Re^6: Perl precedence details

in reply to Re^5: Perl precedence details
in thread Perl precedence details

Yes, I guess you are right.

Anyway, my search and experiments gives me answers:

Yes, if in compile-time perl knows that sub has ($) prototype - it's being treated as a named unary operator.
Seems there is no any list of named unary built-ins, except I found in the book
Optional arguments after first one makes sub a list operator.

Generally it what i wanted

Comment on Re^6: Perl precedence details

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Re^7: Perl precedence details by Anonymous Monk on Jun 20, 2015 at 18:15 UTC
Seems there is no any list of named unary built-ins, except I found in the book Sure there is :p its perlfunc, anything which takes at most one arg is unary http://perl5.git.perl.org/perl.git/blob/HEAD:/t/op/cproto.t anything with "($)" or "(_)" or "(;$)" or "(_;$)" or "(;)" or "()" or "(;+)" is unary http://perl5.git.perl.org/perl.git/blob?f=ext/Pod-Functions/Functions_pm.PL parses perlfunc.pod ... might be able to get the info if its there Yeah, it would be nice if there was a hash just saying this prototype stuff	[reply]
Re^8: Perl precedence details by hurricup (Pilgrim) on Jun 20, 2015 at 19:25 UTC
Thank you!!!	[reply]
Re^8: Perl precedence details by hurricup (Pilgrim) on Jun 20, 2015 at 19:34 UTC
Crap, some unaries has undef proto, like return. But anyway, this is really helpful.	[reply]
Re^9: Perl precedence details by Anonymous Monk on Jun 20, 2015 at 22:37 UTC
return isn't unary, it can take zero or more args ... I only see defined/delete/exists goto/last/redo, that are "unary", the rest like chomp/chop.. can all take more than one arg	[reply]
Re^8: Perl precedence details by hurricup (Pilgrim) on Jun 21, 2015 at 06:54 UTC
And (_;$) is not unary, two arguments	[reply]

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