[tpm] Riddle me this ...

[Tom Legrady]

An operator is a routine

On Fri, Dec 28, 2012 at 10:57 PM, James E Keenan <jkeen at verizon.net> wrote:

> On 12/27/12 11:43 PM, Uri Guttman wrote:
>
>  i want to clarify my statement there. it is the scalar prototype of not
>> that is causing this as well as its low precedence.
>>
>
> I'm familiar with prototypes as an aspect of subroutines in Perl (for
> instance, Test::More::is is, under the hood, prototyped $$;$), but not with
> the concept of prototypes for operators.  Can you point me toward the
> documentation or source code for that?
>
>
> Thank you very much.
> Jim Keenan
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