Why does the outer declaration cause an identifier error, while the inner declaration is fine?
function outer() {
function inner() {
console.log('Executing inner function');
}
var inner = new inner();
}
var outer = new outer();
If you change the last line to var x = new outer();, it'll run fine.
I believe you have found the answer yourself. In this case, there is a confusion in the language flow due to the use of the same name for the function (function outer()), global variable (var outer) and object (new inner()). Be more semantic with the name of the functions, variables and objects you intend to use.
This is because of Local and Global declarations of Variables/Objects.
1. var inner is considered as a local variable and it can have the same name .
2. var outer is declared outside of Outer Function. So it is considered as an non local
variable. So it must have an unique name than the existing ones.
I restructured your code in a way that works with your original idea. Basically I read the MDN documentation about possible errors with the new operator. What happens is that the new operator needs a function that defines a scope so that only afterwards can it be stored in a variable and can return whatever you want. In the documentation of this part there are some examples of errors that I believe would clarify your doubts.
MDN: TypeError: "x" is not a constructor
function showTheOuterFn() {
function fnInner() {
console.log('Executing inner function');
}
let getInner = new fnInner();
}
let outer = new showTheOuterFn();
Disclaimer: This question is purely curiosity driven and has to do a lot with how the javascript works.
I understand why the following code works. Due to closures, foo has access to the scope where a resides. This makes sense.
var a = 10
var foo = function(){
console.log(a);
}
setTimeout(foo,1000)
However, i wonder why the following also works (explained right after).
var a = 10
setTimeout(function(){
console.log(a);
},1000)
The function is defined in the argument of the function receiving it and essentially was never a closure to the scope that contains a. We know that when a function receives an argument, it creates a local variable for that argument so for example
var outerVar="5"
var bar = function(a){
//implicitly, var a = outerVar happens here
console.log(a)
}
bar(something);
So following that logic, the function passed to setTimeout couldnt have access to a and yet it does.
Im suspecting that when a function is defined in the argument space what happens is, it realy is defined before being assigned as an argument but have no proof of that. Any pointers highly appreciated.
Thanks a bunch.
It's not exactly closure, but it's close.
Strictly speaking, closure is when a variable's scope ends, but is still enclosed in an inner function that still lives on:
function createTimer() {
let counter = 0;
return function() {
return counter++;
}
}
const timer = createTimer(); // function() { ... }
console.log(timer(), timer(), timer()); // 0, 1, 2
The function in which counter is defined has returned, the scope ended, and under normal circumstances, counter should have died and garbage collected. But the inner function returned from createTimer() still has a reference to it, from the enclosed scope, that is a closure.
In JavaScript, every function has access to all of the scopes of all of its ancestors, this is what you're seeing here.
The function passed to setTimeout() has access to a because a is defined in the scope around it.
When you look at a javascript code and see how it works, the best way according to me is first understand the javascript engine how it works.
First it traverses the code and assigns all the variables to the scope and further more in the second trace it assigns the values based on the scopes.
So in your first code,
The engine first traverses and assigns
var a to global scope,
var foo as global scope,
Then when setTimeout runs it calls foo function and logs the value of a as that of the global a as it doesnt have any local “a” so checks the lexical scoping.
In your second code,
Var a is again global scoped
and no other declaration this time.
In second traverse it assigns the value 10 to a and interprets the settimeout and prints the value
In your third code,
Same as the second one except the fact that instead what “foo” was giving to the settimeout function, you wrote your callback function then n there itself.
By the time l it executed the setTimeout,
Each of your codes have the value for “a” in the global scope that they are accessing.
Hey i came across this video on youtube http://www.youtube.com/watch?v=KRm-h6vcpxs
which basically explains IIFEs and closures. But what I am not understanding is whether i need to return a function in order to call it a closure.
E.x.
function a() {
var i = 10;
function b() {
alert(i);
}
}
in this case can i call it a closure as it is accessing the 'i' variable from the outer function's scope or do i need to return the function like this
return function b(){alert(i);}
A closure is simply a function which holds its lexical environment and doesn't let it go until it itself dies.
Think of a closure as Uncle Scrooge:
Uncle Scrooge is a miser. He will never let go of his money.
Similarly a closure is also a miser. It will not let go of its variables until it dies itself.
