http://qs1969.pair.com?node_id=561607

I think I understand what is going on in the first of your updated solutions but if I change it to read

my @nums = (\$x =~ /\s(\S+)/)[1] =~ /(\d+)/g;

expecting output of

3 33

it doesn't work unless I also make the first match global like this

my @nums = (\$x =~ /\s(\S+)/g)[1] =~ /(\d+)/g;

I think this is because the round brackets around the match put the match into list context and the [0] subscript grabs the first elements of the match; however, since the match is non-global there will only ever be one element in the list and trying to get more will not work. If we want a second or subsequent element we must make the match global to capture more than one element.

Have I understood this correctly or am I completely missing the point?

Cheers,

JohnGG

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Re^3: regex for multiple capture within boundary
by ikegami (Patriarch) on Jul 17, 2006 at 01:28 UTC

That's exactly it (although there could be 0 elements if the match fails).

my @nums = (\$x =~ /\s(\S+)/)[0] =~ /(\d+)/g;
could also be written as
my @nums = (\$x =~ /\s(\S+)/ ? \$1 : undef) =~ /(\d+)/g;

If you're going to use /g, drop the \s:

# @nums = ('3', '33'); my \$word = 2; my @nums = (\$x =~ /(\S+)/g)[\$word] =~ /(\d+)/g;

or use split:

# @nums = ('3', '33'); my \$word = 2; my @nums = (split(' ', \$x))[\$word] =~ /(\d+)/g;