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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