How variable "str2" is available inside the callback method - javascript

How the variable 'str2' is available inside the callback method passed to display method?
str2 should be visible only inside function 'name'.
a = {
display: function (n){
console.log("I am inside display method");
n();
}
}
function name(a,str2)
{
a.display(function (){
console.log(str2);
})
}
name(a, 'ddd');

Variable scoping in Javascript is hierarchical. Lower scopes have access to all variables of all higher scopes. Here's an example:
function outer() {
var foo = 2;
function inner() {
console.log(foo); // 2
}
}
It doesn't matter if a variable is passed as a parameter or if it was defined as a local var.

Yes str should be visible only inside function 'name' yes it is working same you said, in JavaScript function declarations loads before any code is executed, and scope depend on execution context
in your code scope of str2 is inside name() function and you call a.display() within this scope (within scope of name()) that's why str2 available inside display() at the time of execution.

To evaluate name(a, 'ddd'), the compiler will create a new stack frame and place two slots on it, one a, which will be a reference to the global object of the same name, and str2, which will contain the string literal 'ddd'. Then the body will be evaluated.
In the body, to evaluate a.display(function(){...}), the value of a.display will be resolved to the function with parameter n. To evaluate n(function(){...}), a new stack frame will be created with n assigned to the closure that results from evaluating the anonymous callback (a closure being a combination of a pointer to the static scope of the function and the compiler generated code for the function itself). Then the body of the a.display function will be evaluated.
In the body, the console.log will be called with the given string. Then the callback n() will be evaluated. Since n doesn't take any parameters, it will just be evaluated in the topmost stack frame, so when the time comes to evaluate console.log(str2), str will not be found on the current stack frame, so the compiler will follow the scope chain all the way to the frame where we bound 'ddd' to str and a to the function.
That's a pretty long answer to your simple question. I'll try to shorten it later. Also, corrections are welcome, as I'm being very hand-wavy with the evaluation process.

In Javascript the variable declared on a certain scope is available in every inner scope. In this case it's a concept called "closure". This answer might give you good insight about scoping in Javascript: What is the scope of variables in JavaScript?
Hope this simple example can help you understand its usefulness:
function counter () {
var votes = 0;
this.upvote = function() { votes++; }
this.downvote = function() { votes--; }
this.getValue = function() { return votes; }
}
var counter1 = new counter();
counter1.upvote();
console.log(counter1.getValue()) // Prints 1
counter1.upvote();
counter1.upvote();
console.log(counter1.getValue()) // Prints 3

Related

Why and when an unexecuted closure can capture outer variable in javascript?

I can't figure out why an unexecuted closure can capture the outer variables.
I do read about some articles about execute context, lexical environment, memory management, but none of these can solve my question:
function foo() {
var a = 1;
return function() {
console.log(a);
}
}
var f = foo() // line 7
// HERE variable a was been captured
f = undefined // line 10
// HEAE variable a was been released
When the engine execute to line7, the foo execution context was created above global execution context, but after line7 the closure was never been execute, so the closure execution context was never been created, so was the lexical environment. foo execution context was been popped up, the variable a will be released.
I can't find what's wrong with my point.
So why and when the variable in closure be captured?
function foo() {
var a = 1;
return function() {
console.log(a);
}
}
Function inside function are known as closure.
When one function has another function inside it, and say top level function has some data like 'a' in this case, then all the inner functions will get access to that data, this will happen only if inner function has some reference about those variables.
Let's say if you have 'b' variable beside 'a' and you are not using it anywhere in inner functions this will be simply ignored by javascript. 'Closure' will not hold this value like 'a'.
This is actually where closure gives devs power.Check out below example -
const test = (num1) => (num2) => console.log(num1*num2);//closure is in action here
var s = test(100)
s(2); // this will give output of 200
Hope this helps.
Thanks.
Here is a discussion on bugs.chromium that is related to your question. As per this , even when the else block never executed , reference to an outer object used in function returned by else block is created and stored in heap. It is never garbage collected. Refer to below discussion :-
https://bugs.chromium.org/p/chromium/issues/detail?id=315190
The key point is lexical scope:
Lexical scope is the scope model used by the JavaScript language, which differs to some other languages which use dynamic scope. Lexical scope is the scope defined at lexing time.
Considering this:
var a = 1
console.log(a)
console.log(b)
var b = 2
console.log(c)
You can get the result:
1
undefinded
ReferenceError: c is not defined
So, you can see how JavaScript handling the variables: all variable are defined at lexing time and assignment at runtime. This is what they called hoisting.
Back to the question: closures capture variables at lexing time in which js engine read your code and defined variable and bind them.
Read more about compilation at: https://v8.dev/blog/background-compilation

