (function(){
    var i2 = i;
    setTimeout(function(){
        console.log(i2); //i2 is undefined in JavaScript-No-Closure 
    }, 1000)
  })();
}

therefore this will print undefined 10 times in JavaScript-No-Closure.

Hence the first solution uses closure.

Let's look at the second solution;

for(var i = 0; i < 10; i++) {
  setTimeout((function(i2){
    return function() {
        console.log(i2); //i2 is undefined in JavaScript-No-Closure
    }
  })(i), 1000);
}

therefore this will print undefined 10 times in JavaScript-No-Closure.

Both solutions uses closures.

Edit: It is assumed that these 3 code snippets are not defined in the global scope. Otherwise the variables foo and i would be bind to the window object and therefore accessible through the window object in both JavaScript and JavaScript-No-Closure.

Solution 4

I've never been happy with the way anybody explains this.

The key to understanding closures is to understand what JS would be like without closures.

Without closures, this would throw an error

function outerFunc(){
    var outerVar = 'an outerFunc var';
    return function(){
        alert(outerVar);
    }
}

outerFunc()(); //returns inner function and fires it

Once outerFunc has returned in an imaginary closure-disabled version of JavaScript, the reference to outerVar would be garbage collected and gone leaving nothing there for the inner func to reference.

Closures are essentially the special rules that kick in and make it possible for those vars to exist when an inner function references an outer function's variables. With closures the vars referenced are maintained even after the outer function is done or 'closed' if that helps you remember the point.

Even with closures, the life cycle of local vars in a function with no inner funcs that reference its locals works the same as it would in a closure-less version. When the function is finished, the locals get garbage collected.

Once you have a reference in an inner func to an outer var, however it's like a doorjamb gets put in the way of garbage collection for those referenced vars.

A perhaps more accurate way to look at closures, is that the inner function basically uses the inner scope as its own scope foudnation.

But the context referenced is in fact, persistent, not like a snapshot. Repeatedly firing a returned inner function that keeps incrementing and logging an outer function's local var will keep alerting higher values.

function outerFunc(){
    var incrementMe = 0;
    return function(){ incrementMe++; console.log(incrementMe); }
}
var inc = outerFunc();
inc(); //logs 1
inc(); //logs 2

Solution 5

You are both using closures.

I 'm going with the Wikipedia definition here:

In computer science, a closure (also lexical closure or function closure) is a function or reference to a function together with a referencing environment—a table storing a reference to each of the non-local variables (also called free variables) of that function. A closure—unlike a plain function pointer—allows a function to access those non-local variables even when invoked outside of its immediate lexical scope.

Your friend's attempt clearly uses the variable i, which is non-local, by taking its value and making a copy to store into the local i2.

Your own attempt passes i (which at the call site is in scope) to an anonymous function as an argument. This is not a closure so far, but then that function returns another function that references the same i2. Since inside the inner anonymous function i2 is not a local, this creates a closure.

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