[stack] Re: Properties of Concatenative langauges and Forth

--- In [email protected], "Christopher Diggins"
<cdiggins@...> wrote:
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> Forth without compile-time words isn't Turing-complete as far as I can
> tell.

Yes, it is (ignoring the fact that implementations are finite),
because it's a TWO stack machine. (I'm assuming the memory addressing
operators @ and ! aren't allowed for the purposes of this discussion.)

 I don't think you can just give it a pass. Can you even
> construct a conditional expression in Forth without compile-time
> words?

Yes, actually. You can change the default compiler behaviour to close
secondaries with a branch back to the beginning instead of a return,
so the default is an infinite loop, and then use a conditional return
?R which consumes a true/false flag on the stack. If you get rigorous
about not even allowing the compiler to have a compile time keyword to
terminate it, you can do what I did with Furphy (for just that reason,
to avoid making "special" words fundamental). That uses Reverse Polish
naming, so that the first unrecognised token both terminates and names
a secondary. I took that approach even further, to make Furphy
modeless (more precisely, it has a single compile and go mode where
Forth has distinct execution and compile modes).