Why won't $' work in a trinary operation?

satchboost has asked for the wisdom of the Perl Monks concerning the following question:

Yes, having read all the wonderful stuff up above, I know I shouldn't be using $` and $'. However, I don't have much of a choice because where I work still hasn't upgraded to 5.60. (I do have a request in to ITS, though...) So, I've got the following code:

my ($foo, $bar) = (123, 456);
my $base_string = "#200";
$base_string =~ /^#/ ? $foo = $'
                     : $bar = $base_string;
print "$foo $bar\n";
[download]

This prints out "#200 456" when it should be printing "200 456". Now, I noticed that if I split the trinary up into a if-else, it worked just as expected. Why would that be?

Comment on Why won't $' work in a trinary operation? Download Code

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Re: Why won't $' work in a trinary operation? by japhy (Canon) on Feb 05, 2001 at 20:53 UTC
The problem is that your `?:` line is parsed by Perl as: `($base_string =~ /^#/ ? ($foo = $') : $bar) = $base_string;` [download] Add parens to your expression to get the operations to occur in the right manner: `$base_string =~ /^#/ ? ($foo = $') : ($bar = $base_string);` [download]	[reply] [d/l] [select]
Re: Why won't $' work in a trinary operation? by chipmunk (Parson) on Feb 05, 2001 at 21:03 UTC
This gotcha is explained in the standard documentation for the conditional operator in perlop: `Because this operator produces an assignable result, using assignments without parentheses will get you in trouble. For example, this: $a % 2 ? $a += 10 : $a += 2 Really means this: (($a % 2) ? ($a += 10) : $a) += 2 Rather than this: ($a % 2) ? ($a += 10) : ($a += 2)` [download]	[reply] [d/l]
Re: Why won't $' work in a trinary operation? by Fastolfe (Vicar) on Feb 06, 2001 at 00:39 UTC
To avoid this confusion, never use the trinary operator in a void context (when you have no intention of using the value returned). It's usually clearer to write it out, and avoids messy precedence issues like this. Remember that `?:` is built so that it can be used in expressions as a replacement for a single value in a complex operation. If all you're using it for is to control execution (make one assignment versus another), use a normal if/else construct: `if ($base_string =~ /^#/) { $foo = $'; } else { $bar = $base_string; }` [download] Arguably using `/^#(.*)/` and `$1` instead of `$'` is going to be a bit more efficient. Generally it's best to shy away from $` and `$'`, as these add costly operations to an otherwise straightforward regex match.	[reply] [d/l] [select]
(tye)Re: Why won't $' work in a trinary operation? by tye (Sage) on Feb 06, 2001 at 03:31 UTC
I have a different rule for avoiding this problem: Whenever you do an assignment in an expression, put parens around the assignment. There are lots of things beside ?: that will bite you when you put an assignment inside of an expression, for example, ($new=$old) =~ s/x/y/g; - tye (but my friends call me "Tye")	[reply]
Re (tilly) 1: Why won't $' work in a trinary operation? by tilly (Archbishop) on Feb 06, 2001 at 03:58 UTC
What version of Perl are you running that you have to use $'? Explicit backreferences have worked for ages. Just use a pattern like /^#(\d+)/ (modify to taste) and assign $1. Using $' or $` anywhere in your program slows down every RE possibly dramatically. So Don't Do That.	[reply]