In the below code snippet, i is declared after function f, so how can function f access i's value?
let f;
if (true) {
f = () => {
console.log(i)
}
let i = 1;
}
f();
i is declared after function f
Ehm, yes, but in the same scope (block).
i is declared in the same block as the value of f (an anonymous function that uses i), so it is in scope:
f = () => {
console.log(i)
}
// same scope as the definition of the anonymous function:
let i = 1;
Now at the time of declaration, the anonymous function does not need to know what i is, and if you never declared it, it would just resolve to undefined when you do f().
At the time of execution, i is already declared and assigned a value. The anonymous function "recalls" this variable because it was declared in it's scope. This is called a closure.
let keyword scope.
First of the let keyword scope for block of code in your program 'i' is inside of if block so 'i' value scope is inside of if block.
so,if you create 'n' number of function inside of if block it'll access the 'i' value.
please check my example remove the comment in f1() inside console.log(j) and put debugger you'll understand clear.
Here, you'll get error because 'j' is only access for f() means 'j' scope only inside of f(). If you use inside of f() it'll access. Check f2() it's able to access 'j' value.
<script>
debugger
let f,f1,f2;
if (true) {
f = () => {
console.log(i)
let j=11;
f2=()=>{
console.log(j);
}
}
f1=()=>{
console.log(i);
//console.log(j);
}
let i = 1;
}
f();
f1();
f2();
</script>
You are correct that the variables declared by lexical let declarations may not be accessed without error until after the declaration lexically occurs. (This is in contrast to var declarations, which make their declared variable accessible and referencible starting from the top of their containing scope, regardless of where the declaration occurs within that scope.)
The variables referenced by name in f are not accessed until f is actually called. The JavaScript engine does not ask the question "In what scope is the variable i and what is its value?" until f is executed. At the chronological time f actually is executed, the surrounding block scope has had the variable i made accessible within it, so f can use that variable to resolve the identifier i.
When you refer to i at execution time, the JavaScript engine looks up the scope chain to find the nearest declared variable named i. The declaration let i is an instruction to the JavaScript engine to make available a variable named i to the current block scope. The fact that the surrounding scope has no accessibl i at the time f is defined is immaterial; what matters is that the scope does have a variable i by the time f is executed.
Consider this example:
var f;
if(true) {
f = ()=>{ console.log(i) }
try { f(); } catch(e) { console.log(e); }
let i = 1;
try { f(); } catch(e) { console.log(e);}
}
The first call produces an error, because during the execution of f, no i has yet been made accessible, while the second execution, after the let i statement, runs without error.
As Javascript is an interpreted language, it will execute statements one at a time, line by line, without looking at the entire code first (unlike compiled languages). What this means is that your function will not throw an error at the line console.log(i); before runtime.
At runtime, you are defining and initializing i before calling the function f, so by the time f is called, the value of i is already known.
Regarding scopes, i and f are declared in the same block (not inside a function or anything), therefore f can access i fine. No scope issues there.
I am having trouble understanding why the following code prints 1 and not 10. I thought that console.log would print 10 because if a variable is not declared within a function it is part of the global scope.
If I remove the line, function a(){}, the console.log statement prints 10. Can someone please explain why the presence of this line affects the value of a.
Additionally, if this question can be refined I would appreciate any suggestions
function b() {
a = 10;
return;
function a(){}
}
var a = 1;
b();
console.log(a);
Just add console.log(a); to your function and you will realize that 'a' is a function that has local scope, which means, it inside function b() and can't really be called from outside. Hence, the value of a remains as 1, as nothing really affects it.
function b() {
console.log(a);
a = 10;
return;
function a(){}
}
var a = 1;
b();
console.log(a);
Function declarations create local variables, and all declared variables are hoisted to the top of their containing function scope. Thus, you have something equivalent to:
function b() {
function a(){}
a = 10;
return;
}
which is in turn roughly equivalent to
function b() {
var a = function(){}
a = 10;
return;
}
As you can see in this form, it's clear that a is a local variable, so a = 10 assigns a value to the local a, not the global a.
