Re^2: Why do I need parentheses here?

func ?PATTERN?; is no different than func $x+1;. There's nothing special (or odd) about it.

I think I got it. So func ?PAT? is just another way of writing

func($_ =~ ?PAT?)
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--
Ronald Fischer <ynnor@mm.st>

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Re^3: Why do I need parentheses here? by ikegami (Patriarch) on May 27, 2009 at 14:17 UTC
Yes. `?PAT?` and `$_ =~ ?PAT?` are equivalent, just like `/PAT/` and `$_ =~ /PAT/` are. (Aforementioned `split` is exceptional since it treats `/PAT/` as `qr/PAT/` or similar.)	[reply] [d/l] [select]
Re^4: Why do I need parentheses here? by rovf (Priest) on May 27, 2009 at 14:57 UTC
Is there any difference between `/PAT/` and `qr/PAT/` ??? I thought this is always the same. -- Ronald Fischer <ynnor@mm.st>	[reply] [d/l] [select]
Re^5: Why do I need parentheses here? by ikegami (Patriarch) on May 27, 2009 at 15:01 UTC
`/PAT/` (`m/PAT/`) is the match operator. `qr/PAT/` compiles the pattern and returns the pattern without doing any matching. perlop `$ perl -le'$_="abc"; print /(.)/;' a $ perl -le'$_="abc"; print qr/(.)/;' (?-xism:(.))` [download]	[reply] [d/l] [select]
Re^6: Why do I need parentheses here? by rovf (Priest) on May 27, 2009 at 15:08 UTC
Re^3: Why do I need parentheses here? by Anonymous Monk on May 27, 2009 at 14:01 UTC
Checking ;) `C:\>perl -MO=Deparse,-p -e"sub foo{} foo ?bar? sub foo { } foo(?bar?); -e syntax OK` [download] Yes	[reply] [d/l]