This is related to Javascript a=b=c statements.
I do understand that
foo = foo.x = {n: b}; // console.log(foo) => {n: b}
but
This is because when you write
var foo = {};
foo.x = foo = {n: b} //a=b=c
while the line is being executed, foo is pointing to {}
but when this statement is broken down to
foo.x = (foo = {n: b}) /a=(b=c)
foo's reference has changed from {}
to {n:b}
but foo
in foo.x
(a) is still pointing to old reference of foo
since left hand expression was evaluated before assignment had begun.
As per the spec
If LeftHandSideExpression is neither an ObjectLiteral nor an ArrayLiteral,
a. then Let lref be the result of evaluating LeftHandSideExpression.
Which means before the assignment foo.x
was still having reference to old foo
.
So, if you tweak your example a little bit by doing
var foo = {z:2};
foo.x = foo.n = {n: 1};
In this example, you didn't change the reference to foo
, only assigned new property, so the output now is
Object {z: 2, n: Object, x: Object}
Now, it has retained the reference to old foo
since a new reference was not assigned hence all properties z
, n
and x
are being retained.
It equals
let tmp = foo;
foo = {n: b};
tmp.x = foo;
You could see, that old foo
(stored in z
in this example) was modified:
> z=foo={};
{}
> foo.x = foo = {n: b};
{ n: 10 }
> foo
{ n: 10 }
> z
{ x: { n: 10 } }
I got it.
var foo = {}; // now foo is a reference point to object {}
foo.x = foo = {n:1}; // first foo is refer to a new object {n:1}, then old foo referred object {} set a prop x
// try this to get what you want
var foo = foo1 = {};
foo.x = foo = {n:1};
console.log(foo, foo1) // here foo1 is what you want
With:
foo.x = foo = {n: b};
The leading foo.x
is partially evaluated first, enough to determine the exact target for the assignment, before proceeding to actually assign it.
It behaves more along the lines of:
var oldFoo = foo;
foo = {n: b};
oldFoo.x = foo;
This is mentioned in the standard. The left side of the =
is evaluated (1.a) before the value is placed there (1.f):
AssignmentExpression : LeftHandSideExpression = AssignmentExpression
1) If LeftHandSideExpression is neither an ObjectLiteral nor an ArrayLiteral, then
a) Let lref be the result of evaluating LeftHandSideExpression.
...
f) Perform ? PutValue(lref, rval).