How JS Scope / Closures is Passing Argument

Can someone explain why in the following function, I am able to pass an argument to a nested function? My understanding is that it has to do something with closures and scope(which I thought I had a decent understanding of), but I can seem to follow how exactly this argument is being passed.
The below function outputs 1,2 respectively. But how is the return statement doThis() getting the argument/parameter for "a"? I cant figure where/how this is accessed/passed.
function doSomething(){
var b = 1;
return function doThis(a){
console.log(b);//1
console.log(a);//2
}
}
var exec = doSomething();
exec(2);
The function doSomething() returns another function so when you execute
var exec = doSomething();
You can think of that exec as containing the following function
function doThis(a){ // <- takes an argument
console.log(1); // comes from the b argument in the outer scope
console.log(a); // not set yet
}
Thus when you call exec(2) you are actually calling doThis() with an argument 2 which becomes the value of a.
This is a slightly simplified version. To expand on that, the doSomething() function is described as closing over doThis() creating a closure. Conversely, the function doThis() is closed over or inside a closure. The closure itself is simply a limited state around the function:
function doSomething(){ // --> defines the closure
var b = 1; // variable only visible within doSomething()
return function doThis(a){ //<--> function has access to everything in doSomething(). Also defines another closure
console.log(b); // --> accesses the OUTER scope
console.log(a); // <-- comes from the INNER scope
} // <-- end INNER scope
} // --> end OUTER scope
When you execute doSomething() the returned result still retains access to the scope within it, this is why doThis() has access to the value b - it's simply reachable for it. It's similar how you can do
var foo = 40;
function bar(value) {
return foo + value;
}
console.log(bar(2));
Only in this instance any other code will have acces to foo as it's a global variable, so if you do foo = 100 in a different function, that will change the output of bar(). A closure prevents the code inside from being reachable from outside the closure.
When you assign var exec = doSomething();, exec is basically writing:
var doSomething = function(a) {
console.log(b);
console.log(a);
}
It became its own function. So passing in 2 like so exec(2) works like any normal function except that it has the variable b available to it because of the closure.

JavaScript Function Related Issue [duplicate]

