meaning of 'operator' (was Re: question on max | min op)

TSa wrote:
> Mark J. Reed wrote: 
>
> I'm not sure what the defined difference between function and operator
> is in Perl 6 but I make a very clear distinction. An operator is acting
> an *one* type, that is &op:(::T,T-->T) while a function is a mapping of
> different types &f:(::S,::T-->::R). IOW, Operator does Function but not
> the converse. So overloading an operator with inhomogenous types should
> not change the abstract properties of the operator but be a convenient
> form to upgrade the operants.

I consider an operator to be any pure routine whose invocation does not
carry any side-effects besides possibly updating the values of some of its
arguments; any values that its invocation results in or that any of its
arguments are set to is dependent only on the values of its arguments.

There is no restriction on what the types of its parameters or results are;
they can be of the same types or different types; eg a pure function to
cast a number as a string is an operator.

Whether or not something is an operator has nothing to do with syntax, just
semantics.  Either a function (which results in a value) or a procedure
(which does not) can be an operator.  Infix|prefix|postfix|.meth|func() etc
does not matter, that is all just syntax.  As a trivial case, a procedure
with zero parameters either isn't an operator or is a no-op.

Any routine which looks at or manipulates the environment is not an operator
.

Most language built-ins would qualify as operators, including all pure
functions and ordinary assignment operators.

Built-ins that would not be operators include: print, curr_time, <$*STDIN> e
tc.

-- Darren Duncan