I have a question regarding variable assignment
var foo = {n: 1};
foo.x = foo = {n: 2};
console.log(foo.x) // undefined
Why foo.x is undefined ?
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if you assign 5 to x and y why will it be threeRitesh Karwa– Ritesh Karwa2015-05-19 17:47:44 +00:00Commented May 19, 2015 at 17:47
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sorry, @RiteshK i removed thatvimal1083– vimal10832015-05-19 17:48:58 +00:00Commented May 19, 2015 at 17:48
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If foo is an object, and foo.x is a reference to foo... so foo === foo.x === foo.x.x... You just can't do that ^^Arthur– Arthur2015-05-19 17:49:23 +00:00Commented May 19, 2015 at 17:49
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It has to do with variable declarationRitesh Karwa– Ritesh Karwa2015-05-19 17:53:04 +00:00Commented May 19, 2015 at 17:53
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1RiteshK it's not the question. you can do it on 2 lines instead of one and it's working fineArthur– Arthur2015-05-19 17:55:51 +00:00Commented May 19, 2015 at 17:55
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1 Answer
The = assignment operator is right-associative, meaning that a chain of assignments is evaluated right to left. That is, the language treats this
x = y = x+2;
as if it were written
x = (y = x+2);
So first y is assigned the result of the addition (5), and then x is assigned the value of that assignment, which is also 5.
The first example in your question is the same story, but things are a little more complicated:
foo.x = foo = { n: 2 };
The first assignment that's made is to foo, of that object literal. However, before that point, the language will have determined the reference value of foo.x, which is a reference to a (not-yet existing) property on the original foo object. Thus, when the value of the right-hand assignment is made to that property, it does work, but the property is set on the old object, not the new one!
Let's change the example:
var foo = {n: 1}, foo2 = foo;
foo.x = foo = {n: 2};
console.log(foo2.x); // { n: 2 }
That version preserves another reference to the original value of foo in the variable foo2. Even though foo is overwritten in the double assignment, foo2 will continue to refer to the original foo.
4 Comments
Ritesh Karwa
This question has been removed
Pointy
@RiteshK yes - see the answer update, and I'm adding an example
Arthur
This is why
foo.x = foo = {a:1}; is different to foo = {a:1}; foo.x = foo;. In the first case the reference isn't create, but on second yes.
Praveen
@Pointy is foo.x = foo = {n: 2}; breaks down to foo.x = {n: 2}; foo = {n: 2}; ?