🚀 go-pugleaf

In article ... veit@du9ds3.uni-duisburg.de (Holger Veit) writes:

   In <721458276@sheol.UUCP> throopw@sheol.UUCP (Wayne Throop) writes:

   >: From: hrubin@pop.stat.purdue.edu (Herman Rubin)
   >: Message-ID: <BxI802.7EL@mentor.cc.purdue.edu>
   >: Which languages allow the introduction of new operators?  

   >Lisp.  Forth.  Postscript.  Smalltalk.  Full PL/I had the ability
   >to coin new statement types and keywords (though not new
   >expression syntax).

   You forgot Prolog.

He forgot (if he ever heard of) Pop11, too.

But Herman might not think of those things as ``operators''. Consider the Pop11
definition:

    define 5 ++( x, y ); lvars x, y; x + y + 1 enddefine;

The ``5'' is the precedence (it happens to be the same as ``+'') and makes
``++'' a (nominally infix) operator, so expressions such as

    x ++ y, f(x) ++ g(h(y), z)

work. But (in Pop) an operator is *just* a procedure with a pleasant calling
syntax; writing ``x ++ y'' just means ``push x, push y, call ++'' (Pop is an
open-stack language), ``x * y'' means ``push x, push y, call -'', and ``f(x)''
means ``push x, call f''. *Some* procedure calls are coded in-line, but this is
not a property of being (syntactically) operators; it is a property of being a
particular procedure, such as fast_front, or fi_+.

I *think*, from earlier discussions, that Herman thinks being-an-operator is
something to do with translating-to-an-instruction (with the appropriate
effect). In which case, as usual, He and Us are talking past each other.
--

Regards,    | "See the darkness all around   is coming down on you."
Kers.       | - Renaissnace, ``Running Hard''.