For example:
function getCounter() {
var count = 0;
return function counter() {
return ++count;
};
}
var counter = getCounter();
See that function counter? The one returned by the getCounter function? That function is a miser. It will not let go of the count variable even though the count variable belongs to the getCounter function call and that function call has ended. Hence we call counter a closure.
See every function call may create variables. For example a call to the getCounter function creates a variable count. Now this variable count usually dies when the getCounter function ends.
However the counter function (which can access the count variable) doesn't allow it to die when the call to getCounter ends. This is because the counter function needs count. Hence it will only allow count to die after it dies itself.
Now the really interesting thing to notice here is that counter is born inside the call to getCounter. Hence even counter should die when the call to getCounter ends - but it doesn't. It lives on even after the call to getCounter ends because it escapes the scope (lifetime) of getCounter.
There are many ways in which counter can escape the scope of getCounter. The most common way is for getCounter to simply return counter. However there are many more ways. For example:
var counter;
function setCounter() {
var count = 0;
counter = function counter() {
return ++count;
};
}
setCounter();
Here the sister function of getCounter (which is aptly called setCounter) assigns a new counter function to the global counter variable. Hence the inner counter function escapes the scope of setCounter to become a closure.
Actually in JavaScript every function is a closure. However we don't realize this until we deal with functions which escape the scope of a parent function and keep some variable belonging to the parent function alive even after the call to the parent function ends.
For more information read this answer: https://stackoverflow.com/a/12931785/783743
Returning the function changes nothing, what's important is creating it and calling it. That makes the closure, that is a link from the internal function to the scope where it was created (you can see it, in practice, as a pointer. It has the same effect of preventing the garbaging of the outer scope, for example).
By definition of closure, the link from the function to its containing scope is enough. So basically creating the function makes it a closure, since that is where the link is created in JavaScript :-)
Yet, for utilizing this feature we do call the function from a different scope than what it was defined in - that's what the term "use a closure" in practise refers to. This can both be a lower or a higher scope - and the function does not necessarily need to be returned from the function where it was defined in.
Some examples:
var x = null;
function a() {
var i = "from a";
function b() {
alert(i); // reference to variable from a's scope
}
function c() {
var i = "c";
// use from lower scope
b(); // "from a" - not "c"
}
c();
// export by argument passing
[0].forEach(b); // "from a";
// export by assigning to variable in higher scope
x = b;
// export by returning
return b;
}
var y = a();
x(); // "from a"
y(); // "from a"
The actual closure is a container for variables, so that a function can use variables from the scope where it is created.
Returning a function is one way of using it in a different scope from where it is created, but a more common use is when it's a callback from an asynchronous call.
Any situation where a function uses variables from one scope, and the function is used in a different scope uses a closure. Example:
var globalF; // a global variable
function x() { // just to have a local scope
var local; // a local variable in the scope
var f = function(){
alert(local); // use the variable from the scope
};
globalF = f; // copy a reference to the function to the global variable
}
x(); // create the function
globalF(); // call the function
(This is only a demonstration of a closure, having a function set a global variable which is then used is not a good way to write actual code.)
a collection of explanations of closure below. to me, the one from "tiger book" satisfies me most...metaphoric ones also help a lot, but only after encounterred this one...
closure: in set theory, a closure is a (smallest) set, on which some operations yields results also belongs to the set, so it's sort of "smallest closed society under certain operations".
a) sicp: in abstract algebra, where a set of elements is said to be closed under an operation if applying the operation to elements in the set produces an element that is again an element of the set. The Lisp community also (unfortunately) uses the word "closure" to describe a totally unrelated concept: a closure is an implementation technique for representing procedures with free variables.
b) wiki: a closure is a first class function which captures the lexical bindings of free variables in its defining environment. Once it has captured the lexical bindings the function becomes a closure because it "closes over" those variables.”
c) tiger book: a data structure on heap (instead of on stack) that contains both function pointer (MC) and environment pointer (EP), representing a function variable;
d) on lisp: a combination of a function and a set of variable bindings is called a closure; closures are functions with local state;
e) google i/o video: similar to a instance of a class, in which the data (instance obj) encapsulates code (vtab), where in case of closure, the code (function variable) encapsulates data.
f) the encapsulated data is private to the function variable, implying closure can be used for data hiding.
