Why declared variables in Javascript are non-configurable in the current scope? - javascript

I have created a variable x with the keyword var but when I do the following:
var x = 10;
delete x;
It returns false. basically, I don't want to delete the x variable but my question is that why javascript does not allow to configure the variables declared in the current scope context. This is also mentioned in this documentation, but the question is why?

Because otherwise every x might or might not throw an error or might suddenly refer to another variable:
let x = 2;
{
let x = 3;
if(Math.random() > 0.5) delete x;
console.log(x); // ?!
}
That makes code conpletely error prone and unpredictable and makes it impossible to optimize, every line might suddenly become a syntax error, and thats why it is not possible to delete variables that way.
However there is another way to get this behaviour by adding an object as scope which you can mutate, and thats the reason why no one uses the with statement:
const scope = { b: 2 };
with(scope) {
console.log(b); // 2
delete scope.b;
console.log(b); // reference error
}

You cannot delete a variable if you declared it (with var x;) at the time of first use.
However, if your variable x first appeared in the script without a declaration,
you can delete the variable if you didn't use var keyword.
you can get more information from this resource
http://www.javascripter.net/faq/deleteavariable.htm

Related

Javascript Why can't use let inside eval()

I would like someone to please tell me why Javascript does not complain when I do this:
eval("x = 1");
console.log(x);
output: 1
...However, it complains if I do this:
eval("let x=1");
console.log(x);
output:
> ReferenceError: x is not defined
Thank you.
Note: I know using eval is bad, serious security risks, blah,blah, thank you for that. I just want to understand the theory behind this.
Edit: Now that you've updated your question, I can help a little more specifically.
The reason
eval("let x = 1");
console.log(x);
doesn't work is because let is local to its scope. You cannot access the variable outside of the scope of eval.
You would need to use var to make the variable globally accessible.
To quote MDN:
let allows you to declare variables that are limited to the scope of a block statement, or expression on which it is used, unlike the var keyword, which declares a variable globally, or locally to an entire function regardless of block scope.
The reason your original, unedited question didn't work was because you were assigning the variable without actually doing anything with it.
eval("var x = 1"); // Undefined
eval("const x = 1"); // Undefined
eval("let x = 1"); // Undefined
Now if you give it something to spit back out:
eval("var x = 1; x"); // 1
eval("const x = 1; x"); // 1
eval("let x = 1; x"); // 1
At least that's the way it works in my chromium console.
The way I see it is that in the first example you're simply evaluating the response of the statement x = 1.
However, it doesn't return a response when evaluating because there is no response to return.
It's just an assignment. You are telling the eval that there is this variable called x and you are telling it that it has a value of 1 but you are not telling it what to do with that variable.
It is the equivalent of you creating the following:
doNothing.js
let x = 1;
If you run this program, it will not output anything to the console because all it is doing is assigning a value to x.

Mutating key inside JavaScript's for..in loop [duplicate]

