prolog - YAP PROLOG 超时问题

Question

谁能解释为什么下面的 YAP 代码不会导致变量 Result 与 time_out 统一？

?- time_out(sleep(3),1,Result).
Result = success.
?- time_out(sleep(3),2,Result).
Result = success.
 ?- time_out(sleep(3),1000,Result).
Result = success.
 ?- time_out(sleep(3),2000,Result).

根据documentation ，谓词应按如下方式工作:

time_out(+Goal, +Timeout, -Result)

Execute goal Goal with time limited Timeout, where Timeout is measured in milliseconds. If the goal succeeds, unify Result with success. If the timer expires before the goal terminates, unify Result with time_out.

Answer 1

YAP 的文档还指出:

Last, even though the timer is set in milliseconds, the current implementation relies on alarm/3, and therefore can only offer precision on the scale of seconds.

反过来，sleep/1 的文档指出:

Block the current thread for Time seconds.

但这种情况下的问题可能与时间无关，而是对 sleep/1 的调用阻塞了正在调用 time_out/3 的当前线程自己执行。尽管如此，我尝试了具有其他目标的 time_out/3 谓词，但也无法获得预期的超时时间。因此，它可能存在一些错误。