g) closure in non-functional programming languages: callback with cookie in C is a similar construct, also the glib "closure": a glib closure is a data structure encapsulating similar things: a signal callback pointer, a cookie the private data, and a destructor of the closure (as there is no GC in C).
h) tiger book: "higher-order function" and "nested function scope" together require a solution to the case that a dad function returns a kid function which refers to variables in the scope of its dad implying that even dad returns the variables in its scope cannot be "popup" from the stack...the solution is to allocate closures in heap.
i) Greg Michaelson ($10.15): (in lisp implementation), closure is a way to identify the relationship betw free variables and lexical bound variables, when it's necessary (as often needed) to return a function value with free variables frozen to values from the defining scope.
j) histroy and etymology: Peter J. Landin defined the term closure in 1964 as having an environment part and a control part as used by his SECD machine for evaluating expressions. Joel Moses credits Landin with introducing the term closure to refer to a lambda expression whose open bindings (free variables) have been closed by (or bound in) the lexical environment, resulting in a closed expression, or closure. This usage was subsequently adopted by Sussman and Steele when they defined Scheme in 1975, and became widespread.
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.
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:
Take a function. Let's call it F.
List all the variables of F.
The variables may be of two types:
Local variables (bound variables)
Non-local variables (free variables)
If F has no free variables then it cannot be a closure.
If F has any free variables (which are defined in a parent scope of F) then:
There must be only one parent scope of F to which a free variable is bound.
If F is referenced from outside that parent scope, then it becomes a closure for that free variable.
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:
List the variables:
i2 is a local variable.
i is a free variable.
setTimeout is a free variable.
g is a local variable.
console is a free variable.
Find the parent scope to which each free variable is bound:
i is bound to the global scope.
setTimeout is bound to the global scope.
console is bound to the global scope.
In which scope is the function referenced? The global scope.
Hence i is not closed over by f.
Hence setTimeout is not closed over by f.
Hence console is not closed over by f.
Thus the function f is not a closure.
For g:
List the variables:
console is a free variable.
i2 is a free variable.
Find the parent scope to which each free variable is bound:
console is bound to the global scope.
i2 is bound to the scope of f.
In which scope is the function referenced? The scope of setTimeout.
Hence console is not closed over by g.
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:
List the variables:
i2 is a local variable.
g is a local variable.
console is a free variable.
Find the parent scope to which each free variable is bound:
console is bound to the global scope.
In which scope is the function referenced? The global scope.
Hence console is not closed over by f.
Thus the function f is not a closure.
For g:
List the variables:
console is a free variable.
i2 is a free variable.
Find the parent scope to which each free variable is bound:
console is bound to the global scope.
i2 is bound to the scope of f.
In which scope is the function referenced? The scope of setTimeout.
Hence console is not closed over by g.
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"));
}
}
We know that both lexicalScope and regularFunction aren't closures from the above definition.
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).
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"));
};
}
We know that only closureFunction is a closure from the above definition.
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).
When we execute the program we observe that message is actually being alerted.
What do we infer from this?
JavaScript interpreters do not treat closures differently from the way they treat other functions.
Every function carries its scope chain along with it. Closures don't have a separate referencing environment.
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.
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).
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.
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
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.
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);
}
Closure
A closure is not a function, and not an expression. It must be seen as a kind of 'snapshot' from the used variables outside the function scope and used inside the function. Grammatically, one should say: 'take the closure of the variables'.
Again, in other words: A closure is a copy of the relevant context of variables on which the function depends on.
Once more (naïf): A closure is having access to variables who are not being passed as parameter.
Bear in mind that these functional concepts strongly depend upon the programming language / environment you use. In JavaScript, the closure depends on lexical scoping (which is true in most C-languages).
So, returning a function is mostly returning an anonymous/unnamed function. When the function access variables, not passed as parameter, and within its (lexical) scope, a closure has been taken.
So, concerning your examples:
// 1
for(var i = 0; i < 10; i++) {
setTimeout(function() {
console.log(i); // closure, only when loop finishes within 1000 ms,
}, 1000); // i = 10 for all functions
}
// 2
for(var i = 0; i < 10; i++) {
(function(){
var i2 = i; // closure of i (lexical scope: for-loop)
setTimeout(function(){
console.log(i2); // closure of i2 (lexical scope:outer function)
}, 1000)
})();
}
// 3
for(var i = 0; i < 10; i++) {
setTimeout((function(i2){
return function() {
console.log(i2); // closure of i2 (outer scope)
}
})(i), 1000); // param access i (no closure)
}
All are using closures. Don't confuse the point of execution with closures. If the 'snapshot' of the closures is taken at the wrong moment, the values may be unexpected but certainly a closure is taken!
