Please ignore that this can be re-written without variables. It's just to have a simple example.
window.onload = function() {
var a = document.body, b = function() {console.log(1)};
a.onkeydown = b;
};
I know what happens: it works. But how?
If b was a global variable, the interpreter would store a reference to it. In this example, does the interpreter store a reference to the local variable, only to replace it, with what I assume is a copy of the function, when the local variable is destroyed? Or is the reference to the local variable still stored somewhere behind the scenes, and is then re-purposed?
Functions (and other objects) are always passed by reference. b does not contain the function, rather it points to it. When you assign a.onkeydown = b, you are making a.onkeydown point to the same function object. Then the function ends, so the local b variable is destroyed but the function it points to is still there - it would only be removed by the garbage collector if there were nothing else pointing it to it.
Related
I have read this excellent piece:
How do JavaScript closures work?
But I still have questions...
StackOverflow tags defines "Closures" as "A Closure is a first class function that refers to (closes over) variables from the scope in which it was defined. If the closure still exists after its defining scope ends, the variables it closes over will continue to exist as well".
When my functions end with a return, do all variables defined within them get deleted and memory allocation made free? It appears to me that conditions exist when one could answer yes or no.
Yes, I know memory is plentiful and free but that is not my question.
Think of an intranet based application which ajax's data back and forward during a working day. As data is received by the client, massaged, and no longer required, it can be dispensed with. Poor programming technique could result in a build up of data no longer user or required (or worse, data being used that is out of sync).
Reading the piece on closures above tells me that some variables within a function can still be referenced from outside the function scope. I am guessing that I will need to re-read the closure document several times as perhaps I've not fully understood it.
Can someone share an example on where a variable continue to exist?
And... Do I gain any benefit if, at the begining of each function I declare my variables, and prior to using a "return true;" if I reassign the variables a value of null?
All help appreciated...
num will exist within a function outside of IIFE c which returns a
var c = (function() {
var num = 10;
var a = function a(n) {
console.log(n)
return n * Math.random()
};
return a.bind(null, num)
}());
var res = [c(), c(), c()];
console.log(res)
When you return true from a function you don't have a closure. You have a closure when you return a function from a function, like this:
function makeAdder( firstNumToAdd ){
return function( secondNumToAdd ){
return mainNumToAdd + secondNumToAdd;
}
}
var add3 = makeAdder(3);
var seven = add3(4);
In the above example, the variable firstNumToAdd, although is scoped to the function makeAdder is always accessible to the inner function, even though the function exited after executing the line var add3 = makeAdder(3);
Please correct me if I am wrong, but I don't think that
some variables within a function can still be referenced from outside the function scope
is true.
Closures remember the context of a function call and bind the relevant variables (i.e. this, etc) to the closure's scope. If that is correct, then calling the closure again would cause the old variables' values to get garbage collected and you wouldn't have to worry about memory waste or out of sync data.
obj = {
func: function(){
var test = 'preserved value';
this.func2 = function(){
console.log(test);
};
}
};
obj.func();
obj.func2(); //'preserved value'
obj.func = undefined;
obj.func2(); //'preserved value'
As you can see, even after completely blowing away the original scope (func) the enclosing scope retains the test var.
I'm fairly new to Javascript and wrote the following jQuery code which works:
function updateItems() {
var now = Math.round((new Date()).getTime() / 1000);
$(".something").each(function() {
$(this).html(now.toString());
});
}
updateItems();
Why does this work? One might think that now would not be accessible from inside the function. I guess I could run some tests to see what happens if I try to modify now from inside the function, if I run another each() right after that, etc. But a basic understanding of how the scope works here and generally in Javascript cases like this would be greatly appreciated.
Also, is this type of function accurately called a "dynamic function" or is there a better name for it?
When you use function() { ... }, you create what is technically called a closure. They effectively can capture everything from the enclosing scope.
If you use now in the closure, you would get the value of now when you execute the closure, and (potentially) not the value it had when you created it.
Note that the enclosing scope here is the scope of the outer function, not of the outer block, and you may have to take extra care if you're creating closures in a loop, for instance.
This is just a nested function. It has access to the now variable because functions have access to the variables in scope where they're defined (this is actually how globals work as well). It's called a "closure" (it "closes over" the variables in scope where its defined), which sounds obscure, but don't worry — closures are not complicated (disclosure: my blog) once you know a couple of things about how JavaScript works.