Because both function declarations and variables are treated similarly under the hood.
When a function is called, its contents are first parsed to configure the stack frame ready for its execution.
As part of this process, function declarations and variables are inserted as identifier/value pairs into the same data structure (Environment Record inside the Lexical Environment) inside the stack frame (Execution Context) created when a function is called.
So by declaring a function a inside b, you create an identifier named "a" in the Environment Record created for the call to b, shadowing a in the outer context.
So a in the outer context remains 1.
I faced with strange behaviour when using eval in JS.
var f = function () {
var x = 10;
return function () {
eval('console.log(x);');
window['eval']('console.log(x);');
}
};
f()();
OUTPUT:
10
undefined:1
console.log(x);
^
ReferenceError: x is not defined
Why using eval explicitly captures the x but global['eval'] doesn't?
And even though global['eval'] doesn't capture x, why it's unable to see after eval, which already captured x?
window['eval'] operates at global scope, eval() operates at local scope.
From Mozilla's Javascript reference:
If you use the eval function indirectly, by invoking it via a
reference other than eval, as of ECMAScript 5 it works at global scope
rather than local scope; this means, for instance, that function
declarations create global functions, and that the code being
evaluated doesn't have access to local variables within the scope
where it's being called.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/eval
Your inner function does not actually capture the reference of x, and so it is never directly passed to eval.
eval usually works at the local scope and so the first call succeeds (because the local scope contains the declaration of x).
However, if you invoke eval in such a way that you don't have a direct reference to it, it will invoke itself in the global scope, which var x is not a part of, and it fails.
Just don't use eval.
You can use Function.prototype.bind() to pass x to returned function
var f = function () {
var x = 10;
function y(n) {
eval(`console.log(${n})`);
window["eval"](`console.log(${n})`);
}
return y.bind(this, x)
};
f()();
window.eval work in global scope.
var variable = 1;
(function(){
var variable = 100,
cmd = "++variable";
document.write(eval(cmd)+"\n"); // increment local var 100 and output 101
document.write(window.eval(cmd)+"\n"); // increment global var 1 and output 2
})();
This question already has answers here:
What is the (function() { } )() construct in JavaScript?
(28 answers)
Closed 6 years ago.
from MDN, just wondering why the () around the function, why the extra () after it, and why the var b inside the function doesn't replace the value of the first var b, given it doesn't use the keyword let, which would keep b local to that function, thanks
var a = 1;
var b = 2;
(function() {
var b = 3;
a += b;
})();
a; // 4
b; // 2
When you write var b inside a function, the var keyword makes it a local variable. So, var b inside the function is a local variable and var b at the top which is outside the function is in global scope. You can read more about scoping in MDN. Also, the () after the function is called immediate function invocation which means that the function is given a call soon after it is defined. Also, since a inside the function has no var before its declaration, it will take the global a which is 1 and add 3, local variable b's value it.
The function is simply called immediately. That’s what the last parentheses are for. The extra parentheses around the function declaration are because
function() {
var b = 3;
a += b;
}();
would create a syntax error as the function keyword would not be interpreted as an expression here.
and why the var b inside the function doesn't replace the value of the first var b, given it doesn't use the keyword let, which would keep b local to that function,
You may have heard that the special thing about let is that it is block-scoped. That’s true. But var has always been scoped to the function and since the second var b is inside its own function it’s scoped to that function and does not affect the b from the top scope here.
The template
(function(){})();
Is a template of an anonymous function.
To explain it, let's create a function x() that alerts 'hello':
function x() {
alert('hello');
}
To call x, we will:
x();
Now, it's clear that if we replace an object's name with it's value, the statememt will stay the same (take for example var k = 5; alert(k); alert(5) - the two alerts are the same because variable's name was replaced with it's value).
In our case, x's value is:
function () {
alert('hello');
}
So if we replace x with it's value in the statement x(); ->
(function () {
alert('hello');
})();
And that's the source of the syntax. The result of that will be an immediate call to the specified function. Of course, this syntax works also for functions with parameters and functions with return types.
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.