Today, I got completely surprised when I saw that a global variable has undefined value in a certain case.
Example:
var value = 10;
function test() {
//A
console.log(value);
var value = 20;
//B
console.log(value);
}
test();
Gives output as
undefined
20
Here, why is the JavaScript engine considering global value as undefined? I know that JavaScript is an interpreted language. How is it able to consider variables in the function?
Is that a pitfall from the JavaScript engine?
This phenomenon is known as: JavaScript Variable Hoisting.
At no point are you accessing the global variable in your function; you're only ever accessing the local value variable.
Your code is equivalent to the following:
var value = 10;
function test() {
var value;
console.log(value);
value = 20;
console.log(value);
}
test();
Still surprised you're getting undefined?
Explanation:
This is something that every JavaScript programmer bumps into sooner or later. Simply put, whatever variables you declare are always hoisted to the top of your local closure. So, even though you declared your variable after the first console.log call, it's still considered as if you had declared it before that.
However, only the declaration part is being hoisted; the assignment, on the other hand, is not.
So, when you first called console.log(value), you were referencing your locally declared variable, which has got nothing assigned to it yet; hence undefined.
Here's another example:
var test = 'start';
function end() {
test = 'end';
var test = 'local';
}
end();
alert(test);
What do you think this will alert? No, don't just read on, think about it. What's the value of test?
If you said anything other than start, you were wrong. The above code is equivalent to this:
var test = 'start';
function end() {
var test;
test = 'end';
test = 'local';
}
end();
alert(test);
so that the global variable is never affected.
As you can see, no matter where you put your variable declaration, it is always hoisted to the top of your local closure.
Side note:
This also applies to functions.
Consider this piece of code:
test("Won't work!");
test = function(text) { alert(text); }
which will give you a reference error:
Uncaught ReferenceError: test is not defined
This throws off a lot of developers, since this piece of code works fine:
test("Works!");
function test(text) { alert(text); }
The reason for this, as stated, is because the assignment part is not hoisted. So in the first example, when test("Won't work!") was run, the test variable has already been declared, but has yet to have the function assigned to it.
In the second example, we're not using variable assignment. Rather, we're using proper function declaration syntax, which does get the function completely hoisted.
Ben Cherry has written an excellent article on this, appropriately titled JavaScript Scoping and Hoisting.
Read it. It'll give you the whole picture in full detail.
I was somewhat disappointed that the problem here is explained, but no one proposed a solution. If you want to access a global variable in function scope without the function making an undefined local var first, reference the var as window.varName
Variables in JavaScript always have function-wide scope. Even if they were defined in the middle of the function, they are visible before. Similar phenomena may be observed with function hoisting.
That being said, the first console.log(value) sees the value variable (the inner one which shadows the outer value), but it has not yet been initialized. You can think of it as if all variable declarations were implicitly moved to the beginning of the function (not inner-most code block), while the definitions are left on the same place.
See also
Javascript function scoping and hoisting
Javascript variable declarations at the head of a function
There is a global variable value, but when control enters the test function, another value variable is declared, which shadows the global one. Since variable declarations (but not assignments) in JavaScript are hoisted to the top of scope in which they are declared:
//value == undefined (global)
var value = 10;
//value == 10 (global)
function test() {
//value == undefined (local)
var value = 20;
//value == 20 (local)
}
//value == 10 (global)
Note that the same is true of function declarations, which means you can call a function before it appears to be defined in your code:
test(); //Call the function before it appears in the source
function test() {
//Do stuff
}
It's also worth noting that when you combine the two into a function expression, the variable will be undefined until the assignment takes place, so you can't call the function until that happens:
var test = function() {
//Do stuff
};
test(); //Have to call the function after the assignment
The simplest way to keep access to outer variables (not just of global scope) is, of course, to try to not re-declare them under the same name in functions; just do not use var there. The use of proper descriptive naming rules is advised. With those, it will be hard to end up with variables named like value (this aspect is not necessarily related to the example in the question as this variable name might have been given for simplicity).
If the function might be reused elsewhere and hence there is no guarantee that the outer variable actually defined in that new context, Eval function can be used. It is slow in this operation so it is not recommended for performance-demanding functions:
if (typeof variable === "undefined")
{
eval("var variable = 'Some value';");
}
If the outer scope variable you want access to is defined in a named function, then it might be attached to the function itself in the first place and then accessed from anywhere in the code -- be it from deeply nested functions or event handlers outside of everything else. Notice that accessing properties is way slower and would require you to change the way you program, so it is not recommended unless it is really necessary: Variables as properties of functions (JSFiddle):
// (the wrapper-binder is only necessary for using variables-properties
// via "this"instead of the function's name)
var functionAsImplicitObjectBody = function()
{
function someNestedFunction()
{
var redefinableVariable = "redefinableVariable's value from someNestedFunction";
console.log('--> functionAsImplicitObjectBody.variableAsProperty: ', functionAsImplicitObjectBody.variableAsProperty);
console.log('--> redefinableVariable: ', redefinableVariable);
}
var redefinableVariable = "redefinableVariable's value from someFunctionBody";
console.log('this.variableAsProperty: ', this.variableAsProperty);
console.log('functionAsImplicitObjectBody.variableAsProperty: ', functionAsImplicitObjectBody.variableAsProperty);
console.log('redefinableVariable: ', redefinableVariable);
someNestedFunction();
},
functionAsImplicitObject = functionAsImplicitObjectBody.bind(functionAsImplicitObjectBody);
functionAsImplicitObjectBody.variableAsProperty = "variableAsProperty's value, set at time stamp: " + (new Date()).getTime();
functionAsImplicitObject();
// (spread-like operator "..." provides passing of any number of arguments to
// the target internal "func" function in as many steps as necessary)
var functionAsExplicitObject = function(...arguments)
{
var functionAsExplicitObjectBody = {
variableAsProperty: "variableAsProperty's value",
func: function(argument1, argument2)
{
function someNestedFunction()
{
console.log('--> functionAsExplicitObjectBody.variableAsProperty: ',
functionAsExplicitObjectBody.variableAsProperty);
}
console.log("argument1: ", argument1);
console.log("argument2: ", argument2);
console.log("this.variableAsProperty: ", this.variableAsProperty);
someNestedFunction();
}
};
return functionAsExplicitObjectBody.func(...arguments);
};
functionAsExplicitObject("argument1's value", "argument2's value");
I was running into the same problem even with global variables. My problem, I discovered, was global variable do Not persist between html files.
<script>
window.myVar = 'foo';
window.myVarTwo = 'bar';
</script>
<object type="text/html" data="/myDataSource.html"></object>
I tried to reference myVar and myVarTwo in the loaded HTML file, but received the undefined error.
Long story/day short, I discovered I could reference the variables using:
<!DOCTYPE html>
<html lang="en">
<!! other stuff here !!>
<script>
var myHTMLVar = this.parent.myVar
/* other stuff here */
</script>
</html>