The fun thing here is that it closes over the now variable that's specific to that particular call to updateItems, and the reference it has is live (it's not a copy of now as of when the function was created). Each call to updateItems creates a new anonymous function you're passing to each, and each of those functions has access to the now variable created during the call to updateItems. In your case, you're just using it immediately and returning, but what if you kept a reference to the function? What then? The answer is that's fine, it keeps a reference to the now variable it relates to:
function say(msg) {
var f = function() {
alert(msg);
};
return f; // We return a reference to the function
}
var s1 = say("Hi");
var s2 = say("there");
s1(); // Alerts "Hi" (see below)
s2(); // Alerts "there"
The call to s1 alerts "Hi" because the function created by our call to say still has access to the msg argument we gave to say, even though say has returned. And the same for s2, of course.
And just to prove that it's not a copy as of when the function is created:
function say(msg) {
// Create the function
var f = function() {
alert(msg);
};
// Update `msg`
msg += " (updated)";
// Return the function
return f;
}
var s1 = say("Hi");
var s2 = say("there");
s1(); // Alerts "Hi (updated)"
s2(); // Alerts "there (updated)"
See how the function used the updated version of msg, even though the function was created before we updated it. The function has access to the variable itself, not a copy of the variable's value.
There is one thing in JavaScript which I can't understand.
When we declare a variable like this:
var var_Name
it called Local Variable.
When we declare a variable like this:
var_Name
it called Global Variable.
When exactly the browser resets their values? Once the page refreshes?
What I mean is, when the pages loaded, I declare a variable, and by the time, I increases it's value. When the value is going to set back to it's original initialization? When I refresh the page? or if it's a Local Variables it refreshes right away after stopping using it and if it's Global variable it refreshes only when the page refreshes?
Thanks!
Global declaration:
var_Name = 1;
// is equal to
var var_Name = 1;
// and is also equal to
window.var_Name = 1;
The variable lives as long as the window object does.
So it will be available until you leave/reload the current page.
A local variable inside a function lives as long as the function is running:
function y(){
var x = 1;
console.log(typeof x !== 'undefined');
}
y();
console.log(typeof x === 'undefined');
A global variable can be accessed from any function in any file loaded by the page.
And when the page reloads the variable is unset until some code is run again and assign a value to it.
The Javascript engine does not maintain any variable values over a page loads. Thus, any global variables that were declared and defined during the life of a page will be lost when that page reloads.
Variables declared with var are limited to the scope in which they are defined. For a browser, there is only Function scope and Global scope. As a result, a local variable declared in a function will cease to exist once that function returns, unless it is captured by a closure (a large topic that is worth searching the Internet about).
At no point to variables 'refresh' to their originally assigned values without an explicit action by the program. They are either in scope or out of scope and when in scope, they have whatever value was last assigned to them.
It's worth noting that executing the following outside of a function:
var foo = 1;
will still create a Global variable, since you're in the Global scope if you're not in a Function's scope.
Global variable is visible on all document. Local variable is visible only in block you define it. Variables will not be visible on the other documents.
When you refresh page all the variables are newly set that mean you won't get any variable's value until you save it (for example in cookie or via ajax).
It is said, javascript clears a variable from memory after its being referenced last.
just for the sake of this question i created a JS file for DEMO with only one variable;
//file start
//variable defined
var a=["Hello"]
//refenence to that variable
alert(a[0]);
//
//file end
no further reference to that variable, so i expect javascript to clear varaible 'a'
Now i just ran this page and then opened firebug and ran this code
alert(a[0]);
Now this alerts the value of variable, If the statement "Javascript clears a variable after there is no further reference it" is true how come alert() shows its value.
Is it because all variable defined in global context become properties of window object, and since even after the execution file window objects exist so does it properties.
Of course the variable HAS to be kept around in this example - after all, you keep the ENVIRONMENT it has been defined in around. As long as you COULD still access it is has to be kept, so if you can make this test and it ever fails - the JS interpreter is "kaputt".
Do this instead:
(function () {
var a = 1;
}());
// and NOW, at THIS point there is nothing any more referencing "a"
Of course, from out "here" you cannot test that - you'd have to look at the internal memory structures of the JS interpreter/compiler (today in the days of V8 it's more like a "interpiler" or "compreter" aynway).
In addition though, like all garbage collecting (GC) languages, the memory is not freed immediately but depending on GC strategy used by the particular implementation of Javascript, which also takes current memory usage and demand into account. GC is a costly operation and is delayed, if possible, and/or run when it does not take away CPU resources from the real app.
By the way, slight variation:
(function () {
var a = 1;
//empty function assigned to global namespace,
//in browsers "window" is the global object
window.example_function = function () {};
}());
In THIS example the result is the same as in yours: as long as you view that page the memory for "a" is NEVER released: the function assigned to property example_function of the global object still exists, so all of the environment is was defined in has to be kept! Only after
delete window.example_function
which deletes that property - which in the end is a pointer to the (empty) function expression - from the global object, could variable "a" be released. Also note that "a" is not even used by that function, see Google and search for "lexical scoping", one of the most defining properties of the Javascript language.
