CODEWITHSUNDEEP
javascriptjquery
Om3ga
Om3ga

Reputation: 32893

Calling a function before declaring it using function expression works. Why?

I declared two functions using function expression innerone and innertwo. I first declared innerone and after that innertwo. Inside innerone I am calling innertwo function. But my concern is that I am declaring innertwo after innerone using function expression which means innertwo is not hoisted. So why these functions work in this order? Is it mandatory to change their order?

Here is code

var one = function () {
    var innerone = function () {
        innertwo();
    },

    innertwo = function () {
        console.log('innertwo');
    };

    return {
        innerone: innerone
    };
};

var o = new one();
o.innerone();

Upvotes: 0

Views: 103

Answers (2)

jfriend00
jfriend00

Reputation: 707706

You asked in your comment for simpler ways to write this code, so here are a couple simpler ways. The first still returns the structure and can be just called as a function without new. The second adds a property to the function object, but still preserves the private function innertwo(). The second option looks the cleanest to me.

Option 1:

function one() {
    function innertwo() {
        console.log('innertwo');
    }

    return {
        innerone: function() {
            innertwo();
        }
    };
}

var o = one();    // new is not needed here
o.innerone();

Option 2:

function one() {
    function innertwo() {
        console.log('innertwo');
    }

    this.innerone = function () {
        innertwo();
    };
};


var o = new one();
o.innerone();

Upvotes: 0

user1796666
user1796666

Reputation:

It's working because innerone is called only when you call it. And by the time it's called innertwo is defined.

Upvotes: 5

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