in reply to regexp word break help

Here's one way to do it: ($str) = $str =~ /(.{1,45})\b/s;

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Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a (from-home) job
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

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Re: Re: regexp word break help
by George_Sherston (Vicar) on May 01, 2002 at 11:42 UTC
    I really like that, partly because I don't understand quite why it works. Without the brackets $str would be set to "1", which I take to be the return value of the successful pattern matching.

    But why do the brackets stop this happening? I should have thought that $str and ($str) wd behave the same way.

    And then why do they not perform this function in ($str) = $str =~ s/(.{1,45})\b/foo/s;, in which $str gets set to "1"?

    /me backs respectfully away from the shrine...

    § George Sherston
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      I should have thought that $str and ($str) wd behave the same way.

      This node explains the difference between list context and scalar context very well. Many of the replies have great links to some interesting articles on the subject.

      And then why do they not perform this function in ($str) = $str =~ s/(.{1,45})\b/foo/s;, in which $str gets set to "1"?

      The perlop manpage explains the return values of the m/// and s/// operators in list and scalar context. The s/// operator returns the number of substitutions in either context.

      buckaduck

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