述語クエリの実行中用語

Question

に部分項の出現を見つけるために、Nの値が1の代わりに2であることが出てきた：

出現（X、S（X、T（X、Y））、 N）。 N = 1

occurrences(Sub,Term,R) :- occurrences(Sub,Term,0,R). 
occurrences(Sub,Sub,Acc,R) :- R is Acc + 1. 
occurrences(Sub,Term,Acc,R) :- functor(Term,_,N),occurrences(N,Sub,Term,Acc,R). 
occurrences(N,Sub,Term,Acc,R4) :- N > 0, arg(N,Term,R),  occurrences(Sub,R,Acc,R2), 
           N1 is N-1, occurrences(N1,Sub,Term,R2,R4). 
occurrences(N,_,_,Acc,Acc) :- N < 1. 

Answer 1

あなたは既にコードで正しい場所を見ています。これで最後までコピーするだけです。

http://eu.swi-prolog.org/pldoc/doc/swi/library/occurs.pl?show=src#occurrences_of_term/3

ライセンスと著作権情報のため、ファイルの先頭を参照してください。

http://eu.swi-prolog.org/pldoc/doc/swi/library/occurs.pl?show=src

occurrences_of_term(Sub, Term, Count) :- 
    count(sub_term(Sub, Term), Count). 

%! occurrences_of_var(+SubTerm, +Term, ?Count) 
% 
% Count the number of SubTerms in Term 

occurrences_of_var(Sub, Term, Count) :- 
    count(sub_var(Sub, Term), Count). 

%! sub_term(-Sub, +Term) 
% 
% Generates (on backtracking) all subterms of Term. 

sub_term(X, X). 
sub_term(X, Term) :- 
    compound(Term), 
    arg(_, Term, Arg), 
    sub_term(X, Arg). 

%! sub_var(-Sub, +Term) 
% 
% Generates (on backtracking) all subterms (==) of Term. 

sub_var(X0, X1) :- 
    X0 == X1. 
sub_var(X, Term) :- 
    compound(Term), 
    arg(_, Term, Arg), 
    sub_var(X, Arg). 


       /******************************* 
       *    UTIL   * 
       *******************************/ 

%! count(:Goal, -Count) 
% 
% Count number of times Goal succeeds. 

:- meta_predicate count(0,-). 

count(Goal, Count) :- 
    State = count(0), 
    ( Goal, 
     arg(1, State, N0), 
     N is N0 + 1, 
     nb_setarg(1, State, N), 
     fail 
    ; arg(1, State, Count) 
    ).