如何在swi-prolog中导入运算符?

use*_*669 4 module prolog swi-prolog

我有以下文件:

:- use_module(library(clpfd)).

isPowTwo(N) :- N #> 0, N #= 2^_.
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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.
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现在,我只想导入两个使用的运算符:

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

isPowTwo(N) :- N #> 0, N #= 2^_.
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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)
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不用括号#>#=没有区别。

Pau*_*ura 5

您的use_module/2指令中存在语法错误(因此会Syntax error: Operator expected出现错误消息)。您可以通过以下方式更正它:

:- use_module(library(clpfd), [(#>)/2, (#=)/2]).
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但是,您还需要导入相应的运算符以在isPowTwo/1谓词的定义中使用:

:- use_module(library(clpfd), [(#>)/2, (#=)/2, op(700,xfx,(#>)), op(700,xfx,(#=))]).
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  • 这个想法是,您可以像在op(_,_,#>)中那样通过模式导入运算符,因此您无需重复优先级。这坏了。为此推送了一个修复程序,因此它应在7.3.27中工作。 (2认同)