2

我从Prolog开始，我有点困惑......

我有一个简单的程序：

sum(0, []).
sum(Total, [Head|Tail]) :- sum(Sum, Tail), Total is Head + Sum.

当我调试时，我可以看到 Prolog 首先用 Head 和 Tail 拆分列表，所以结果是 0 + 空列表，然后它开始对数字求和并再次将其添加到列表中。

有人可以解释为什么它没有Total is Head + Sum. 先出现然后再次将列表拆分为 Head 和 Tail 吗？

编辑：这是跟踪：

[trace]  ?- sum(X, [1,2,3]).
Call: (6) sum(_G345, [1, 2, 3]) ? creep
Call: (7) sum(_G424, [2, 3]) ? creep
Call: (8) sum(_G424, [3]) ? creep
Call: (9) sum(_G424, []) ? creep
Exit: (9) sum(0, []) ? creep
Call: (9) _G430 is 3+0 ? creep
Exit: (9) 3 is 3+0 ? creep
Exit: (8) sum(3, [3]) ? creep
Call: (8) _G433 is 2+3 ? creep
xit: (8) 5 is 2+3 ? creep
Exit: (7) sum(5, [2, 3]) ? creep
Call: (7) _G345 is 1+5 ? creep
Exit: (7) 6 is 1+5 ? creep
Exit: (6) sum(6, [1, 2, 3]) ? creep
X = 6.
4

2 回答 2

5

您的定义将添加添加到堆栈中。避免加法的优化将是一种称为尾递归的通用技术的特例。

以下定义可以使用尾递归：

sum(X,L):-sum(0,L,X).
sum(X,[],X).
sum(N, [Head|Tail],Y) :- N1 is Head + N, sum(N1,Tail,Y).

它为部分和的值引入了一个累加器，并将其与列表的尾部一起携带。这是sum(X,[1,2,3])查询执行的跟踪。

?- trace, sum(S,[1,2,3]),notrace,nodebug.
   Call: (7) sum(_G584, [1, 2, 3]) ? creep
   Call: (8) sum(0, [1, 2, 3], _G584) ? creep
^  Call: (9) _G792 is 1+0 ? creep
^  Exit: (9) 1 is 1+0 ? creep
   Call: (9) sum(1, [2, 3], _G584) ? creep
^  Call: (10) _G795 is 2+1 ? creep
^  Exit: (10) 3 is 2+1 ? creep
   Call: (10) sum(3, [3], _G584) ? creep
^  Call: (11) _G798 is 3+3 ? creep
^  Exit: (11) 6 is 3+3 ? creep
   Call: (11) sum(6, [], _G584) ? creep
   Exit: (11) sum(6, [], 6) ? creep
   Exit: (10) sum(3, [3], 6) ? creep
   Exit: (9) sum(1, [2, 3], 6) ? creep
   Exit: (8) sum(0, [1, 2, 3], 6) ? creep
   Exit: (7) sum(6, [1, 2, 3]) ? creep
S = 6 .
于 2012-07-09T18:36:52.980 回答
-2

这是另一个版本。我一直在使用概念映射软件来帮助设计代码，我无法在脑海中做复杂的事情。

sum(A, [], A).
sum(Total, [Head|Tail], AuxNum) :-
    Sum is Head+AuxNum,
    sum(Total, Tail, Sum).


   1 ?- trace,sum(Total,[1,2,3],0).
   Call: (7) sum(_G2090, [1, 2, 3], 0) ? creep
   Call: (8) _G2221 is 1+0 ? creep
   Exit: (8) 1 is 1+0 ? creep
   Call: (8) sum(_G2090, [2, 3], 1) ? creep
   Call: (9) _G2224 is 2+1 ? creep
   Exit: (9) 3 is 2+1 ? creep
   Call: (9) sum(_G2090, [3], 3) ? creep
   Call: (10) _G2227 is 3+3 ? creep
   Exit: (10) 6 is 3+3 ? creep
   Call: (10) sum(_G2090, [], 6) ? creep
   Exit: (10) sum(6, [], 6) ? creep
   Exit: (9) sum(6, [3], 3) ? creep
   Exit: (8) sum(6, [2, 3], 1) ? creep
   Exit: (7) sum(6, [1, 2, 3], 0) ? creep
Total = 6.
于 2013-06-04T16:59:35.167 回答