序言:介于/3之间，末尾有一个列表 [英] Prolog: between/3 with a list at the end

问题描述

我的问题因这个问题而出现 您能在纯序言中在/3之间写作吗?

是否可以在/3和第三个参数之间建立一个列表，所以如果您问

between(2,6,X).

它来了

X=[2,3,4,5,6]

不喜欢

X=2
X=3
X=4
....

我不知道该如何工作(我的所有解决方案均无效.) 我是Prolog的初学者，所以我不知道.

对不起，英语不好.

感谢您的帮助:)

解决方案

首先进入图书馆并获得一本好书，例如Sterling和Shapiro的序言的艺术".

两种方式:

?- findall(X, between(2, 6, X), Xs).
Xs = [2, 3, 4, 5, 6].

您还应该查看bagof/3和setof/3.

对于直接方法numlist/3，请参见例如 SWI-Prolog的实现.没有参数检查，它可以归结为:

numlist(U, U, List) :- !,
    List = [U].
numlist(L, U, [L|Ns]) :-
    L2 is L+1,
    numlist(L2, U, Ns).

有几种方法可以打破谓词的原状.

?- numlist(1,0,L).

不会终止.您需要先检查参数，然后再将其传递给此特定版本的numlist/3:

must_be(integer, L),
must_be(integer, U),
L =< U

这些检查并入链接的SWI-Prolog实现的库谓词numlist/3中.

my question comes up because of this question Can you write between/3 in pure prolog?

would it be possible to make between/3 and the third argument is a list so if you ask

between(2,6,X).

it comes

X=[2,3,4,5,6]

and not like

X=2
X=3
X=4
....

I can’t figure out how this must work (all my solutions don’t work..) I’m a Prolog beginner so I have no idea..

sorry for the bad English..

Thanks for your help :)

解决方案

Start by going to the library and getting a good book, for example "The Art of Prolog" by Sterling and Shapiro.

Two ways:

?- findall(X, between(2, 6, X), Xs).
Xs = [2, 3, 4, 5, 6].

You should also take a look at bagof/3 and setof/3.

For a direct way, numlist/3, see for example the SWI-Prolog implementation. Without argument checking it comes down to:

numlist(U, U, List) :- !,
    List = [U].
numlist(L, U, [L|Ns]) :-
    L2 is L+1,
    numlist(L2, U, Ns).

There are several ways to break the predicate as it stands.

?- numlist(1,0,L).

will not terminate. You need to either check the arguments before passing them to this particular version of numlist/3:

must_be(integer, L),
must_be(integer, U),
L =< U

These checks are incorporated in the library predicate numlist/3 from the linked SWI-Prolog implementation.

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