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Re: Perl Idioms Explained - @ary = $str =~ m/(stuff)/g

by bart (Canon)
on Sep 17, 2003 at 10:15 UTC ( [id://292089]=note: print w/replies, xml ) Need Help??


in reply to Perl Idioms Explained - @ary = $str =~ m/(stuff)/g

$num_matches_stuff = $str =~ m/(stuff)/g; @all_the_matches = $str =~ m/(stuff)/g;
In scalar context we get the count of the matches, in array context we get the matches.
No, the former just is not right. You haven't actually tried it, have you?
$str = "stuff stuff stuff"; $num_matches_stuff = $str =~ m/(stuff)/g; print $num_matches_stuff;
Result:
1

The /g modifier in scalar context is very special. It is intended to be used in a loop, something like this:

$str = "stuff stuff stuff"; while($str =~ m/(stuff)/g) { print "Got one!\n"; }
Result:
Got one!
Got one!
Got one!
So in scalar context, it will match at most once at a time — next time around, it'll continue where it left off last time.

Therefore, the returned valued of //g when used in scalar context is either 0, or 1.

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Re: Re: Perl Idioms Explained - @ary = $str =~ m/(stuff)/g
by tachyon (Chancellor) on Sep 17, 2003 at 13:15 UTC

    Hmm it is of course as you say. I must have got myself confused with the behaviour of s///.

    $data = 'stuff stuff stuff'; $num = $data =~ s/(stuff)//g; print $num, $/; __DATA__ 3

    You can force it to do as I said by cheating thusly....

    $num = () = 'stuff stuff stuff' =~ m/(stuff)/g; print $num; __DATA__ 3

    cheers

    tachyon

    s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

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