Bart asked:
> So to what does "123foo" evaluate in numeric context? Now, it produces
> 123. Which is probably useful...
Yes. And I would expect that it will continue to do so.
Perl is primarily about convenience, not consistency ;-)
Damian
On Sun, 7 Oct 2001 12:27:17 +1000 (EST), Damian Conway wrote:
>The step you're missing is that the non-numeric string "hello",
>when evaluated in a numeric context, produces NaN. So:
>
> "hello" == 0 && 0 != NaN
>
>is:
>
> Nan == 0 && 0 != NaN
>
>which is false.
So to what does "
On 10/6/01 10:27 PM, Damian Conway wrote:
>> Doesn't that mean:
>>
>> "hello" == 0 && 0 != NaN
>>
>> will evaluate to true?
>
> No. The step you're missing is that the non-numeric string "hello",
> when evaluated in a numeric context, produces NaN. So:
>
>"hello" == 0 && 0 != NaN
> $a == $b != NaN
>
> really means this:
>
> $a == $b && $b != NaN
>
> But "$a == $b != NaN" is supposed to "[solve] the problem of numerical
> comparisons between non-numeric strings." Well, what if:
>
> $a = 'hello';
> $b = 0;
>
> Do
Okay, so this:
100 < -s $filepath <= 1e6
really means this:
100 < -s $filepath && -s $filepath <= 1e6
which means that this:
$a == $b != NaN
really means this:
$a == $b && $b != NaN
But "$a == $b != NaN" is supposed to "[solve] the problem of numerical
comparisons between
