How to understand the output of my function in JavaScript

Question

I was browsing around the internet and came upon this JavaScript function

   function foo(){
        function bar() {
            return 3;
        }
        return bar();
        function bar() {
            return 8;
        }
    }
    console.log(foo());
    =>8

It's really confusing why this function is returning 8 when it is called. I would assume that this code would run down the contents of the function and stop at return bar();and return 3. What am I missing here?

This is because function declarations get hoisted to the top of the block. — zxqx
– zxqx, Commented Oct 27, 2014 at 20:52
functions get hoisted - doesn't matter if it was declared after the statement, it can still be ran. The last bar() overwrites the first. — tymeJV
– tymeJV, Commented Oct 27, 2014 at 20:52
Just curious, is this a question of purely academic value? Why would someone write code like that? Seriously, is there a good reason? — Logic Artist
– Logic Artist, Commented Oct 27, 2014 at 20:55
@LogicArtist -- Was probably used in an example to demonstrate hoisting. — tymeJV
– tymeJV, Commented Oct 27, 2014 at 21:00

Travis J · Accepted Answer · 2014-10-27 20:53:20Z

3

This is a result of variable hoisting. function definitions go before anything else. It really looks like this

function foo(){
    var bar;
    bar = function() {
        return 3;
    }
    bar = function() {
        return 8;
    }
    return bar();
}

"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." var^MDN

answered Oct 27, 2014 at 20:53

Travis J

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1 Comment

Ingo Bürk Over a year ago

It might be worth mentioning that using var bar = function () { … } in the original code would actually make a difference here.

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