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Access local variable inside a callback function

var inner = function() { console.log(x); }
// test 1
(function(cb) { var x = 123; cb(); })(inner);
// test 2
(function(cb) { var x = 123; cb.apply(this); })(inner);
// test 3
(function(cb) { var x = 123; cb.bind(this)(); })(inner);
// test 4
(function(cb) { cb.bind({x: 123})(); })(inner);
All tests result in:
ReferenceError: x is not defined
Do someone know how it is possible to access 'x' as a local variable inside the callback?

Fact: when you do var inner = function() { console.log(x); } in your first line, x is not defined. Why? Because, inside your inner function, there's no local declaration of x (which would be done with var x = something). The runtime will then look up in the next scope, that is the global scope. There isn't, also, a declaration of x, so x is also not defined there.
The only places where there is a variable called x are inside each one of your 4 IIFEs following. But inside the IIFEs, each x is a different variable, in a different scope. So, if what you want is to console.log() the x defined inside each IIFE, you are taking the wrong approach.
Keep in mind that, when you define inner, you are capturing the environment inside the function's closure. It means that, whatever value could x have there (in the declaration of the function), would be the available value to the x variable later, when the inner function would be used. The fact that your x there is not defined is only an accessory, and is not what is causing the undesired behavior.
So, what happens is that when you call your inner function inside any of your IIFEs, the x referred to inside the inner function declaration is a captured value of what x had as a value when the function was defined, not the value that x has now in the scope where the function is currently being called. This is what is called lexical scope.
To solve this, you would have to pass the value that you want to console.log() inside the inner function as a parameter to the inner function, as so:
var inner = function(x) { console.log(x); }
// test 1
(function(cb) { var x = 123; cb(x); })(inner);

The only way to access the local variable x in the callback, is to pass it as an argument:
var inner = function(some_var) { console.log(some_var); }; //logs 123
(function(cb) { var x = 123; cb(x); })(inner);
OR
var inner = function(some_var) { console.log(some_var); }; //logs 123
(function(cb) { var x = 123; cb.apply(this,[x]); })(inner);
OR
var inner = function(some_var) { console.log(some_var); }; //logs 123
(function(cb) { var x = 123; cb.call(this,x); })(inner);
FURTHER
Because JS is lexically scoped, trying to reference the local variable after the anonymous function has finished executing is impossible by any other means. If you don't pass it as an argument to make it available elsewhere, JS will see it as non-reachable and it will be eligible for garbage collection.

You could redefine the callback function in the current scope:
var inner = function() { console.log(x); }
(function(cb) { var x = 123; eval('cb = ' + cb.toString()); cb(); })(inner);
// or
(function(cb) { var x = 123; eval('(' + cb.toString() + ')')(); })(inner);
This will not work if the function relies on anything in the scope in which it was originally defined or if the Javascript file has been minified. The use of eval may introduce security, performance, and code quality issues.

Have you tried using events? Emit an event inside the anonymous function, then subscribe to it in your own function somewhere else that carries out your logic.

Javascript module pattern and function invocation

Hello I have started to learn some javascript and I can't wrap my head around this in "normal" functions. I don't understand why the following two functions output different results. f2 outputs 5, while f1 outputs 1. Why is that?
var f1 = function(){
var x= 1;
var add = function(){
x=5;
};
var result = function(){
console.log(x);
};
return {
add: add,
result: result
};
};
f1().add();
f1().result();
var f2= (function(){
var x= 1;
var add = function(){
x=5;
};
var result = function(){
console.log(x);
};
return {
add: add,
result: result
};
})();
f2.add();
f2.result();

The first example shows two invocations of f1().
The first invocation of f1() creates a new variable scope with x set to 1, and returns the object with the methods. The .add() then sets that x to 5.
The second invocation of f1() creates another new variable scope with x again set to 1, and returns the object with the methods. The .result() then returns that x which is still 1.
The second example only invokes f2() once, so there's no new variable scope with x and new methods being created.
So basically f1's two invocations initializes x with each call, and returns two different objects with methods that closer over the two different x variables.
The f2 is invoked once so there's one x variable shared by the one object with methods returned. Therefore the .add() call and the .result() call are using the same x variable.

Let me outline what happens in your code:
// Declares a function named f1.
var f1 = function () {
// Searches each scope (the scope of f1, the scope containing f1, etc.) for
// a variable named x. If found, it will reassign it to 1. If the search reaches
// the global scope and no variable is found it will declare and initialize
// a global variable.
x = 1;
// Declares a local variable named add to a function that reassigns the value of x.
var add = function () {
x = 5;
};
// Declares a local variable named result to a function that logs the value of x.
var result = function () {
console.log(x);
};
// Returns an object containing each function.
return {
add: add,
result: result
};
};
// Calls f1 and the returned add function.
f1().add();
// Calls f1 again (which reassigns x) and calls the returned result function.
f1().result();
// Creates a variable named f2 and assigns to the result of applying an anonymous
// function. f2 now references the returned object.
var f2 = (function () {
// Reassigns x, does not create a new variable.
x = 1;
var add = function () {
x = 5;
};
var result = function () {
console.log(x);
};
return {
add: add,
result: result
};
})();
// The difference now is that the function that contains add and result is only
// called once, during the initialization of f2. Had you written var g = f1();
// g.add(); g.result(); you would have gotten the exact same results.
f2.add();
f2.result();

Lexical scope/closures in javaScript

I understand functions in 'js' have lexical scope (i.e. functions create their environment (scope) when they are defined not when they are executed.)
function f1() {
var a = 1;
f2();
}
function f2() {
return a;
}
f1(); // a is not defined
When I run just 'f()' it returns the inner function. Which I get, that's what 'return' does!
function f() {
var b = "barb";
return function() {
return b;
}
}
console.log(b); //ReferenceError: b is not defined
Why do you get 'ReferenceError: b is not defined?'
But doesn't the inner function above have access to it's space, f()'s space etc. Being that 'b' is being returned to the global space, wouldn't the console.log() work?
However when I assign 'f()' to a new variable and run it:
var x = f();
x();// "barb"
console.log(b); //ReferenceError: b is not defined
This returns 'b' which is "barb", but when you run console.log() again you'll get 'ReferenceError: 'b' is not defined'; Isn't 'b' in the global scope now since it has been returned? SO why didn't 'x()' also return the inner function just like 'f()' did?

You, my friend, are thoroughly confused. Your very first statement itself is wrong:
functions create their environment (scope) when they are defined not when they are executed
Actually it's the opposite. Defining a function doesn't create a scope. Calling a function creates a scope.
What's a scope?
To put it simply, a scope is the lifespan of a variable. You see, every variable is born, lives and dies. The beginning of a scope marks the time the variable is born and the end of the scope marks the time it dies.
In the beginning there's only one scope (called the program scope or the global scope). Variables created in this scope only die when the program ends. They are called global variables.
For example, consider this program:
const x = 10; // global variable x
{ // beginning of a scope
const x = 20; // local variable x
console.log(x); // 20
} // end of the scope
console.log(x); // 10
Here we created a global variable called x. Then we created a block scope. Inside this block scope we created a local variable x. Since local variables shadow global variables when we log x we get 20. Back in the global scope when we log x we get 10 (the local x is now dead).
Block Scopes and Function Scopes
Now there are two main types of scopes in programming - block scopes and function scopes.
The scope in the previous example was a block scope. It's just a block of code. Hence the name. Block scopes are immediately executed.
Function scopes on the other hand are templates of block scopes. As the name suggests a function scope belongs to a function. However, more precisely, it belongs to a function call. Function scopes do not exist until a function is called. For instance:
const x = 10;
function inc(x) {
console.log(x + 1);
}
inc(3); // 4
console.log(x); // 10
inc(7); // 8
As you can see every time you call a function a new scope is created. That's the reason you get the outputs 4, 10 and 8.
Originally, JavaScript only had function scopes. It didn't have block scopes. Hence if you wanted to create a block scope then you had to create a function and immediately execute it:
const x = 10; // global variable x
(function () { // beginning of a scope
const x = 20; // local variable x
console.log(x); // 20
}()); // end of the scope
console.log(x); // 10
This pattern is called an immediately invoked function expression (IIFE). Of course, nowadays we can create block scoped variables using const and let.
Lexical Scopes and Dynamic Scopes
Function scopes can again be of two types - lexical and dynamic. You see, in a function there are two types of variables:
Free variables
Bound variables
Variables declared inside a scope are bound to that scope. Variables not declared inside a scope are free. These free variables belong to some other scope, but which one?
Lexical Scope
In lexical scoping free variables must belong to a parent scope. For example:
function add(x) { // template of a new scope, x is bound in this scope
return function (y) { // template of a new scope, x is free, y is bound
return x + y; // x resolves to the parent scope
};
}
const add10 = add(10); // create a new scope for x and return a function
console.log(add10(20)); // create a new scope for y and return x + y
JavaScript, like most programming languages, has lexical scoping.
Dynamic Scope
In contrast to lexical scoping, in dynamic scoping free variables must belong to the calling scope (the scope of the calling function). For example (this is also not JS - it doesn't have dynamic scopes):
function add(y) { // template of a new scope, y is bound, x is free
return x + y; // x resolves to the calling scope
}
function add10(y) { // template of a new scope, bind y
var x = 10; // bind x
return add(y); // add x and y
}
print(add10(20)); // calling add10 creates a new scope (the calling scope)
// the x in add resolves to 10 because the x in add10 is 10
That's it. Simple right?
The Problem
The problem with your first program is that JavaScript doesn't have dynamic scoping. It only has lexical scoping. See the mistake?
function f1() {
var a = 1;
f2();
}
function f2() {
return a;
}
f1(); // a is not defined (obviously - f2 can't access the `a` inside f1)
Your second program is a very big mess:
function f() {
var b = "barb";
return function() {
return b;
}
}
console.log(b); //ReferenceError: b is not defined
Here are the mistakes:
You never called f. Hence the variable b is never created.
Even if you called f the variable b would be local to f.
This is what you need to do:
function f() {
const b = "barb";
return function() {
return b;
}
}
const x = f();
console.log(x());
When you call x it returns b. However that doesn't make b global. To make b global you need to do this:
function f() {
const b = "barb";
return function() {
return b;
}
}
const x = f();
const b = x();
console.log(b);
Hope this helped you understand about scopes and functions.

You get, "ReferenceError: b is not defined" because "b" is not defined where your console.log() call is. There's a "b" inside that function, but not outside. Your assertion that "b is being returned to the global space" is false.
When you invoke the function returned by your "f()" function, that will return a copy of the value referenced by that closure variable "b". In this case, "b" will always be that string, so the function returns that string. It does not result in the symbol "b" becoming a global variable.

But doesn't the inner function above have access to it's space, f()'s space etc.
Yes it has. It accesses the b variable and returns its value from the function.
Being that 'b' is being returned to the global space
No. Returning a value from a function is not "making a variable available in the caller scope". Calling the function (with f()) is an expression whose result is the value that the function returned (in your case, the unnamed function object). That value can then be assigned somewhere (to x), a property of it can be accessed or it can be discarded.
The variable b however stays private in the scope where it was declared. It is not [getting] defined in the scope where you call console.log, that's why you get an error.
What you want seems to be
var x = f();
var b = x(); // declare new variable b here, assign the returned value
console.log( b ); // logs "barb"

function f1() {
var a = 1;
f2();
}
function f2() {
return a;
}
f1(); // a is not defined
f2(); does not knows about the a,because you never passed 'a' to it,(That's Scope are
created when the functions are defined).Look function f2() would have been able to acess
a if it was defined inside f1();[Functions can access the variables in same scope in
which they are "DEFINED" and NOT "CALLED"]
function f() {
var b = "barb";
return function(){
return b;
}
}
console.log(b);
First of all You Need to Call f(); after executing f(); it would return another function
which needs to be executed. i.e
var a=f();
a();
it would result into "barb" ,In this case you are returning a function not the var b;
function f() {
var b = "barb";
return b;
};
console.log(f());
This would print barb on screen

Javascript variable declarations at the head of a function

I've been told that javascript variables should all come before they are used in a function, such that:
function contrived() {
var myA, myB;
myA = 10;
myB = 20;
return myA + myB;
}
Is prefered over:
function furtherContrivance() {
var myA = 10;
var myB = 20;
return myA + myB;
}
Is this the case? And why is that?

