- --- In jslint_com@yahoogroups.com, "pauanyu" <pcxunlimited@...> wrote:
> return this[prop] = (this[prop] === val1) ? val2 :

The thing that JSLint did not like was the use of an assignment as an expression. It will accept

> (this[prop] === val2) ? val1 : this[prop];

>

>

> I think you'd be hard-pressed to say the ?: version is less elegant and readable than the first two. There's just one problem: it uses nested ?: operators, so JSLint doesn't like it.

this[prop] = (this[prop] === val1) ? val2 :

(this[prop] === val2) ? val1 : this[prop];

return this[prop];

or

return (this[prop] = (this[prop] === val1) ? val2 :

(this[prop] === val2) ? val1 : this[pro - --- In jslint_com@yahoogroups.com, "Douglas Crockford" <douglas@...> wrote:
>

Oh, I see! So this works:

> The thing that JSLint did not like was the use of an assignment as an expression. It will accept

>

> this[prop] = (this[prop] === val1) ? val2 :

> (this[prop] === val2) ? val1 : this[prop];

> return this[prop];

>

> or

>

> return (this[prop] = (this[prop] === val1) ? val2 :

> (this[prop] === val2) ? val1 : this[pro

>

return (this[prop] = (this[prop] === val1) ? val2 :

(this[prop] === val2) ? val1 : this[prop]);

Thanks for your help. Nice to know that JSLint handles nested ?: operators. - --- In jslint_com@yahoogroups.com, Michael Lorton <mlorton@...> wrote:
>

I did a little tinkering, and here's what I came up with:

> Well, val1 and val2 are the only possible values, you could do this:

>

> var next = {};

> next[val1]=val2;

> next[val2]=val1;

> return this[prop] = next[this[prop]] ;

>

> If the function is to be executed over and over, the first three statement are only needed at initialization time. Plus, this algorithm generalizes to any fixed, repeating sequence of values. Say you wanted to loop through 1, 4, 9 over and over (starting at 1 if the property is not already in the sequence):

>

> return this[prop] = { 1: 4, 4: 9, 9: 1} [this[prop]] || 1;

>

> Was this the question or did you want to talk about the ternary operator?

>

> M.

>

return (this[prop] = ({ val1: val2, val2: val1 })[this[prop]]);

What a fantastic construct. Create an anonymous object, then use it immediately! Unfortunately this doesn't work, because val1 and val2 are variables that are determined at run-time, so this will fail.

Nonetheless, you have my gratitude for pointing out a wonderful construct that I plan to use more often in my code. For now, I'll stick with the ?: operator. - Yeah, I did exactly that first too, but { val1 : val2 } is equivalent to { "val1" : val2 }. The braces notation only works if the sequence is expressed as constants. Of course, most sequences you run across express an obvious relationship and it's better to encode the relationship. If you want true-false-true-false, return (this[prop] = !this[prop]); is much more sensible than return (this[prop] = { true: false, false: true}this[prop]).

Not terrible on topic, but how many people know about the !! operator? It converts truthy and falsy values to actual booleans. So !!1yields true and !!0yields false.

OK, I'm lying, there's no !! operator -- it's just the ! operator twice, but it does work.

M.

________________________________

From: pauanyu <pcxunlimited@...>

To: jslint_com@yahoogroups.com

Sent: Thursday, July 9, 2009 9:38:16 AM

Subject: Re: [jslint] Nested ?: operator values.

--- In jslint_com@yahoogroups.com, Michael Lorton <mlorton@...> wrote:

>

> Well, val1 and val2 are the only possible values, you could do this:

>

> var next = {};

> next[val1]=val2;

> next[val2]=val1;

> return this[prop] = next[this[prop]] ;

>

> If the function is to be executed over and over, the first three statement are only needed at initialization time. Plus, this algorithm generalizes to any fixed, repeating sequence of values. Say you wanted to loop through 1, 4, 9 over and over (starting at 1 if the property is not already in the sequence):

>

> return this[prop] = { 1: 4, 4: 9, 9: 1} [this[prop]] || 1;

>

> Was this the question or did you want to talk about the ternary operator?

>

> M.

>

I did a little tinkering, and here's what I came up with:

return (this[prop] = ({ val1: val2, val2: val1 })[this[prop]]);

What a fantastic construct. Create an anonymous object, then use it immediately! Unfortunately this doesn't work, because val1 and val2 are variables that are determined at run-time, so this will fail.

Nonetheless, you have my gratitude for pointing out a wonderful construct that I plan to use more often in my code. For now, I'll stick with the ?: operator.

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