Re: true false ? : expression thingy

"Anno Siegel" <anno4000@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:dcaenn$d40$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Bigus <someone@xxxxxxxxxxxxx> wrote in comp.lang.perl.misc:
[..]>> my $incsubs;
>> $line =~ /\ts/i ? $incsubs = 0 : $incsubs = 1;
>>
[..]
> Your problem is one of precedence. Perl parses your expression like
> this:
>
> ( ( $line =~ /\ts/i) ? ( $incsubs = 0) : $incsubs) = 1;
>
> So if you have a match, $incsubs is set to 0 and $incsubs is returned.
> If you don't have a match, $incsubs is not set to 0, but returned as is.
> In either case what is returned is $incsubs, and 1 is assigned to it
> unconditionally. Hence the result you see.

Ahhh, so this would work:

my $incsubs;
$line =~ /\ts/i ? ($incsubs = 0) : ($incsubs = 1);

> Generally, the "?:" construct (sometimes called "ternary conditional",
> btw) should not be used for operations that have a side effect, like
> the assignments you are doing. Use a normal if/else for that, even if
> it's a bit longer.
>
> Using "?:", your conditional assignment can be written:
>
> my $incsubs = $line =~ /\ts/i ? 0 : 1;
>
> Now the assignment is outside the "?:". But now it becomes apparent
> that this is (almost) equivalent to
>
> my $incsubs = $line !~ /\ts/i;

OK, that sounds good. I 'll use that in this case. The previous syntax would
be useful in situations where I want to make a non boolean assigment
resulting from a regexp match.

Thanks

Bigus


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