Javascript object initialization and evaluation order [duplicate]

Question

If I write

var a = [1,2];
var b = {
  foo: a.pop(),
  bar: a.pop()
};

What is the value of b, according to the specification?

(By experiment, it's {foo: 2, bar: 1}, but I worry whether this is implementation-specific.)

Why would it be implementation-specific? pop mutates the original array. — elclanrs
– elclanrs, Commented Jul 3, 2013 at 0:37
@elclanrs, I think what the question is getting at is whether the values of the properties of object b will always be evaluated in the order in which they are listed in the code. — go-oleg
– go-oleg, Commented Jul 3, 2013 at 0:42
Oh, I see... Now that's actually a good question, but while the order of properties in an object is trivial, the order of evaluation is not, so I'd say this is non-issue but I may be wrong... — elclanrs
– elclanrs, Commented Jul 3, 2013 at 0:43

hopper · Accepted Answer · 2013-10-11 14:55:47Z

12

See ECMAScript section 11.1.5 defining how the ObjectLiteral production is parsed.

In particular:

PropertyNameAndValueList , PropertyName : AssignmentExpression is evaluated as follows:

Evaluate PropertyNameAndValueList.

Evaluate PropertyName.

Evaluate AssignmentExpression.

...

Where (1) is a recursive definition.

This means the leftmost item in an object literal will get evaluated first, and so {foo: 2, bar: 1} is indeed spec-mandated.

edited Oct 11, 2013 at 14:55

hopper

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answered Jul 3, 2013 at 1:08

bobince

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Comments

user2540812 · Accepted Answer · 2013-07-03 00:53:02Z

1

They are evaluated in the order they are written.

answered Jul 3, 2013 at 0:53

user2540812

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