module - 如何在 swi-prolog 中导入运算符？

Question

我有以下文件:

:- use_module(library(clpfd)).

isPowTwo(N) :- N #> 0, N #= 2^_.

一切正常:

?- [importTest].
%     library(pairs) compiled into pairs 0.00 sec, 22 clauses
%    library(lists) compiled into lists 0.01 sec, 122 clauses
%    library(occurs) compiled into occurs 0.00 sec, 14 clauses
%   library(apply_macros) compiled into apply_macros 0.01 sec, 168 clauses
%   library(assoc) compiled into assoc 0.01 sec, 103 clauses
%  library(clpfd) compiled into clpfd 0.15 sec, 2,808 clauses
% importTest compiled 0.16 sec, 2,813 clauses
true.

现在我只想导入两个使用过的运算符:

:- use_module(library(clpfd), [(#>)/2, (#=)/2]).

isPowTwo(N) :- N #> 0, N #= 2^_.

它不起作用:

?- [importTest].
%     library(pairs) compiled into pairs 0.00 sec, 22 clauses
%    library(lists) compiled into lists 0.01 sec, 122 clauses
%    library(occurs) compiled into occurs 0.00 sec, 14 clauses
%   library(apply_macros) compiled into apply_macros 0.01 sec, 168 clauses
%   library(assoc) compiled into assoc 0.01 sec, 103 clauses
%  library(clpfd) compiled into clpfd 0.16 sec, 2,808 clauses
ERROR: .../importTest.pl:3:17: Syntax error: Operator expected
% importTest compiled 0.16 sec, 2,812 clauses
true.

?- isPowTwo(1).
ERROR: toplevel: Undefined procedure: isPowTwo/1 (DWIM could not correct goal)

在 #> 和 #= 两边保留括号没有什么区别。

Answer 1

您的 use_module/2 指令中存在语法错误(因此出现 Syntax error: Operator Expected 错误消息)。您可以通过编写以下内容来更正它:

:- use_module(library(clpfd), [(#>)/2, (#=)/2]).

但是您还需要导入相应的运算符，以便在 isPowTwo/1 谓词的定义中使用:

:- use_module(library(clpfd), [(#>)/2, (#=)/2, op(700,xfx,(#>)), op(700,xfx,(#=))]).