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Minor edits.

This commit is contained in:
Simon Forman 2020-01-31 12:01:13 -08:00
parent a14a55defb
commit cfab1cf3c8
1 changed files with 19 additions and 19 deletions
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thun/thun.pl
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@ -72,9 +72,6 @@ joy(InputString, StackIn, StackOut) :-
██║ ██║██╔══██╗██╔══██║██║╚██╔╝██║██║╚██╔╝██║██╔══██║██╔══██╗
╚██████╔╝██║ ██║██║ ██║██║ ╚═╝ ██║██║ ╚═╝ ██║██║ ██║██║ ██║
╚═════╝ ╚═╝ ╚═╝╚═╝ ╚═╝╚═╝ ╚═╝╚═╝ ╚═╝╚═╝ ╚═╝╚═╝ ╚═╝
The grammar of Joy is very simple. A Joy expression is zero or more Joy
terms separated by blanks, and terms can be either integers, Booleans,
@ -170,27 +167,30 @@ thun(bool(C), [A|B], D, E) :- thun(A, B, [bool(C)|D], E).
thun(list(A), [], B, [list(A)|B]).
thun(list(C), [A|B], D, E) :- thun(A, B, [list(C)|D], E).
% Partial reduction for func/3 cases works.
% Partial reduction works for func/3 cases.
thun(symbol(A), [], B, C) :- func(A, B, C).
thun(symbol(A), [C|D], B, F) :- func(A, B, E), thun(C, D, E, F).
% Combinators look ok too.
% Combinators look ok too.
thun(symbol(A), D, B, C) :- combo(A, B, C, D, []).
thun(symbol(A), C, B, G) :- combo(A, B, F, C, [D|E]), thun(D, E, F, G).
% thun(symbol(A), D, B, C) :- combo(A, B, C, D, []).
% thun(symbol(A), C, B, G) :- combo(A, B, F, C, [D|E]), thun(D, E, F, G).
% I think the order of these two rules should be reversed because
% it will be pretty rare for a combinator to result in a null
% expression and we don't want to process combo/5 twice just to
% notice that, eh? Swap the rules and add green cuts after combo/5
% and that should make it more efficient.
% However, in this case, I think the original version will be more
% efficient.
% Ach! THe "green" cuts would mess up multi-rule combinators! D'oh!
thun(symbol(Combo), E, Si, So) :- combo(Combo, Si, S, E, Eo), thun(Eo, S, So).
% Neither functions nor definitions can affect the expression so we
% don't need to do similar gardening to those rules. The unification
% of the head clauses will distinguish the cases for them.
% In the reduced rules Prolog will redo all the work of the combo/5
% predicate on backtracking through the second rule. It will try combo/5,
% which usually won't end in Eo=[] so the first rule fails, then it will
% try combo/5 again in the second rule. In this form, after combo/5 has
% completed Prolog has computed Eo and can index on it for thun/3.
%
% Neither functions nor definitions can affect the expression so this
% doesn't apply to those rules. The unification of the head clauses will
% distinguish the cases for them.
% Definitions don't work though (See "Partial Reducer" section below.)
% I hand-wrote the def/3 cases here. (Maybe if append/3 was reduced?)
@ -1104,9 +1104,9 @@ thunder([ % Source code for thun/4.
(thun( int(I), E, Si, So) :- thun(E, [ int(I)|Si], So)),
(thun(bool(B), E, Si, So) :- thun(E, [bool(B)|Si], So)),
(thun(list(L), E, Si, So) :- thun(E, [list(L)|Si], So)),
(thun(symbol(Def), E, Si, So) :- def(Def, Body), append(Body, E, [T|Eo]), thun(T, Eo, Si, So)),
(thun(symbol(Func), E, Si, So) :- func(Func, Si, S), thun(E, S, So)),
(thun(symbol(Combo), E, Si, So) :- combo(Combo, Si, S, E, Eo), thun(Eo, S, So))
% (thun(symbol(Def), E, Si, So) :- def(Def, Body), append(Body, E, [T|Eo]), thun(T, Eo, Si, So)),
(thun(symbol(Func), E, Si, So) :- func(Func, Si, S), thun(E, S, So))
% (thun(symbol(Combo), E, Si, So) :- combo(Combo, Si, S, E, Eo), thun(Eo, S, So))
]).
/*