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From: John-Paul Stewart <jpstewart@sympatico.ca>
Newsgroups: comp.lang.javascript
Subject: Re: Why does (a,b,c) return only c?
Date: Mon, 25 May 2015 13:58:39 -0400
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On 25/05/15 07:43 AM, Steven D'Aprano wrote:
> Consider:
>
> js> var x = (1,2,3);
> js> x
> 3
>
>
> I expected an error, or an array, or something other than just 3. What's the
> explanation for only the last value being returned? I presume it's a
> deliberate design feature.

Yes, it's a design feature.  See, for example:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comma_Operator

"The comma operator evaluates each of its operands (from left to right) 
and returns the value of the last operand."

> Under what circumstances would I use it?
>
> I came across this trying to swap two variables. This was my first attempt,
> but I'm afraid that I don't understand what is happening here:
>
> js> var a = 1;
> js> var b = 3;
> js> a, b = b, a;  // swap a and b around
> 1
> js> print(a, b);
> 1 3
>
>
> On the swap line, if the right hand side evaluates as only a single value, I
> expected any of these behaviours:
>
> (1) an error (can't assign two targets on the left to one on the right);
>
> (2) the "a," on the left gets dropped, and the "b," on the right gets
> dropped, so we end up with "b = a" only;
>
> (3) a gets assigned to undefined, and b to 1;
>
> (4) or a to 1 and b to undefined.
>
> Obviously all of my guesses are wrong. What is actually happening there?

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence

If you read up on operator precedence, you'll see the comma has the 
lowest precedence of all operators.  So your attempt at swapping is 
actually parsed as 3 statements:  "a", "b = b", "a".  That should make 
it clear why the values don't change and why there's no error.