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.

Answer 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.

Answer 2

Consider the following. This creates and recreates a function f that closes on i, but different ones!:

i=100;

f=function(i){return function(){return ++i}}(0);
alert([f,f(),f(),f(),f(),f(),f(),f(),f(),f(),f()].join('\n\n'));

f=function(i){return new Function('return ++i')}(0);        /*  function declarations ~= expressions! */
alert([f,f(),f(),f(),f(),f(),f(),f(),f(),f(),f()].join('\n\n'));

while the following closes on "a" function "itself"
( themselves! the snippet after this uses a single referent f )

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

or to be more explicit:

for(var i = 0; i < 10; i++) {
    console.log(    f = new Function( 'console.log('+i+')' )    );
    setTimeout( f,  1000 );
}

NB. the last definition of f is function(){ console.log(9) } before 0 is printed.

Caveat! The closure concept can be a coercive distraction from the essence of elementary programming:

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

x-refs.:
How do JavaScript closures work?
Javascript Closures Explanation
Does a (JS) Closure Require a Function Inside a Function
How to understand closures in Javascript?
Javascript local and global variable confusion

Answer 3

You and your friend both use closures:

A closure is a special kind of object that combines two things: a function, and the environment in which that function was created. The environment consists of any local variables that were in-scope at the time that the closure was created.

MDN: https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Closures

In your friend's code function function(){ console.log(i2); } defined inside closure of anonymous function function(){ var i2 = i; ... and can read/write local variable i2.

In your code function function(){ console.log(i2); } defined inside closure of function function(i2){ return ... and can read/write local valuable i2 (declared in this case as a parameter).

In both cases function function(){ console.log(i2); } then passed into setTimeout.

Another equivalent (but with less memory utilization) is:

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

Answer 4

Let's look at both ways:

(function(){
    var i2 = i;
    setTimeout(function(){
        console.log(i2);
    }, 1000)
})();

Declares and immediately executes an anonymous function that runs setTimeout() within its own context. The current value of i is preserved by making a copy into i2 first; it works because of the immediate execution.

setTimeout((function(i2){
    return function() {
        console.log(i2);
    }
})(i), 1000);

Declares an execution context for the inner function whereby the current value of i is preserved into i2; this approach also uses immediate execution to preserve the value.

Important

It should be mentioned that the run semantics are NOT the same between both approaches; your inner function gets passed to setTimeout() whereas his inner function calls setTimeout() itself.

Wrapping both codes inside another setTimeout() doesn't prove that only the second approach uses closures, there's just not the same thing to begin with.

Conclusion

Both methods use closures, so it comes down to personal taste; the second approach is easier to "move" around or generalize.

Answer 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.

Answer 6

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