Re: Programmatic regex disjunction

This should work:

use re "eval";
my @pats = ( qr/one/, qr/two/, qr/three/ );
my $joined = "(??{" . join("})|(??{", map "\$pats[$_]", 0..$#pats) . "
+})";
my $any    = qr/$joined/;
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Comment on Re: Programmatic regex disjunction Download Code

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Re^2: Programmatic regex disjunction by Aristotle (Chancellor) on Nov 22, 2005 at 09:01 UTC
I really loathe the “join with right+left delimiter” idiom for paired delimiters. Since you don’t need a map for this, let me demonstrate that extreme to show why I don’t like it: `my $joined = "(??{ \$pats[" . join( "] })\|(??{ \$pats[", 0 .. $#pats ) + . "] })";` [download] Uhm, oh dear. Instead, I’d always prefer a map to put the delimiters in place before joining – it’s clearer, with less repetition of literals to boot. You’ve recognised the need for a map to clarify anyway – so take it to conclusion: `my $joined = join '\|', map "(??{ \$pats[$_] })", 0 .. $#pats;` [download] I think that looks a lot friendlier (than your code; to say nothing of my demonstration of atrocity). Makeshifts last the longest.	[reply] [d/l] [select]

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Re^2: Programmatic regex disjunction
by Aristotle (Chancellor) on Nov 22, 2005 at 09:01 UTC

I really loathe the “join with right+left delimiter” idiom for paired delimiters. Since you don’t need a map for this, let me demonstrate that extreme to show why I don’t like it:

my $joined = "(??{ \$pats[" . join( "] })|(??{ \$pats[", 0 .. $#pats )
+ . "] })";
[download]

Uhm, oh dear.

Instead, I’d always prefer a map to put the delimiters in place before joining – it’s clearer, with less repetition of literals to boot. You’ve recognised the need for a map to clarify anyway – so take it to conclusion:

my $joined = join '|', map "(??{ \$pats[$_] })", 0 .. $#pats;
[download]

I think that looks a lot friendlier (than your code; to say nothing of my demonstration of atrocity).

Makeshifts last the longest.

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