And even more fun:
Had I written
//define a variable in global namespace
var example_function;
(function () {
var a = 1;
//empty function assigned to global namespace,
//in browsers "window" is the global object
example_function = function () {};
}());
I could not use "delete" to remove it, in fact I could not ever release it. That's because in the first example example_function becomes a new property of the global object (emphasis), while in this last example it is a variable, and "delete" works with object properties only. (I used "windows.example_function" in the earlier example, but without the "window." it would have been just the same, example_function would have been created as property in the global object. Only the "var" statement creates variables.)
Is it because all variable defined in global context become properties of window object, and since even after the execution file window objects exist so does it properties.
Yes. You would either have to close the window, or explicitly remove a from the window properties.
A side effect of your script is that there's a property name "a" defined in the global object ( window ), so there is actually still a global reference to a.
(function()
{
var a = 12;
alert(a);
})();
If we wrap the declaration of a in self-executing function like here, a will be inaccessible after the function has exited.
Yes, although you can mitigate this behavior by declaring and acting on your variables within the scope of a function:
( function myfunc(a){
var b = 100;
alert(a); // 200
alert(b); // 100
}(200));
alert(a); // undefined
alert(b); // undefined
Referencing a variable does not mean using it in some manner. As long as there is at least one reference to the variable which can still be accessed, it has to be kept in memory.
Looking at your example, the following line has no effect on whether the variable gets cleared or not. Even if a wasn't being used anywhere, it would still have been kept intact.
//refenence to that variable
alert(a[0]);
Although an object may be created inside a function, its lifetime may extend beyond the function itself. The only thing to remember is whether there are still any valid references to the object in question. Consider this example,
(function() {
var a = ["Hello"];
var copy = a;
a = null;
setTimeout(function() {
// a: null, copy: ["Hello"]
console.log("a: %o, copy: %o", a, copy);
}, 2000);
})();
The variable a is created inside an anonymous function, and another variable copy is referencing it. a is set to null right after, but we still have 1 reference to it from copy. Then we create a timeout which makes another reference to the same variable. Now even though this anonymous function will go out of scope after it executes the setTimeout, we still left 1 reference to it in the timeout function. So this copy variable will still be intact when the timeout function is run.
When we have code like:
function a(){
var x =0;
this.add=function(){
alert(x++);
}
}
var test = new a();
test.add(); // alert 0
test.add(); // alert 1
test.add(); // alert 2
How does this work?
Doesn't that the value of 'x' in a() should be 'gone' as soon as test = new a() is complete? The stack contains x should also be gone as well, right? Or, does javascript always keep all the stacks ever created in case they will be referenced in future? But that wouldn't be nice, would it...?
The word you're looking for is “closure”.
Creating a function inside another function gives the inner function a (hidden) reference to the local scope in which the outer function was running.
As long as you keep a copy of your test, that has an explicit reference to the add function, and that function has an implicit reference to the scope created when calling the a constructor-function. That scope has an explicit reference to x, and any other local variables defined in the function. (That includes the this value and the constructor's arguments — though you can't access them from inside add as that function's own this/arguments are shadowing them.)
When you let go of test, the JavaScript interpreter can let go of x, because there's no way to get a reference back to that variable.
What you're seeing is the effect of a closure. The function being defined within the other function gets access to all of the variables and such in scope where it is — even after the outer function returns. More here, but basically, the variables (and arguments) in the function all exist as properties on an object (called the "variable object") related to that function call. Because the function you've bound to this.add is defined within that context, it has an enduring reference to that object, preventing the object from being garbage-collected, which means that that function can continue to access those properties (e.g., the variables and arguments to the function).
You normally hear people saying that the function closes over the x variable, but it's more complex (and interesting) than that. It's the access to the variable object that endures. This has implications. For instance:
function foo() {
var bigarray;
var x;
bigarray = /* create a massive array consuming memory */;
document.getElementById('foo').addEventListener('click', function() {
++x;
alert(x);
});
}
At first glance, we see that the click handler only ever uses x. So it only has a reference to x, right?
Wrong, the reference is to the variable object, which contains x and bigarray. So bigarray's contents will stick around as well, even though the function doesn't use them. This isn't a problem (and it's frequently useful), but it emphasizes the underlying mechanism. (And if you really don't need bigarray's contents within the click handler, you might want to do bigarray = undefined; before returning from foo just so the contents are released.)