JavaScript 'hoisting' [duplicate]
Asked Answered
C

5

89

I came across JavaScript 'hoisting' and I didn't figure out how this snippet of code really functions:

var a = 1;

function b() {
    a = 10;
    return;

    function a() {}
}

b();
alert(a);

I know that function declaration like ( function a() {} ) is going to be hoisted to the top of the function b scope but it should not override the value of a (because function declarations override variable declarations but not variable initialization) so I expected that the value of the alert would be 10 instead of 1!!

Cell answered 9/3, 2013 at 13:22 Comment(2)
Hoisting is JavaScript's default behavior of moving declarations to the top. (function declarations are "moved" from where they appear in the flow of the code to the top of the code. This gives rise to the name "Hoisting".) Read moreSiler
Here is a link with easy explanation about "Function expressions vs. function declarations and Hoisting" ... gomakethings.com/function-expressions-vs-function-declarationsRamonaramonda
L
118
  1. The global a is set to 1
  2. b() is called
  3. function a() {} is hoisted and creates a local variable a that masks the global a
  4. The local a is set to 10 (overwriting the function a)
  5. The global a (still 1) is alerted
Leo answered 9/3, 2013 at 13:26 Comment(6)
thank you @Leo I was missing the fact number 3 (that function a() {} creates a local variable that masks the global a)Cell
@Quentin: Can you please show with code how the function would look with hoisting?Biracial
@Brown_Dynamite — It looks like it looks in the question. It functions in the order described in this answer.Leo
So if a was not hoisted, calling function b() would set a = 10 globally, alert(a) would then output 10? If I'm understanding correctly?Average
The best way to understand hoisting is to talk like a compiler, GrateSiler
@Leo thanks for clear explanation about quirk.Seeseebeck
E
6

It's because the order of compilation/interpretation in this example is somewhat misleading. The function a () {} line is interpreted before any of the rest of the function is executed, so at the very beginning of the function, a has the value of function a () {}. When you reassign it to 10, you are reassigning the value of a in the local scope of function b(), which is then discarded once you return, leaving the original value of a = 1 in the global scope.

You can verify this by placing alert()s or the like in the appropriate places to see what the value of a is at various points.

Electricity answered 9/3, 2013 at 13:35 Comment(0)
A
5

(1) JavaScript does not have block statement scope; rather, it will be local to the code that the block resides within.

(2) Javascript's declaration of variables in a function scope, meaning that variables declared in a function are available anywhere in that function, even before they are assigned a value.

(3) Within the body of a function, a local variable takes precedence over a global variable with the same name. If you declare a local variable or function parameter with the same name as a global variable, you effectively hide the global variable.

you code is same as: (read comment)

<script>
var a = 1;          //global a = 1
function b() {
    a = 10;         
    var a = 20;     //local a = 20
}
b();
alert(a);           //global a  = 1
</script>

reference:
(1) JavaScript Variable Scope:
(2) A Dangerous Example of Javascript Hoisting
(3) Variable scope

So in your code:

var a = 1;          //global a = 1  
function b() {
    a = 10;         
    return;
    function a() {} //local 
}
b();
alert(a);           //global a = 1  
Adao answered 9/3, 2013 at 16:40 Comment(6)
where you wrote "local a Undefined" - that's actually not true. function declarations are parsed before any content of the parent scope, so in this case when you assign a = 10, you are already overriding function a() {} with value 10. See Quentin's answerDarcie
@rochal No, please read this article Javascript hoisting can be especially nefarious in loops Also check code on safari browser...let me know And Yes its a=10 on most browser ..Actually Undefined behaviour not ERRORAdao
read point 4. from Quentin's answer. At this point a already exists - it's a function, not an Undefined. How would Safari be different to i.e. Chrome..?Darcie
@rochal variables declared in a function are available anywhere in that function, even before they are assigned a value I also checked it, and giving 10 But according to three tutorials (I given link) it should be undefined and 10 is result because of undefined. Quentin answered according to output but I think it may be not 10 for some browsers..Adao
once again, in your example code: a = 10; //local a Undefined is incorrect a = 10; //local a is function(){} is correct because as soon as you check the value of a inside function b() it already has been overridden by the local function a(){} so your comment that "local a Undefined" is not valid. And browser has nothing to do with it, it's a language feature.Darcie
@rochal Yes actually you are Correct it shouldn't be a browser Issue. Corrected The link: confused me A Dangerous Example of Javascript Hoisting Do you understands what it says by Undefined ?Adao
F
2
  1. function declaration function a(){} is hoisted first, hence in local scope a is created.
  2. If you have two variable with same name (one in global another in local), local variable always get precedence over global variable.
  3. When you set a=10, you are setting the local variable a , not the global one.

Hence, the value of global variable remain same and you get, alerted 1

Frazzled answered 17/5, 2014 at 13:30 Comment(0)
C
0

When I read the same article you did JavaScript Scoping and Hoisting, I was confused as well because the author never showed what the two opening example codes are interpreted as in the compiler.

Here is example you provided, and the second on the page:

var a = 1;
function b() {
    function a() {} // declares 'a' as a function, which is always local
    a = 10;
    return;
}
b();
alert(a);

and here is the first example on the page:

var foo = 1;
function bar() {
    var foo; // a new local 'foo' variable
    if (!foo) {
        foo = 10;
    }
    alert(foo);
}
bar();

Hope this helps

Coleorhiza answered 6/11, 2013 at 5:30 Comment(1)
I believe that article explains the hoisting topic quite good 2n.pl/blog/javascript-part-4Haffner

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