Re: true false ? : expression thingy

Bigus <someone@xxxxxxxxxxxxx> wrote in comp.lang.perl.misc:
> I'm not sure what this method is called, which is perhaps why I've had no
> luck when searching around the web for info on it, but when you want a
> 1-line alternative to an if...else.. construct, you can use the following
> syntax:
>
> $ctr > 3 ? $text = "yes" : $text = "no";
>
> that works fine, however, when I try this:
>
> my $incsubs;
> $line =~ /\ts/i ? $incsubs = 0 : $incsubs = 1;
>
> it doesn't work.. well, $incsubs equals 1 regardless whereas it works in the
> standard if...else.. construct.
>
> I guess this is because the regular expression doesn't return true/false, or
> sth like that.. is there a "trick" you can use to make this type of method
> work with regexps?

Wrong diagnosis. A regex returns a boolean value in scalar context,
that part of your expression is quite all right.

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.

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;

which is what I would use.

Anno
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