How would one get the LHS (in the above case var) so I can use it in whatever implementation I decide to use?

If you want a real infix in operator, you will have to hack the perl parser, namely perly.y, toke.c and related files.

You could overload an existing infix operator with a in subroutine, but in your programs you would still use the overloaded operator (not in), which then just behaves as in, and your $var has to be blessed to that end.

There's a notation which at least looks similar to infix - method call: 

$var->in(@list)

[download]

- but $var has to be blessed here, too, into the package into which the in method has been compiled:

package in {
    sub new { bless \$_[1],$_[0] }
    sub in  { my $t = shift; scalar grep { $$t eq $_ } @_ }
};
my $var = in->new(7);
print "yup\n" if $var->in( 0 .. 8 );
__END__
yup

[download]

Not very tingly. Sorry about that.

perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'

In reply to Re: Create a new operator, get LHS by shmem
in thread Create a new operator, get LHS by stevieb

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