在硬件设备的命令和控制上下文中，我需要一个无限循环“获取-详细-发布”并记住当前状态，以观察 bool 值等输入的演变这“成为”真实。
我编写了一个模型，即以下程序，它产生了奇怪的行为(对我来说)，我很惊讶，因为我设置了有界变量Previous而没有失败，为什么？
我期望出现类似“变量 Previous 已绑定(bind)”之类的错误消息，但事实并非如此，回溯是为了解决约束(很多次!)。
#!/usr/bin/swipl
init( Previous, AtStart ) :-
get_time( AtStart ),
Previous is 0.
run( AtStart, Previous ) :- % I want this to be executed only once for each period!
get_time( Now ),
Elapsed is Now - AtStart,
Current is Previous + random( 20 ),
format( "~w: Previous = ~w~n", [Elapsed, Previous] ),
format( "~w: Current = ~w~n", [Elapsed, Current ] ),
Previous is Current.
periodicTask :-
init( Previous, AtStart ),
repeat,
run( AtStart, Previous ),
sleep( 1.0 ).
:-
periodicTask.
它会以这种输出无限期地运行:
?- periodicTask.
?- periodicTask.
0.0231266: Previous = 0
0.0231266: Current = 10
0.243902: Previous = 0
0.243902: Current = 16
............................ A lot of lines
0.934601: Previous = 0
0.934601: Current = 0
true ;
2693.8: Previous = 0
2693.8: Current = 19
............................ A lot of lines
2694.65: Previous = 0
2694.65: Current = 0
true ;
3694.98: Previous = 0
3694.98: Current = 2
............................ A lot of lines
3695.17: Previous = 0
3695.17: Current = 0
true ;
4695.47: Previous = 0
4695.47: Current = 10
............................ A lot of lines
4695.55: Previous = 0
4695.55: Current = 0
true
为什么？
如何编写无限循环代码来重置(解除绑定(bind))一些变量并保留全局上下文？
“通过递归”的答案似乎在这里不适用，不是吗？
最佳答案
这是一个用尾递归调用编写的简单循环，它将激活 T 时的“状态”传输到激活 T+1 时的“状态”，这很自然(正如哥德尔在每个人都开始使用之前的意图) for 和 while):
periodicTask :-
get_time(AtStart),
run(AtStart,0).
stopCriteriumFulfilled :- fail. % TO BE DONE
run(AtStart,Previous) :-
!, % Possibly optional: drop open chociepoints in infinite loop
get_time(Now),
Elapsed is Now - AtStart,
Current is Previous + random( 20 ),
format( "~w: Previous = ~w~n", [Elapsed, Previous] ),
format( "~w: Current = ~w~n", [Elapsed, Current ] ),
%
% Now call yourself with a new context in which the
% variable "Current" ("here", in this activation)
% becomes the variable "Previous" ("there", in the next activation)
% But first sleep a bit (N.B. numbers are numbers, one can use 1
% instead of 1.0)
% One may want to add a stop criterium here to succeed the predicate
% instead of performing the tail-recursive call unconditionally.
%
(stopCriteriumFulfilled
-> true
; (sleep(1),
run(AtStart,Current))). % RECURSIVE TAIL CALL
:- periodicTask.
如果需要更多的“状态性”而不是将当前状态传递到下一个激活，请查看以下内容:
关于prolog - 如何重复积累(永远)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/64856890/