Neither result is consistent with my understanding of the 5.22 documentation. 
    while (<>) {
        ...                     # code for each line
    }

is equivalent to the following Perl-like pseudo code:

    unshift(@ARGV, '-') unless @ARGV;
    while ($ARGV = shift) {
        open(ARGV, $ARGV);
        while (<ARGV>) {
            ...         # code for each line
        }
    }

[download]

I would expect the open to fail silently.(There is no explicit test for errors.) The read on the unopened filehandle should return a false value. The loop would terminate without ever executing the print statement.

Bill

In reply to Re^2: while(<>) in perl 5.22 by BillKSmith
in thread while(<>) in perl 5.22 by xorl

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