I guess some people might prefer the former style because that's how it works inside. All local variables exist for the entire lifetime of the function, even if you use var to declare them in the middle of the function.
There's nothing wrong with declaring variables later in the function, syntax-wise, it might just be confusing as the variables will then exist before the line that declares them. Hence this function:
function bar() {
alert(foo); // Alerts "undefined". Not an error because the variable does exist.
var foo = 10;
alert(foo); // Alerts the value 10.
}
Is equivalent to this:
function bar() {
var foo;
alert(foo);
foo = 10;
alert(foo);
}
Another related fact is that nested function definitions (done using function foo() { ... }) will get moved to the top of the containing function as well, so they will be available even if the code that calls them comes before them.

Yes, the variable declaration should come at the top of the function:
function foo() {
var a, b;
}
However, initializing variables can be part of the declaration:
function foo() {
var a = 10, b = 20;
}
The reasoning behind declaring all variables at the top of the function where they are used is to avoid scope confusion.
Here is an example of bad code:
function foo() {
var b;
for (var i = 0; i < 5; i++) {
var a;
a = b = i;
setTimeout(function(){
console.log(a, b);
}, 1000);
}
}
If you execute the code, it will log 4, 4 5 times, rather than counting up. This is because only functions act as closures and introduce new scope. In JavaScript, any var declaration within a function gets executed at the beginning of the function.
This makes the above error much more visible:
function foo() {
var a, b, i;
for (i = 0; i < 5; i++) {
a = b = i;
setTimeout(function(){
console.log(a, b);
}, 1000);
}
}

There is no difference in this case between this two. I'd go with:
function furtherContrivance() {
var myA = 10,
myB = 20;
return myA + myB;
}
which is knows as single var pattern in javascript.
What you really need to take care of is defining your variables in the beginning of your functions. There is a thing in javascript called variables hoisting which means that variable definitions used in function "raise" on top. It's best described by an example:
var x = 'global'; // global (bounded to a global object which is window in browsers)
function func() {
alert(x); // undefined (you expected 'global', right?)
var x = 'local';
alert(x); // local
}
func();
what really happens is called (as I said) variables hoisting (definition of x raises on top), so the code above is actually the same as:
var x = 'global';
function func() {
var x; // definition of `x` raised on top (variables hoisting)
alert(x); // undefined in a local scope
x = 'local';
alert(x);
}
What a javscript interpreter does is it looks inside a function, gathers locally defined variables and raises them on top - this might be a good reason why you should use single var pattern.

In the example you give this is absolutely not the case. In a language like Javascript, it will be more of a developer preference, but it won't have any impact on the result.

Yes, place them at the top. It adds to code clarity.
Try this example:
var x = 1;
(function() {
x++;
alert( x ); // What will this alert show?
var x = 'done';
alert( x );
})();
Looks like it should alert 2, but it alerts NaN.
This is because the variable declaration is hoisted to the top, but the initialization stays in the same place.
So what is actually happening is:
var x = 1;
(function() {
var x;
x++;
alert( x ); // What will this alert show? NaN
x = 'done';
alert( x );
})();
...which makes the NaN expected.

For readability, it's definitely preferred.
However, Javascript "hoists" declarations. Hoisting means that vars and functions will be automatically moved to the top of their scope. This allows you to do things such as use a function before it's declared:
function myScope()
{
test();
function test()
{
//...
}
}
This can lead to some confusion, especially if variables within block scopes are declared. For example:
for(var i in foo)
{
var e = myFunc();
}
The declaration of e will be hoisted to the top of the closure, and e will be initialized to undefined. This allows for some interesting non-intuitive situations, such as:
if(!foo) //Will not throw reference error because foo is declared already
{
var foo = {};
}
So, regardless of how you declare your variables, they'll all get "moved up" to the top of the function anyway.
Hope this helps!

How does this JavaScript closure work?

