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我正在尝试callto_status(Goal, Status)根据调用的结果定义一个始终成功并统一 Status的关系Goal（换句话说，我想实现一个具体化的版本call_with_inference_limit/3）。call_with_inference_limit/3我的实现使用具有相同接口的SWI call_with_time_limit/3（在这种情况下也应该可以工作）。的实施call_with_..._limit不会回溯，所以我认为最好不要给人以报告答案替代目标的印象。

我介绍了辅助谓词derivable_st以提高可读性。它处理成功和超时情况，否则失败。

% if Goal succeeds, succeed with Status = true,
% if Goal times out, succeed with Status = timeout
% if Goal fails, fail
derivable_st(Goal, Status) :-
    T = 10000,                               % set inference limit
%    copy_term(Goal, G),                    % work on a copy of Goal, we don't want to report an answer substitution
    call_with_inference_limit(G, T, R),     % actual call to set inference limit
    (  R == !
    -> Status = true                        % succeed deterministically, status = true
    ;  R == true
    -> Status = true                        % succeed non-deterministically, status = true
    ;  (  R == inference_limit_exceeded     % timeout
       -> (
              !,                            % make sure we do not backtrack after timeout
              Status = timeout              % status = timeout
          )
       ;  throw(unhandled_case)             % this should never happen
       )
    ).

主要谓词环绕derivable_st并处理失败情况和可能抛出的异常（如果有的话）。我们可能想要找出堆栈溢出（在推理限制太高的情况下发生），但现在我们只报告任何异常。 

% if Goal succeeds, succeed with Status = true,
% if Goal times out, succeed with Status = timeout
% if Goal fails, succeed with Status = false
% if Goal throws an error, succeed with Status = exception(The_Exception)
% Goal must be sufficiently instantiated for call(Goal) but will stay unchanged
callto_status(Goal, Status) :-
    catch((  derivable_st(Goal, S)            % try to derive Goal
          -> Status = S                       % in case of success / timeout, pass status on
          ;  Status = false                   % in case of failure, pass failure status on, but succeed
          ),
          Exception,
          Status = exception(Exception)       % wrap the exception into a status term
    ).

谓词适用于一些简单的测试用例：

?- callto_reif( length(N,X), Status).
Status = true.

?- callto_reif( false, Status).
Status = false.

?- callto_reif( (length(N,X), false), Status).
Status = timeout.

我现在的问题有点含糊：这个谓词是否像我声称的那样做？您是否看到任何错误/改进点？我很感激任何意见！

编辑：正如@false 所建议的，注释掉了copy_term/2

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1 回答 1

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这是一个较短的解决方案：

callto_status(Goal, Status) :-
    call_with_inference_limit(Goal, 10000, Status0),
    (Status0 = ! -> !, Status = true; Status = Status0).
callto_status(_, false).

你看看原来的用处有多大！status是避免不必要的选择点：

?- callto_status(member(X,[1,2,3]), Y).
X = 1,
Y = true 
X = 2,
Y = true 
X = 3,
Y = true.

?- callto_status(fail, Y).
Y = false.

您当然也可以仅替换Status0 = ! -> !, Status = true为Status0 = ! -> Status = true。然后，您将始终获得剩余的选择点：

?- callto_status(member(X,[1,2,3]), Y).
X = 1,
Y = true 
X = 2,
Y = true 
X = 3,
Y = true 
Y = false.

从问题中不清楚你到底想要什么。

于 2019-07-27T12:57:15.453 回答