Shapes (Hidden Classes)
The best way to explain object shapes is through examples. It all begins with the root shape.
The root shape is where the transition chain starts from, it has a pointer to a PropertyTable
,
we will explain what it is and does later on!
NOTE: We will annotate the shapes with S
followed by a number.
If we have an example of JavaScript code like:
let o = {};
The following chain is created:
We transition, the object o
has S1
shape. The root shape does not have a prototype. So we transition into a shape that has the
Object.prototype
as __proto__
. We can see that the shapes inherited the PropertyTable
from the root
.
Ok, Let us add a property x
:
o.x = 100; // The value is not important!
Then this happens:
The object o
has shape S2
shape now, we can see that it also inherited the PropertyTable
, but it's property count is 1
and
an entry has been added into the PropertyTable
.
We can see that the property added is writable
, configurable
, and enumerable
, but we also see Slot 0
,
this is the index into the dense storage in the object itself.
Here is how it would look with the o
object:
Let's define a getter and setter y
// What the getter/setter are not important!
Object.defineProperty(o, "y", {
enumerable: true,
configurable: true,
get: function () {
return this.x;
},
set: function (value) {
this.x = value;
},
});
We can see that the property has been added into the property table, it has the has_get
and has_set
flags set,
in the object there are two elements added, the first is the get
function and the second is the set
function.
Slots are varying in length, two for accessor properties and one for data properties, the index points to the first value in the object storage.
What would happen if an object had S2
shape and we tried to access a property y
how does it know if it
has or doesn't have a property named y
? By the property count on the shape, it has property count 1
,
all the object in the PropertyTable
are stored in a map that preserves the order and and can be indexed.
When we do a lookup the on property table, if the index of the property is greater than the property count (1
),
than it does not belong to the shape.
Now, Let's create a new object o2
, with property x
:
let o2 = { x: 200 };
After this o2
would have the S2
shape.
How does the shape know that it can reuse S1
then to go to S2
? This is not the real structure!
Every shape has pointers to forward transitions that happened, these are weak pointers so we don't keep
alive unused shapes. The pointers have been omitted, so the diagrams are clearer (too many arrows).
Ok, now let us define a property z
instead of y
:
o2.z = 300;
The following changes accure to the shape tree:
Now o2
has S4
shape. We can also see that PropertyTable
has been forked, because we can no longer add a property at position 1
.
What would happen if we wanted to delete a property x
from object o
:
delete o.x;
NOTE:: o2
and its shape have been omitted from the diagram.
When a deletion happens, we find the node in the chain where we added the property, and get it's parent (base
),
we also remember that what transitions happened after the property insertion, then we apply them
one by one until we construct the chain and return the last shape in that chain.