variable scope and javascript execution context

I have some trouble to understand the Javascript execution context
see below code:
<script >
var global_var;
first();
function first() {
var first_var = 'a';
second();
}
function second() {
var second_var = 'b';
console.log(first_var);
console.log(second_var);
}
console.log('in the outer');
//second();
</script>
Based on my understanding, when first function call the second function, the second function is inside first function's execution context, so, the second can also access first function's properties, here is first_var
But, the actually output in function second display "first_var is not defined"
I am confused, could some one give me some explanation?
Thanks in advance.
Scoping is based in the lexical structure of the code, not the dynamic runtime relationship between functions (the "thread of execution"). The lexical structure is the static organization of the code; the "nesting" of function inside function.
In your case, the "second" function is declared outside of the "first" function, so the local variable in "first" is not visible to the code in "second".
If you were to move the "second" function inside "first":
function first() {
function second() {
var second_var = 'b';
console.log(first_var);
console.log(second_var);
}
var first_var = 'a';
second();
}
then the variable would be visible.
JavaScript has two scopes: global and function. (ES6 will introduce block scope with let and const, but for the sake of discussion, just assume the first two.
What this means is that variables defined in a function are only visible within that function. This has nothing to do with execution context.
So, in your example, first_var is only visible in first, second_var is only visible in second, global_var is visible globally because its not defined in a function.
Execution context comes into play with how this is defined, but that is another question and another topic entirely.

Javascript closure

The following program returns "local" and, according to the tutorial Im reading, it is designed to demonstrate the phenomenon ofclosure`
What I don`t understand is why, at the end, in order to call parentfunction, it assigns it to the variable "child" and then calls "child."
Why doesn`t it work by just writing parentFunction(); at the end?
var variable = "top-level";
function parentFunction() {
var variable = "local";
function childFunction() {
print(variable);
}
return childFunction;
}
var child = parentFunction();
child();
parentFunction() returns another function which you assign to var child. Then, you call child() to invoke the function returned by the call to parentFunction().
Running just parentFunction(); at the end wouldn't do anything useful because you would just discard its return value which is a function. But this would work:
parentFunction()();
See this fiddle: http://jsfiddle.net/USCjn/
Update: A simpler example:
function outer() { // outer function returns a function
return function() {
alert('inner function called');
}
}
x = outer(); // what is now in x? the inner function
// this is the same as saying:
// x = function() {
// alert('inner function called');
// }
x(); // now the inner function is called
See this fiddle: http://jsfiddle.net/bBqPY/
Functions in JavaScript can return functions (that can return functions (that can return functions ...)). If you have a function that returns another function then it means that when you call the outer function what you get is the inner function but it is not called yet. You have to call the value that you got as a function to actually run the body of the inner function. So:
x = f();
means - run a function f and store what it returns (which may be a string, a number, an object, an array, or a function) in x. But this:
x = f()();
means - run a function f, expect it to return a function and run that returned function as well (the second parentheses) and store in x what the returned function returned.
The function f here is a higher order function because it returns another function. Functions can also take another functions as arguments. One of the most powerful ideas of functional programming languages in general and JavaScript in particular is that functions are just normal values like arrays or numbers that can be returned and passed around.
You have to first grasp the idea of higher order functions to understand closures and the event system in JavaScript.
2016 Update
Note that currently this:
function outer() {
return function() {
alert('inner function called');
}
}
can be written as:
let outer = () => () => alert('inner function called');
using the ES6 arrow function syntax.
The amazing part about closures is that an inner function (in this case, childFunction) can refer to variables outside of its scope (in this case, variable). parentFunction doesn't return the result of childFunction, but an actual reference to the function!
This means that when you do the following...
var child = parentFunction();
...now, child has a reference to childFunction, and childFunction still has access to any variables it had when the function was created, even if they no longer exist.
In order to have parentFunction call childFunction, you'd need to change your code as follows:
From...
return childFunction;
To:
return childFunction();
Douglas Crockford (Pioneer of JSON, among other things) has a whole article devoted to closures, and scoping in javascript, and it would be well worth it to check out his other articles on javascript.
The point that is being demonstrated is that the function that was returned and assigned to child is still referencing the variable that was declared inside parentFunction instead of the one that was declared outside where child() is being invoked.
The only way to create a variable scope in javascript is in a function body. Normally the variable inside the parentFunction would have been discarded after the function returned.
But because you declared a function inside parentFunction that referenced variable in that scope and passed it out of parentFunction, the variable in the parentFunction is retained via the reference made in the new function.
This protects variable from outside manipulation except by functions that closed around it inside parentFunction.

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