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Re: falsy and truthy

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tlbaxter99@yahoo.com wrote:

> I thought I understood falsy and truthy until I ran this code:
> 
> var falsyValues = [false, 0, null, "", undefined, NaN];
> 
> for (var index in falsyValues)
>    if (!falsyValues[index])
>       console.log("FALSE");
>    else
>       console.log("TRUE");
> 
> As I expected, "FALSE" was printed for all 6 values. No problem.
> 
> Now, when I modify the if() statement to this:
>    if (false == falsyValues[index])
> 
> I see that null, undefined and NaN do NOT compare as being equal to false,
> yet when using the not operator(!) a shown earlier, they ARE considered
> false.
> 
> Can someone explain this? Using the not operator (!) is apparently not the
> same as comparing directly against false.

That is true :)  The “!” operator internally converts its operand, and the 
“if” statement internally converts its expression to a primitive Boolean 
value before evaluating that.  The “==” operator acts as type-converting 
operator as well, but follows different, much more complex rules.

See the ECMAScript Language Specification, 5.1 Edition, §§12.5 (“The if 
Statement”), 11.4.9 (“Logical NOT Operator ( ! )”), and 11.9.1 (“The Equals 
Operator ( == )”) for details.  (AFAIK this has not changed since the first 
Edition.)

Get a real name.  Do not use Google Groups for posting, at least not until 
they fixed the many bugs.

-- 
PointedEars

Twitter: @PointedEars2
Please do not Cc: me. / Bitte keine Kopien per E-Mail. 

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falsy and truthy tlbaxter99@yahoo.com - 2013-07-05 20:01 -0700
  Re: falsy and truthy Denis McMahon <denismfmcmahon@gmail.com> - 2013-07-06 07:20 +0000
  Re: falsy and truthy Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2013-07-06 10:24 +0200
    Re: falsy and truthy "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2013-07-06 11:22 +0200
  Re: falsy and truthy "Joao Rodrigues (J.R.)" <groups_jr-1@yahoo.com.br> - 2013-08-10 09:03 -0300

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