序言中的递归(在列表上) [英] recursion in prolog (on lists)

问题描述

有人可以帮助我了解执行递归序言函数的基础知识吗..

can someone please help me just w/ the basics on performing recursive prolog functions..

append([],X,X). % base
append([X|Y],Z,[X|W]) :- append(Y,Z,W). %recursive

% base case
addup([], 0). % sum of the empty list of numbers is zero

% recursive case: if the base-case rule does not match, this one must:
addup([FirstNumber | RestOfList], Total) :-
    addup(RestOfList, TotalOfRest),   % add up the numbers in RestOfList
    Total is FirstNumber + TotalOfRest.

有人可以用英语或 C/C++/Java 解释一下……步骤如何.我实际上更喜欢看到类似 append 或 reverse 之类的东西..我主要是在操作变量列表而不是整数..(我已经尝试了 10 次这样的 append .. 呃).

Can someone explain either in English or in C/C++/Java whatever.. how the steps. I actually would prefer to see something like append or reverse.. I'm mostly just manipulating lists of variables instead of integers.. (I've tried to work through append like 10 times.. ugh).

推荐答案

免费的在线书籍Learn Prolog Now"有一节专门解释 append 执行的步骤:

The free online book "Learn Prolog Now" has a section dedicated to explaining the steps that append performs:

http:///cs.union.edu/~stregnk/learn-prolog-now/html/node47.html#subsec.l6.defining.append

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