&a(-beer=>none);
sub a
{
    print join("=>",@_);
}
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use strict;
&a(-beer=>'none');
sub a
{
    print join("=>",@_);
}
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Sorry to reply 4 years later. The - has nothing to do with the string or bareword. The => says qoute whats on my left and pretend i'm a comma. so a(-beer=>none) is the same as a("beer", none); Perhaps this is what the original author meant but I wasn't sure and thought i might clarify a bit.


___________
Eric Hodges

I assume that "is the same as a('beer', none)" is a typo, ... and that you meant "is the same as a('-beer', 'none')"

And since we are clarifying ... the '-' is part of the key (or the first argument, depending on how you look at it).

bash$ perl -w -Mstrict -MData::Dumper -e \
"print Dumper( { -name => 'foo', -value => 'bar' } );"
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Results:

$VAR1 = {
          '-name' => 'foo',
          '-value' => 'bar'
        };
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