Why does my code execute a function defined after the return statement?

Question

i have this piece of code:

(function f(){
    function f(){ return 1; }
    return f();
    function f(){ return 2; }
 })();

why does this code print '2'?

That looks like the typical example from a tutorial or quiz. Where did you find it? — cookie monster
– cookie monster, Commented Jul 9, 2014 at 16:17

Quentin · Accepted Answer · 2014-07-09 16:11:13Z

6

Function declarations are hoisted, so both are processed before the return statement is evaluated.

The second function overwrites the first because they have the same name.

answered Jul 9, 2014 at 16:11

Quentin

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Comments

ckersch · Accepted Answer · 2014-07-09 16:13:08Z

4

In javascript, function definition is hoisted to the top of its containing function.

Your function is interpreted by the browser like this:

(function f(){
    //Functions defined first
    function f(){ return 1; }
    function f(){ return 2; } //<- Hoisted to top, now this is what f is

    //Now the rest of the code runs
    return f();

 })();

answered Jul 9, 2014 at 16:13

ckersch

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Comments

Claudiu Creanga · Accepted Answer · 2014-07-09 16:10:59Z

2

Because functions are hoisted and processed before the return statement, so your last function f() returns 2 and it overwrites the first one.

answered Jul 9, 2014 at 16:10

Claudiu Creanga

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