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Re^2: Why does eof have a prototype?

by Anonymous Monk
on Aug 08, 2022 at 19:04 UTC ( #11146036=note: print w/replies, xml ) Need Help??


in reply to Re: Why does eof have a prototype?
in thread Why does eof have a prototype?

the prototype was inadequate

how so?

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Re^3: Why does eof have a prototype?
by ikegami (Patriarch) on Aug 09, 2022 at 13:11 UTC

    That certain operators (eof, open, etc.) have prototypes that aren't adequate for representing the syntax of the operator is the premise of the question.

    For example,

    • An overridden eof operator can't distinguish whether it was was called as eof or eof() while the real operator does.

    • open( STDERR , ">&=", STDOUT ) is perfectly valid with the real operator, but throws a strict error with an overridden open.

      $ perl <<'.' use strict; open( STDERR , ">&=", STDOUT ); BEGIN { *CORE::GLOBAL::open = sub (*;$@) { } } open( STDERR , ">&=", STDOUT ); . Bareword "STDOUT" not allowed while "strict subs" in use at - line 4. Execution of - aborted due to compilation errors.

    Upd: Fixed accidental double negation.

      Filehandles are still too voodoo for me, I hoped something like this could fix that problem...

      use constant STDERR => \*STDERR; use constant STDOUT => \*STDOUT;

      Cheers Rolf
      (addicted to the Perl Programming Language :)
      Wikisyntax for the Monastery

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