I had been through array destructuring syntax, which is well understood.

What exactly are we doing below, when we say var {p, q} = o;?

Is p and q in var {p, q} different from properties of o i.e., 'p' and 'q'? If yes,

why var {a, b} = o; does not work?

> var o = {p: 42, q: true};
    undefined
> p
    ReferenceError: p is not defined
> q
    ReferenceError: q is not defined
> o['p']
    42
> o['q']
    true
> var {p, q} = o;
    undefined
> p
    42
> q
    true
> var {a, b} = o;
    undefined
> a
    undefined
> b
    undefined

*Note: I learnt that, dictionary keys are string literals in javascript.*

I had been through array destructuring syntax, which is well understood.

What exactly are we doing below, when we say var {p, q} = o;?

Is p and q in var {p, q} different from properties of o i.e., 'p' and 'q'? If yes,

why var {a, b} = o; does not work?

> var o = {p: 42, q: true};
    undefined
> p
    ReferenceError: p is not defined
> q
    ReferenceError: q is not defined
> o['p']
    42
> o['q']
    true
> var {p, q} = o;
    undefined
> p
    42
> q
    true
> var {a, b} = o;
    undefined
> a
    undefined
> b
    undefined

*Note: I learnt that, dictionary keys are string literals in javascript.*

• javascript
• ecmascript-6
• destructuring

• var {p, q} = o; === var p = o.p; and var q = o.q; this will only work when the variables have the same name. – Patsy Issa Commented Aug 10, 2015 at 9:10
• @Kitler Is this not a peculiar concept? Which programming language concept is derived to provide such facility? Is there any other language that does such weird stuff? I think destructuring is inspried from python – overexchange Commented Aug 10, 2015 at 9:13
• 2 Weird is very subjective, it is a mon pattern to assign values to attributes from an object where the name matches, this concept is taken from coffeescript iirc. – Patsy Issa Commented Aug 10, 2015 at 9:14
• @Kitler variables have same name? stackoverflow./questions/31915621/… – overexchange Commented Aug 10, 2015 at 9:22
• 3 The same name as the key on the object... It feels like you aren't even trying tbh, give learn es 2015 a read. – Patsy Issa Commented Aug 10, 2015 at 9:24

2 Answers 2

    var o = {p: 42, q: true};
     var {p, q} = o;

Here, var {p,q} = o is just a shorthand for var {p:p , q:q} = o

Consider this.

      var o = { key : "value" };
      var { key : local_var } = o ;
      local_var === o["key"] // true

If you omit the local_var, and write var {key} = o; a new variable key will be created with the idenifier "key"., same like doing var key = o["key"]

So in your example that's like doing

      var p =  o["p"] ;  //42
       var q = o["q"];   //true
       var a = o["a"];  // undefined
       var b = o["b"];   //undefined

This may not be exactly true, but should help you understand it.
It's kind of something like Pattern Matching that other languages provide, but it's different.

That's because using your syntax for object destructing, the names used in the LHS expression ({a, b}) are used as keys into the RHS expression (o). Since a and b are not properties of o, that fails (returns undefined).

The relevant section in the spec for this is in Runtime Semantics: DestructingAssignmentEvaluation under AssignmentProperty : IdentifierReference Initializer (2nd from last). AssignmentProperty are your a (and b... separately), and. The StringValue of a is 'a' and that is used as a key to get a value from o (equivalent o['a'] in this case).

It would work if you'd do:

var {p:a, q:b} = o;

which uses AssignmentProperty : PropertyName : AssignmentElement (last entry) which uses a propery name (p) AND an assignment element (a).