in reply to Re: Re: spaces in filenames
in thread spaces in filenames

dvergin said:

A scalar value in a variable will be interpreted as a pattern also.

I beg to differ. Take a look at the code below:

bash-2.05a$ perl -e 'print scalar(split("\W", "one two three")), "\n";'
1
bash-2.05a$ perl -e 'print scalar(split("\\W", "one two three")), "\n";'
3
bash-2.05a$ perl -e 'print scalar(split(/\W/, "one two three")), "\n";'
3

As you see, both versions produce a different result. My interpretation of your quote to The Camel (and to the fact that the original code compiles at all), is that split() is using the literal string as exactly that, a literal string that must occur verbatim.

This might have changed at some time, though. I have vague recollections of using a literal with split(), but that was way back I guess. I am running 5.8.0 here, BTW.

++ for dvergin for researching his answer :)

Update: As pointed out by MarkM, the literal string is indeed interpreted as a regular expression. It's just that \ must be escaped in my example, so my conclusion turned out to be wrong. ++ to MarkM for showing me the light...

Best regards

-lem, but some call me fokat

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Re: Re: Re: Re: spaces in filenames
by MarkM (Curate) on Feb 03, 2003 at 03:46 UTC

    You are correct that "\W" produces a different result from /\W/ as a first argument to split(). You are not correct in concluding that this means split() is assuming that a quoted string is a literal. Examples:

    $ perl -e '$t = "abc,def"; @t = split("\\W", $t); print "@t\n"'
abc def

    [download]
    $ perl -e '$a = "abc,def"; @a = split("[^a-z]", $a); print "@a\n"'
abc def

    [download]

    In both cases, the string is being interpretted as a regular expression. As a string, a different set of interpolation rules apply.

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