如何使用 SWI-Prolog ./2 功能? [英] How to use SWI-Prolog ./2 function?

问题描述

需要使用 SWI-Prolog 的示例 ./2.

Need example using SWI-Prolog ./2.

签名是

.(+Int,[])

此外，如果有此运算符的名称，我会很高兴知道.搜索."毫无意义.

Also if there is a name for this operator it would be nice to know. Searching for '.' is senseless.

最接近名称的是 SWI 文档第 F.3 节 算术函数

The closest to a name is from SWI documentation section F.3 Arithmetic Functions

List of one character: character code

我尝试了什么

?- X is .(97,[]).

结果

ERROR: Type error: `dict' expected, found `97' (an integer)
ERROR: In:
ERROR:   [11] throw(error(type_error(dict,97),_5812))
ERROR:    [8] '<meta-call>'(user:(...,...)) <foreign>
ERROR:    [7] <user>
ERROR: 
ERROR: Note: some frames are missing due to last-call optimization.
ERROR: Re-run your program in debug mode (:- debug.) to get more detail.

也试过了

?- X is .(97,['a']).

结果相同.

推荐答案

在标准 Prolog 中，'.' 函子是实际的列表项函子.具体来说，[H|T] 等价于 '.'(H, T).例如，如果你运行 GNU Prolog，你会得到:

In standard Prolog, the '.' functor is the actual list term functor. Specifically, [H|T] is equivalent to '.'(H, T). If you run GNU Prolog, for example, you will get this:

| ?- write_canonical([H|T]).
'.'(_23,_24)

yes
| ?- X = .(H,T).

X = [H|T]

yes
| ?-

is/2 运算符将允许您直接从由单个字符代码组成的列表中读取字符代码.所以以下在 Prolog 中有效:

The is/2 operator will allow you to read the character code directly from a list consisting of a single character code. So the following works in Prolog:

| ?- X is [97].

X = 97

yes
| ?- X is .(97,[]).

X = 97

yes

开始变得有趣的是，在 SWI Prolog 中，他们引入了 dictionaries 依赖于 '.' 作为指定字典键的方法.这会与用作列表仿函数的 '.' 仿函数产生冲突.这在字典的链接中进行了解释.因此，当您尝试将 '.' 函子用于列表时，您会得到如下结果:

Where it starts to get interesting is that, in SWI Prolog, they've introduced dictionaries that rely on the '.' as a means of designating dictionary keys. This creates a conflict with the '.' functor being used as a list functor. This is explained in the link on dictionaries. Thus, when you attempt to use the '.' functor for a list, you get something like this:

1 ?- X is [97].
X = 97.

2 ?- X is .(97, []).
ERROR: Type error: `dict' expected, found `97' (an integer)
ERROR: In:
ERROR:   [11] throw(error(type_error(dict,97),_5184))
ERROR:    [9] '$dicts':'.'(97,[],_5224) at /usr/local/lib/swipl-7.4.2/boot/dicts.pl:46
ERROR:    [8] '<meta-call>'(user:(...,...)) <foreign>
ERROR:    [7] <user>
ERROR:
ERROR: Note: some frames are missing due to last-call optimization.
ERROR: Re-run your program in debug mode (:- debug.) to get more detail.

3 ?- X = .(H,T).
ERROR: Arguments are not sufficiently instantiated
ERROR: In:
ERROR:   [11] throw(error(instantiation_error,_6914))
ERROR:    [9] '$dicts':'.'(_6944,_6946,_6948) at /usr/local/lib/swipl-7.4.2/boot/dicts.pl:46
ERROR:    [8] '<meta-call>'(user:(...,...)) <foreign>
ERROR:    [7] <user>
ERROR:
ERROR: Note: some frames are missing due to last-call optimization.
ERROR: Re-run your program in debug mode (:- debug.) to get more detail.

因此，SWI 定义了另一个仿函数 [|] 来服务于 '.' 传统上所服务的目的，即列表仿函数.

SWI has, therefore, defined another functor, [|], to serve the purpose that '.' traditionally served, which is the list functor.

4 ?- X is '[|]'(97, []).
X = 97.

5 ?- X = '[|]'(H, T).
X = [H|T].

6 ?-

奇怪的是，你可能会认为这会发生:

Oddly, you might think then that this would happen:

7 ?- write_canonical([H|T]).
'[|]'(_, _)

然而，发生的事情是这样的:

However, what happens is this:

7 ?- write_canonical([H|T]).
[_|_]
true.

所以对于 SWI Prolog，列表表示 [H|T] 已经是列表的规范表示.

So to SWI Prolog, the list representation [H|T] is already the canonical representation of the list.

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