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.
I wrote this a while ago to remind myself of what a closure is and how it works in JS.
A closure is a function that, when called, uses the scope in which it was declared, not the scope in which it was called. In javaScript, all functions behave like this. Variable values in a scope persist as long as there is a function that still points to them. The exception to the rule is 'this', which refers to the object that the function is inside when it is called.
var z = 1;
function x(){
var z = 2;
y(function(){
alert(z);
});
}
function y(f){
var z = 3;
f();
}
x(); //alerts '2'
After inspecting closely, looks like both of you are using closure.
In your friends case, i is accessed inside anonymous function 1 and i2 is accessed in anonymous function 2 where the console.log is present.
In your case you are accessing i2 inside anonymous function where console.log is present. Add a debugger; statement before console.log and in chrome developer tools under "Scope variables" it will tell under what scope the variable is.
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
I would like to share my example and an explanation about closures. I made a python example, and two figures to demonstrate stack states.
def maker(a, b, n):
margin_top = 2
padding = 4
def message(msg):
print('\n’ * margin_top, a * n,
' ‘ * padding, msg, ' ‘ * padding, b * n)
return message
f = maker('*', '#', 5)
g = maker('', '♥’, 3)
…
f('hello')
g(‘good bye!')
The output of this code would be as follows:
***** hello #####
good bye! ♥♥♥
Here are two figures to show stacks and the closure attached to the function object.
when the function is returned from maker
when the function is called later
When the function is called through a parameter or a nonlocal variable, the code needs local variable bindings such as margin_top, padding as well as a, b, n. In order to ensure the function code to work, the stack frame of the maker function which was gone away long ago should be accessible, which is backed up in the closure we can find along with the function message object.
Wanting to get something straight here...so I have 2 questions
The function below creates a closure.
function Foo(message){
var msg = message;
return function Bar(){
this.talk = function(){alert(msg); }
}
};
Q: Which function is the closure, Foo or Bar?
I always thought the closure to be Foo, because it closes-over Bar once Bar is returned.
Next...
Below is the definition of an anonymous function:
()();
Q: Is the inner-function within this anonymous function also a closure?
(function(){ /* <-- Is this function also a closure? */ })();
You need to use first principles here. Javascript uses lexical scoping. This means the scope of an execution context is determined by how the code is defined (lexical).
I would say the definition of the function Bar is what causes the closure to be created, because msg is "closed-in" in the function.
The actual creation of the closure happens at runtime (which is somewhat of a tautological statement, since nothing in a computer program happens until it is run), because in order to determine the value of msg, in Bar, when Bar is executed, the interpreter needs to know the value of the variable when Foo is executed, and so on up the chain.
I'll give two answers to your question. The pedantic answer is: neither function by itself is the closure. It's the definition of variables within functions, combined with the execution context of functions when they are run, that is defines the closure. The common answer is: any function which closes over a variable is a closure (Bar in your case).
Consider the problem everyone encounters when using Javascript.
function A(x) {
var y = x, fs = [];
for (var i = 0; i < 3; i++) {
fs.push(function(){
console.log (i + " " + x);
})
}
fs.forEach(function(g){g()})
}
A('hi')
Most people would say this would produce the output 'hi 1' followed by 'hi 2' followed by 'hi 3'. However, it produces 'hi 3' 3 times. If just the definition of the function being added to the array, while using variables defined in the outer function, created the closure, how can this be?
It's because you need the execution context to define the closure, which doesn't happen until runtime. At the execution of the functions in the array, i has the value 3. In the forEach statement, that's the execution context, which is why the output always uses 3.
Bar is the closure.
We say that Bar closes over the variable msg in its environment.
Most often the word closure means: a function that uses at least one variable that’s defined in an enclosing scope, in an enclosing function.
To answer your second question: (function(){ ... })() is just what it looks like: one anonymous function, not two. Unless it’s nested inside another function, you generally wouldn’t call it a closure. However, functions nested inside that anonymous function can be closures (and often are).