What's the correct way to write a for-in loop in JavaScript? The browser doesn't issue a complaint about either of the two approaches I show here. First, there is this approach where the iteration variable x is explicitly declared:
for (var x in set) {
...
}
And alternatively this approach which reads more naturally but doesn't seem correct to me:
for (x in set) {
...
}
Use var, it reduces the scope of the variable otherwise the variable looks up to the nearest closure searching for a var statement. If it cannot find a var then it is global (if you are in a strict mode, using strict, global variables throw an error). This can lead to problems like the following.
function f (){
for (i=0; i<5; i++);
}
var i = 2;
f ();
alert (i); //i == 5. i should be 2
If you write var i in the for loop the alert shows 2.
JavaScript Scoping and Hoisting
The first version:
for (var x in set) {
...
}
declares a local variable called x. The second version:
for (x in set) {
...
}
does not.
If x is already a local variable (i.e. you have a var x; or var x = ...; somewhere earlier in your current scope (i.e. the current function)) then they will be equivalent. If x is not already a local variable, then using the second will implicitly declare a global variable x. Consider this code:
var obj1 = {hey: 10, there: 15};
var obj2 = {heli: 99, copter: 10};
function loop1() {
for (x in obj1) alert(x);
}
function loop2() {
for (x in obj2) {
loop1();
alert(x);
}
}
loop2();
you might expect this to alert hey, there, heli, hey, there, copter, but since the x is one and the same it will alert hey, there, there, hey, there, there. You don't want that! Use var x in your for loops.
To top it all off: if the for loop is in the global scope (i.e. not in a function), then the local scope (the scope x is declared in if you use var x) is the same as the global scope (the scope x is implicitly declared in if you use x without a var), so the two versions will be identical.
You really should declare local variables with var, always.
You also should not use "for ... in" loops unless you're absolutely sure that that's what you want to do. For iterating through real arrays (which is pretty common), you should always use a loop with a numeric index:
for (var i = 0; i < array.length; ++i) {
var element = array[i];
// ...
}
Iterating through a plain array with "for ... in" can have unexpected consequences, because your loop may pick up attributes of the array besides the numerically indexed ones.
edit — here in 2015 it's also fine to use .forEach() to iterate through an array:
array.forEach(function(arrayElement, index, array) {
// first parameter is an element of the array
// second parameter is the index of the element in the array
// third parameter is the array itself
...
});
The .forEach() method is present on the Array prototype from IE9 forward.
Actually, if you dislike declaration within for heading, you can do:
var x;
for (x in set) {
...
}
As mentioned in other answers to this question, not using var at all produces unnecessary side-effects like assigning a global property.
Use the one where you declare the loop variable with var. Implicitly declared variables have a different scope that's probably not what you intended.
for(var i = 0; ...)
is a commonly seen pattern but it's different from
for(int i; ...)
in C++ in that that the variable isn't scoped to the for block. In fact, the var gets hoisted to the top of the enclosing scope (function) so a local i will be effectively available both before the for loop (after the beginning of the current scope/function) and after it.
In other words, doing:
(function(){ //beginning of your current scope;
//...
for(var i in obj) { ... };
})();
is the same as:
(function(){ //beginning of your current scope;
var i;
//...
for(i in obj) { ... };
})();
ES6 has the let keyword (instead of var) to limit the scope to the for block.
Of course, you SHOULD be using local variables (ones declared with either var or let or const (in ES6)) rather than implicit globals.
for(i=0; ...) or for(i in ...) will fail if you use "use strict"; (as you should) and i isn't declared.
Using var is the cleanest way, but both work as described here: https://developer.mozilla.org/en/JavaScript/Reference/Statements/for...in
Basically, by using var you ensure that you create a new variable. Otherwise you might accidentally use a previously defined variable.
I think var is good for performance reasons.
Javascript won't look through the whole global scope to see if x already exists somewhere else.
From a general point of view, first version will be for an index that must live within loop's scope, while the other one would be any variable in the scope where loop's constructor got invoked.
If you're going to use loop's index inside for loop and this won't be required by others in next lines, better declare the variable with "var" so you'll be sure "x" is for loop's index initialized with 0, while the other one, if other "x" variable is available in this context, this will get overwritten by loop's index - that's you'll have some logical errors -.
I always use the block scoped let introduced in ES2015.
for (let x in set) {
...
}
Additional reading and examples

Access variable in if block

var func = function () {
var i: number = 0;
if (i == 0) {
var y: number = 1;
}
document.body.innerHTML = y.toString(); // js/ts should complain me in this line
};
func(); // output: 1
As you can see, I've declared variable y inside if block. So, I think it couldn't be referenced outside the scope.
But, when I've tried to run the code, the output is 1.
Is it an issue in typescript/javascript?
Variables in Javascript are hoisted, meaning that var y is moved to the top of the function as if it were declared there.
Initialization, however is not hoisted, so if you change i to be something other than 0, the variable y will still exist, but it will have the value undefined.
This means that the function is exactly equivalent to:
var func = function () {
var i: number = 0;
var y: number;
if (i == 0) {
y = 1;
}
document.body.innerHTML = y.toString(); // js/ts should complain me in this line
};
To get the behavior you expect, you need to use let, which is a part of ECMAScript 2015 (ES6). It is block scoped, as you expect. It will also effectively work so that it is accessible only from the point of definition onwards, which is also probably as you would expect.
If you re-declare a JavaScript variable, it will not lose its value.
The second reference might pave way for a new variable syntax. Actually if you recall variable declaration is not neccessary in javascript. Simpe
y=1;
also works.
The second time when you reference y, outside if block, in my opinion, it tries a re-declaration and retains the old value.
Reference - http://www.w3schools.com/js/js_variables.asp
& http://www.w3schools.com/js/js_scope.asp
Javascript has function scope afaik. Any variable declared within a function, should be accessible from anywhere within the function. So, if you have a function checking if i==0, then you can achieve what you are trying to achieve.
This is as it is supposed to be. Javascript scopes by function, not by block. (This does make it unlike many popular languages)
Variables declared like var myVar = 10; will seep into nested functions but not the other way around. Variables declared like myVar = 10; will go global.
I couldn't find anything which suggested that typescript was any different.
Variables declared inside of an if statement are not scoped to the if statement. They're scoped to the current execution context. There's the Global execution context and then when a function is run, it creates it's own execution context. Inside of your function, you created the variables y and i. It doesn't matter that y was created inside of the if statement, because once it runs, y is created in the scope of the function. So then you do y.toString(), which can access y because it's scoped to the function not the if statement. That's why you get the output of 1. This is not an error, it's by design.

