Question

如何使用＆＃39; Univ＆＃39; （=../2）prolog谓词，使用Logtalk对象方法作为参数？

考虑以下代码：

baz(foo(X)) :-
  write(predicate), write(X), nl.

run :-
    Term =.. [baz, foo(testfoo)],
    write(Term), nl, Term,nl,    
    TermLgt =.. [bar::baz, foo(testfoo2)],
    write(TermLgt), nl, Term,nl.    

:- object(bar).

  :- public(baz/1).
  baz(foo(X)) :-
    write(method), write(X), nl.

:- end_object.

:- object(main).

  :- public(run/0).
  run :-
    Term =.. [baz, foo(testfoo)],
    write(Term), nl, Term,nl,    
    TermLgt =.. [bar::baz, foo(testfoo2)],
    write(TermLgt), nl, Term,nl.

:- end_object.

我将获得：

?- {myfile}.
% (0 warnings)
true.

?- run.
baz(foo(testfoo))
predicatetestfoo

ERROR: =../2: Type error: `atom' expected, found `bar::baz' (a compound)

?- main::run.
baz(foo(testfoo))
ERROR: Undefined procedure: baz/1
ERROR:   However, there are definitions for:
ERROR:         baz/1

用于良好解释/编译的解决方法是什么？似乎问题与swi-prolog构建谓词相同，如predsort/3（predsort/3 doc）。

Answer 1

标准=../2谓词在从列表构造术语时期望第一个列表参数是原子，但bar::baz是带有仿函数::/2的复合词（两者都是定义为谓词 - 用于顶级查询 - 并在加载Logtalk时作为运算符）。解决方案是改为编写：

baz(foo(X)) :-
    write(predicate), write(X), nl.


run :-
    Term =.. [baz, foo(testfoo)],
    write(Term), nl, call(Term), nl,    
    TermLgt =.. [::, bar, Term],
    write(TermLgt), nl, call(Term), nl.    


:- object(bar).

    :- public(baz/1).
    baz(foo(X)) :-
        write(method), write(X), nl.

:- end_object.


:- object(main).

    :- public(run/0).
    run :-
        Term =.. [baz, foo(testfoo)],
        write(Term), nl, Term,nl,    
        TermLgt =.. [::, bar, Term],
        write(TermLgt), nl, Term,nl.

:- end_object.

通过此更改，您将获得：

?- {univ}.
% [ /Users/pmoura/Desktop/univ.lgt loaded ]
% (0 warnings)
true.

?- run.
baz(foo(testfoo))
predicatetestfoo

bar::baz(foo(testfoo))
predicatetestfoo

true.

使用＆＃34; univ＆＃34;使用Logtalk对象方法作为参数的Prolog谓词

1 个答案: