JavaScript code trick: What's the value of foo.x

foo.x = foo = {n: 2};

determines that foo.x refers to a property x of the {n: 1} object, assigns {n: 2} to foo, and assigns the new value of foo – {n: 2} – to the property x of the {n: 1} object.

The important thing is that the foo that foo.x refers to is determined before foo changes.

See section 11.13.1 of the ES5 spec:

1. Let lref be the result of evaluating LeftHandSideExpression.

2. Let rref be the result of evaluating AssignmentExpression.

The assignment operator associates right to left, so you get:

foo.x = (foo = {n: 2})

The left hand side is evaluated before the right hand side.

foo.x = foo = {n: 2};

Here foo refers to {n:1} object before assignment i.e. before the statement is executed.

The statement can be re-written as foo.x = (foo = {n:2});

In object terms the above statement can be re-written as {n:1}.x = ( {n:1} = {n:2} );

Since assignment happens from right to left only. So here we just have to keep a check that foo is referring to which object before execution starts.

On solving the R.H.S: foo = {n:2}; Now foo is referring to {n:2};

Coming back on the problem we are left with:

foo.x = foo;

Now foo.x on L.H.S is still {n:1}.x whereas foo on R.H.S is {n:2}.

So after this statement gets executed {n:1} will become { n:1, x:{n:2} } with bar still referring to it. Where as foo will now be referring to {n:2}.

So on execution foo.x gives undefined as there is only 1 value in foo which is {n:2}.

But if you will try executing bar.x it will give {n:2}. Or if you will just execute bar the result will be

Object {n: 1, x: Object}

As I understand expression :

foo.x = foo = {n: 2};

just exactly the same as:

foo.x = {n: 2} ; 
foo = {n: 2};

And after this it's became obvious that:

 bar=={n: 1, x: {n:2}};
 foo=={n:2};
 foo.x==undefined