Here's some JavaScript:
linkElem.click(function () {
var data = linkElem.data();
alert(''+data.mls + ' ' + data.id);
});
It works.
linkElem is a local variable that I create in a loop inside a function. I assign some data to it with jQuery's .data(). If I did not call .click(), linkElem would be reassigned during the loop and then recycled after the function returns. However, I have created an anonymous function which references linkElem. So I am no longer sure what is going on.
My guess is that all of the anonymous functions and linkElems created during the loop are given UIDs of some kind and moved to persistent/global scope. Is this correct? Gratuitous detail would be much appreciated.

Yes, your description is pretty close. The local storage for a Javascript function call is just a block of memory allocated for local variables. If you "capture" that by creating another function inside a called function, then the storage is retained and the local variables carry on with their lives, unaware that the function that gave them birth might be long dead.
It's important to keep in mind that only functions create such storage — things like brace-enclosed loop bodies are not separate storage areas. Thus a common error is to declare a variable in a function and re-use it among several functions created in a loop. That's not inherently wrong, but the effect can be surprising:
function whatever() {
for (var i = 0; i < 3; ++i) {
setTimeout(function() { alert(i); }, 5000);
}
}
If you run that, you'll see three alerts that all say "3". Why? Because they all share the same "i" variable. You can avoid that by introducing another function layer:
function whatever() {
for (var i = 0; i < 3; ++i) {
setTimeout((function(private_i) { return function() { alert(private_i); }; })(i), 5000);
}
}
The "wrapper" function is just there to provide a local variable (the parameter "private_i") whereto the loop variable "i" can be copied.

However, I have created an anonymous function which references linkElem. So I am no longer sure what is going on.
It still gets reassigned, unless you are wrapping it in another level of scope (NB: another function).
Consider the following:
for (var j = 0;j < 10;j += 1) {
arrayOfLinks[j].onclick = function () {
alert(j);
};
}
In this case, all of those links would alert 10 when clicked, because j is outside of the scope and is being updated.
If you're creating linkElem in the same way, you are likely to only get the result of the last linkElem in the loop.
This is a better way:
linkElem.click(function () {
var data = $(this).data(); // no longer dependent on `linkElem` reference
alert(''+data.mls + ' ' + data.id);
});

Please refer to this How do JavaScript closures work?
This may help you understanding closures.
Whenever you see the function keyword within another function, the inner function has access to variables in the outer function.
function foo(x) {
var tmp = 3;
function bar(y) {
alert(x + y + (++tmp));
}
bar(10);
}
foo(2)
This will always alert 16, because bar can access the x which was defined as an argument to foo, and it can also access tmp from foo.
That is a closure. A function doesn't have to return in order to be called a closure. Simply accessing variables outside of your immediate lexical scope creates a closure.
function foo(x) {
var tmp = 3;
return function (y) {
alert(x + y + (++tmp));
}
}
var bar = foo(2); // bar is now a closure.
bar(10);
The above function will also alert 16, because bar can still refer to x and tmp, even though it is no longer directly inside the scope.
However, since tmp is still hanging around inside bar's closure, it is also being incremented. It will be incremented each time you call bar.
The simplest example of a closure is this:
var a = 10;
function test() {
console.log(a); // will output 10
console.log(b); // will output 6
}
var b = 6;
test();
When a Javascript function is invoked, a new execution context is created. Together with the function arguments and the parent object, this execution context also receives all the variables declared outside of it (in the above example, both 'a' and 'b').
It is possible to create more than one closure function, either by returning a list of them or by setting them to global variables. All of these will refer to the same x and the same tmp, they don't make their own copies.
[you]: Fascinating, tell me more!
Here the number x is a literal number. As with other literals in JavaScript, when foo is called, the number x is copied into foo as its argument x.
On the other hand, JavaScript always uses references when dealing with Objects. If say, you called foo with an Object, the closure it returns will reference that original Object!
function foo(x) {
var tmp = 3;
return function (y) {
alert(x + y + tmp);
x.memb = x.memb ? x.memb + 1 : 1;
alert(x.memb);
}
}
var age = new Number(2);
var bar = foo(age); // bar is now a closure referencing age.
bar(10);
As expected, each call to bar(10) will increment x.memb. What might not be expected, is that x is simply referring to the same object as the age variable! After a couple of calls to bar, age.memb will be 2! This referencing is the basis for memory leaks with HTML objects.