•Re: Elegant way to parse an optional prefix with default?

$name =~ s/^[$@%&<*]//;
my $sigil= $1 || '&';
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Actually, that doesn't work. You don't have a $1 there. If you added the parens to make it work, I still don't quite see how to get it into one line. Nor would I worry about that.

-- Randal L. Schwartz, Perl hacker
Be sure to read my standard disclaimer if this is a reply.

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Re: •Re: Elegant way to parse an optional prefix with default? by John M. Dlugosz (Monsignor) on Nov 08, 2002 at 05:32 UTC
Hmm, it looks like (with the parens, sorry about the typo) $1 is not undef but left to an old value if there is no match?! So I do need to put something on the first line to test for success failure, at least. (or update the regex to make $1 success but blank) My ugly one line is: `my $sigil= ($name =~ s/^([$@%&<*])//) ? $1 : '&';` [download] as I couldn't get it to work with `my ($sigil)= ...` in list context to return matches. I guess that doesn't work with replacements as opposed to matches. —John	[reply] [d/l] [select]

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Re: •Re: Elegant way to parse an optional prefix with default?
by John M. Dlugosz (Monsignor) on Nov 08, 2002 at 05:32 UTC

So I do need to put something on the first line to test for success failure, at least. (or update the regex to make $1 success but blank)

My ugly one line is:

 my $sigil= ($name =~ s/^([$@%&<*])//) ? $1  : '&';
[download]

my ($sigil)= ...

—John

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