scope and availability of a variable inside a function

I am learning javascript and my question might not be used practically anywhere but for interview purpose I want to understand this clearly. below is the code. the first alert alerts 'undefined' and the second one is '4'. the second one is understandable. I want to know why the first alert doesn't alert '5' and undefined? whats the concept behind the same?
Thanks.
var x = 5;
function check(){
alert(x);
var x = 4;
alert(x);
}
check();
var is always hoisted (moved) to the begining of the scope. Your code is equivalent to this:
var x = 5;
function check(){
var x;
alert(x);
x = 4;
alert(x);
}
check();
As you can clearly see, the local x hides the global x, even if the local one doesn't have a value yet (so the first alert shows undefined).
A little extra: conditionals or other blocks don't influence hoisting the var declaration.
var x = 1;
(function() {
x = 3;
if (false) {
var x = 2; // var will be moved to the begining of the function
}
})()
console.log(x) // 1
JavaScript has some concepts that take a little getting used to for sure and sometime the answer for what seems to be a simple question is long winded but i will try to be as quick and concise as possible.
In JavaScript, variables are scoped by functions and you have two places that a variable can be scoped(have context) either 'globally' or 'locally'. The global context in a web browser is the 'window' so in your code example var x = 5; is a global variable. The local context in your code is within your check function so var x = 4; is a local variable to the check function. If you had more functions those would have their own function scope(context) and so on. As a word of caution if you forget the word "var" when declaring a variable inside a function it will become a 'global' variable, be careful. Now that we have the 'global', 'local' scope down what happens before a Javascript programs runs?
A few things happen in the Javascript Engine before your code is executed but we will only look at what happens to variables. Variables are said to be 'hoisted' to the top of their functional context or scope. What that means is that the variable declaration is 'hoisted or moved to the top and assigned the value 'undefined' but not initialized the initialization remains in the same spot they were. This hoisting also makes the variable available anywhere within the function it was defined.
So it looks something like this:
//to be accurate this is hoisted to its functional scope as well
var x;
x = 5;
function check() {
//declaration of 'x' hoisted to the top of functional scope,
//and assigned undefined
var x;
//this will alert undefined
alert(x);
//initialization of variable remains here
x = 4;
//so after the initialization of 'x' this will alert 4
alert(x);
}
check();
Happy Coding!!!
As an aside: They have a new feature named 'let' that you can implement to add block-level scoping in I believe EMCAScript 6 but is not implemented in all the browsers as of yet so that is why i placed it in an aside.

about eval statement constructs and delete

If I run the following code in google chrome console I get the following results
var x = 1;
alert(delete x); // false
eval('var y = 2');
alert(delete y); // true
Why in the first example the variable is not deleted and in the second example it is deleted?
From the Mozilla JS Docs for delete:
delete is only effective on an object's properties. It has no effect
on variable or function names.
The example provided is similar to yours.
x = 42; // creates the property x on the global object
var y = 43; // declares a new variable, y
delete x; // returns true (x is a property of the global object and can be deleted)
delete y; // returns false (delete doesn't affect variable names)
So, why does alert(delete y); work? I couldn't nail down the exact answer, but basically you cannot rely on the scope of eval.
I think eval('var y = 2'); does not get declared as a variable and is treated as a property, but I haven't found evidence for that yet other than the results of our tests. I'm going to continue researching to see if I find the exact reason.
Other articles on eval weirdness:
http://wingolog.org/archives/2012/01/12/javascript-eval-considered-crazy
http://brownplt.github.com/2012/10/21/js-eval.html
http://blog.rakeshpai.me/2008/10/understanding-eval-scope-spoiler-its.html
EDIT 0
Based on #Xavier Holt's comment, I looked up hoisting and scope in regards to eval. This scope cheatsheet from Mozilla Dev docs had the following:
eval may capture assignments, but not var declarations
eval'd vars hoist normally, so evals may capture assignments similar
to with:
function f() { {
let x = "inner";
eval("var x = 'outer'");
print(x); // "outer" }
}
If I'm reading this correctly, then my earlier assumption was right. eval() does not evaluate var declarations by declaring a variable. It must create a property or be treated as a property in order for delete to work.

Categories

Resources