Javascript function variable should generate error instead of undefined

Question

I am learning javascript by myself.Today I came across with a code example which is quite confusing.

function fool(a, b){

    if(b)
      var c = "Mary"
    alert(c);
}

fool(1, true); //Returns "Mary"
fool(1, false); //Returns undefined instead of error

On the other hand if I do this

function fool(a, b){

    //if(b)
    //  var c = "Mary"
    alert(c);
}

fool(1, true); //Firebug error: ReferenceError: c is not defined
fool(1, false);

An error is occurring on the first chance. How is so ?

Read about Variable hoisting

Rayon
– Rayon

2016-03-30 10:35:12 +00:00
Commented Mar 30, 2016 at 10:35 — Rayon
– Rayon, Commented Mar 30, 2016 at 10:35

Eloims · Accepted Answer · 2016-03-30 10:40:08Z

Citing MDN

var hoisting

Because variable declarations (and declarations in general) are processed before any code is executed, declaring a variable anywhere in the code is equivalent to declaring it at the top. This also means that a variable can appear to be used before it's declared. This behavior is called "hoisting", as it appears that the variable declaration is moved to the top of the function or global code.

Example:

bla = 2
var bla;
// ...

// is implicitly understood as:

var bla;
bla = 2;

For your particular case:

// Your code
function fool(a, b){
    if (b)
      var c = "Mary"
    alert(c);
}

// Equivalent code
function fool(a, b){
    var c = undefined;
    if (b)
        c = "Mary";

    alert(c);
}

As a side note, Javascript evolves.

The let keyword was introduced, which works like you expect, but is not supported in all browsers yet, so you can only use it when you control the javascript VM (like using NodeJS, nw.js, ...)

Francesco Pezzella · Accepted Answer · 2016-03-30 10:46:37Z

1

This is how the JS engine will rearrange your first example:

function fool(a, b){
    // statement always executed, no matter what
    var c;

    if (b)
       c = "Mary"

    alert(c);
}

Regardless of whether the if statement gets executed or not the var c statement gets hoisted to the top of the function body and therefore be declared in the current function scope.

answered Mar 30, 2016 at 10:46

Francesco Pezzella

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1 Comment

TEST Over a year ago

always use ccuryl brackets!

Abhay Naik · Accepted Answer · 2016-03-30 10:43:16Z

0

You have commented the variable 'c' and trying to display it.It will give you an error, since there is no variable defined as 'c'.

There is a typo in code.

try this

if(b){
   alert(c);
}

edited Mar 30, 2016 at 10:43

answered Mar 30, 2016 at 10:38

Abhay Naik

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1 Comment

Marymon Over a year ago

But there is also no declaration for first code and 'fool(1, false);' calling. As 'b' is false it wont enter into the if block.

Roli Agrawal · Accepted Answer · 2016-03-30 10:55:14Z

There are two things which i noticed in code 1) If scope: When you dont use bracket({}) it will take only fist sentence under If block. Because else interpreter will not got to know till when if condition is. So yours code which is

 if(b)
      var c = "Mary"
 alert(c);

Should be like

if(b){
      var c = "Mary"
    alert(c);
}

if you want alert(c) executed only on true of if condition.

2) Hoisting: Now second question is even if yours condition is false and sentence doesn't execute how come it alert undefined the reason is variable hoisting which is variable declaration executed before any code execution. So even if you do code like

var k="something";

it will first declare k variable like

var k=undefined;
k="something";

So in yours code first c is declared then execution of code is happening .

I hope it will help

Mogsdad · Accepted Answer · 2016-04-01 02:23:15Z

0

function fool(a, b){

    if(b)
      var c = "Mary"
    alert(c);
}

fool(1, true); //Returns "Mary"
fool(1, false); //Returns undefined instead of error

Let's break down -

You are invoking a function which is passing b as false it goes to the if loop and consider only 1st line as part of if and continue to execute next line which is alert(c). As b='false' so it won't assign Marry to the c var. So, now you are alerting c which is undefined as if won't process it until it get b as true in function parameter.

edited Apr 1, 2016 at 2:23

Mogsdad

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answered Mar 30, 2016 at 10:40

Kalpesh Singh

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2 Comments

Rayon Over a year ago

If anything is not defined.. It should throw Reference Error

Kalpesh Singh Over a year ago

Because c is defined, but not executed?

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