Multiple assignments var a = b = b || {} in javascript

Question

Reading leaflet.js 's code, I came across a method with this line which I don't fully understand:

var events = this._leaflet_events = this._leaflet_events || {};

which can be simplified as

var a = b = b || {};

From what I understand this instruction is a multiple left-handed assignment that is right associative which means that first, JavaScript will run

b = b || {} //if b exists, returns b, else return an empty object

, then

a = b // returns the output of the preceding instruction

Which does not make sense to me. Why not write instead:

a = b || {};

Full context:

addEventListener: function( /*string*/ type, /*function */ fn, /*(optional) object*/ context){
    var events = this._leaflet_events = this._leaflet_events || {};
        events[type] = events[type] || {};
        events[type].push({
        action: fn,
        context: context || this
        });
    return this;
}

I suspect a reference trick since I don't see how this._leaflet_events gets modified by the method otherwise.

---

Thinking about it, writing var a = b = b || {} is actually be a trick to assign var a a reference to b, no matter whether b is defined or not. Modifying a now modifies b.

Back to Leaflet. With

    var events = this._leaflet_events = this._leaflet_events || {};

this._leaflet_events either exists or is initialized to {}. events is assigned this._leaflet_events by reference. The reference's value might be {} but it is still this._leaflet_events that is being modified when modifying events.

On the contrary, writing

    var events = this._leaflet_events || {};

would be a mistake , since if this._leaflet_events is not defined, events will now point to a newly created object whose value will be {}. Modifying events will change the new object but it won't change this._leaflet_events's value.

Same appearent values, different references. Here is the thing.

Answer 1

The statement var a = b = b || {}; does two things:

• It initializes b to {} if it was undefined.
• It sets a to the same as b.

The expression a = b || {}; does not not modify b so it is not equivalent.

Answer 2

The shorter expression will not set anything as b's value

a = b = b || {}; //set b's b value to {} if b is uncdefined, then set a's value to b

a = b || {}; //set a's value to b, or {} if b is undefined

the first statement is in fact equivalent to

b = b || {};
a = b;