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yap-6.3/CHR/chr/examples/type.pl
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Prolog

% rational tree handler with OZ type constraints, see also tree.pl
% ECRC 1993, 950519; LMU 980211, 980312 Thom Fruehwirth
:- use_module( library(chr)).
handler type.
constraints (~)/2, (':<')/2, ('&&')/2.
% T1 ~ T2 means: term T1 is syntactically equal to term T2
% T1 :< T2 means: term T1 has type T2, types are not to be cyclic (infinite)
% Z~X&&Y means: the type Z is the intersection of types X and Y
operator(100,xfx,(~)). % equality
operator(100,xfx,(':<')). % type constraint
operator(110,xfx,('&&')). % type intersection
% need global order on variables for equality with infinite (cyclic) terms
:- use_module( library('chr/ordering'), [globalize/1,var_compare/3]).
% var is smaller than any non-var term
lt(X,Y):- (var(X),var(Y) -> globalize(X),globalize(Y),var_compare(<,X,Y) ; X@<Y).
le(X,Y):- (var(X) -> true ; X@=<Y).
% equality ~ -----------------------------------------------------------------
ident @ T ~ T <=> true.
decompose @ T1 ~ T2 <=> nonvar(T1),nonvar(T2) |
same_functor(T1,T2),
T1=..[F|L1],T2=..[F|L2],
equate(L1,L2).
orient @ T ~ X <=> lt(X,T) | X ~ T.
simplify @ X ~ T1 \ X ~ T2 <=> le(T1,T2) | T1 ~ T2.
same_functor(T1,T2):- functor(T1,F,N),functor(T2,F,N).
equate([],[]).
equate([X|L1],[Y|L2]):- X ~ Y, equate(L1,L2).
% type constraint :< ---------------------------------------------------------
type_identity @ XT :< XT <=> true.
type_decompose @ T1 :< T2 <=> nonvar(T1),nonvar(T2) |
same_functor(T1,T2),
T1=..[_|L1],T2=..[_|L2],
contain(L1,L2).
type_simplify1 @ X ~ T1 \ X :< T2 <=> var(X) | T1 :< T2.
type_simplify2 @ X ~ T1 \ T2 :< X <=> var(X) | T2 :< T1.
type_intersect @ X :< T1, X :< T2 <=> nonvar(T1),nonvar(T2) |
same_functor(T1,T2),
T1=..[F|L1],T2=..[F|L2],
type_intersect(L1,L2,L3),
T3=..[F|L3],
X :< T3.
type_transitiv @ T1 :< Y, Y :< T2 ==> var(Y) | T1 :< T2.
contain([],[]).
contain([X|L1],[Y|L2]):- X :< Y,
contain(L1,L2).
type_intersect([],[],[]).
type_intersect([X|L1],[Y|L2],[Z|L3]):- Z~X&&Y,
type_intersect(L1,L2,L3).
% X~Y&&Z parses as (X~Y)&&Z, therefore it cannot match X~T
type_functional @ Z1~X&&Y \ Z2~X&&Y <=> Z1=Z2.
type_functional @ Z1~Y&&X \ Z2~X&&Y <=> Z1=Z2.
type_propagate @ Z~X&&Y ==> Z :< X, Z :< Y.
/*
% Examples
:- f(a,b):<f(X,X). % succeeds - X is a "top" ('a hole')
a:<X,b:<X.
:- Y~f(U),Z~f(X),X:<Y,X:<Z. % succeeds
Y~f(U),Z~f(X),UX~X&&U,X:<f(UX),UX:<X,UX:<U,UX:<f(UX)
:- Y~f(U),U~a,Z~f(X),X:<Y,X:<Z. % fails
:- X:<Y,X~f(X),X:<f(Y).
X~f(X), f(X):<Y % simplifies nicely
:- X:<Y,Y~f(U),U~a,Z~f(X),X:<Z. % fails
:- X~Y,U:<X,Z:<a,U:<Z,Y:<b. % fails
:- X:<Y,X:<Z,Y~a,Z~b. % fails
:- X:<Y,X:<Z,Y~f(Y,U),Z~f(Z,V),U~a,V~b. % fails
:- X:<f(X,Y), X~f(X1,U), X1~f(X11,U1), U1~g(U), a:<U, b:<U. % succeeds
:- X~ f(X,Y), X~f(X1,U), X1~f(X11,U1), U1~g(U), a:<U, b:<U. % fails
*/
% end of handler type =======================================================