in reply to The mystery of $->[1]

G'day hurricup,

++ Good catch!

The code itself parses as legal Perl:

$ perl -MO=Deparse,-p -e 'my $l = ref $a eq "ARRAY" ? ($a->[0] || $->[ +1]) : $a;' (my $l = ((ref($a) eq 'ARRAY') ? ($a->[0] || ($- > [1])) : $a)); -e syntax OK

And, as you can see, your analysis is correct: '$-' is being compared with '[1]'.

There's actually two adjacent lines like that in the source code:

# Deal with the case of an item actually being an array ref to 1 o +r 2 # hashrefs. Don't assign to $a or $b, as they're aliases to the or +ignal my $l = ref $a eq 'ARRAY' ? ($a->[0] || $->[1]) : $a; my $r = ref $b eq 'ARRAY' ? ($b->[0] || $->[1]) : $b;

I suspect those should be:

my $l = ref $a eq 'ARRAY' ? ($a->[0] || $a->[1]) : $a; my $r = ref $b eq 'ARRAY' ? ($b->[0] || $b->[1]) : $b;

although, I haven't delved deeply enough into the code to say for certain.

I suggest you raise a bug report for this.

-- Ken

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Re^2: The mystery of $->[1]
by hurricup (Pilgrim) on Jul 11, 2015 at 21:05 UTC

    Thought about typo, but last my "discoveries" shown that I'm John Snow in Perl :) And thought it may be some unknown magic.

    Will report a bug.