Hi,

Juerd wrote:
>  Ingo Blechschmidt skribis 2005-06-15 20:18 (+0200):
> > >>     say join ",", @words;     # "hi,my,name,is,ingo";
> > > Following the logic that .words returns the words, the words are no
> > > longer individual words when joined on comma instead of
> > > whitespace...
> > sorry, I don't quite get that.
>  
>      "foo bar baz".words.join(',').words.join(':') ne 'foo:bar:baz';
>  
>  so somewhere in that process, foo, bar and baz managed to no longer be
>  words by the definition of words used by an on whitespace splitting
>  words method - they're one word, together, when they're joined on
>  comma.

ah! I understand :)

So maybe we should allow words() (or however we'll end up calling it) to
take an optional parameter specifying what's considered a wordchar,
with a default of rx/\w+/:

  say "foo bar baz".words()           .join(":");    # same as
  say "foo bar baz".words(rx/\w+/)    .join(":");    # "foo:bar:baz"

  say "foo,bar,baz".words()           .join(":");    # same as
  # "," doesn't match /\w+/
  say "foo,bar,baz".words(rx/\w+/)    .join(":");    # "foo:bar:baz"

  # Now "," is considered to be part of words:
  say "foo,bar,baz".words(rx/[\w|,]+/).join(":");    # "foo,bar,baz"

  say "foo bar baz".words(rx/b../)    .join(":");    # "bar:baz"

Then your example...
  say "foo bar baz".words.join(',').words.join(':'); # "foo bar baz";

And:
  say "foo bar baz".words.join(",").words(rx/[\w|,]+/).join(":");
     # "foo,bar,baz"


I hope these examples make sense...

--Ingo

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