SWI-Prolog (SWISH):无权修改静态过程`(=)/2' [英] SWI-Prolog (SWISH): No permission to modify static procedure `(=)/2'

问题描述

在 SWI-Prolog (SWISH) 中使用不同的列表谓词时，我试图检查原子 a 是否是列表 List1 的一部分，我在程序中定义为 List1 = [a,b,c,d].

我将我的查询公式化为 member(a, List1).，期待一些简单的是"(正如它在这个 youtube 视频 at 59:25)，但我收到了警告

<块引用>

单变量:[List1]

和一个错误

<块引用>

无权限修改静态过程`(=)/2'

根据我在网上查找的了解，警告在这里并不那么重要.但是，我不明白为什么在 a 显然是 List1 的成员时会收到错误消息.

我以两种不同的方式尝试了这个:

1) 通过在程序中添加 List1 = [a,b,c,d]. 并使用 member(a,List1). 进行查询(结果是上面的错误);

2) 通过将 List1 = [a,b,c,d] 直接传递给解释器，然后使用相同的查询 ( member(a,List1).)，这导致了无数的结果，其中 a 移动了列表头部的位置，如下所示:

List1 = [a|_1186]List1 = [_1062, a|_1070]列表 1 = [_1062, _1068, a|_1076]列表 1 = [_1062, _1068, _1074, a|_1082]列表 1 = [_1062, _1068, _1074, _1080, a|_1088]

这是关于我正在使用的特定 Prolog 版本，还是我遗漏了一些非常简单的东西?

编辑

我知道在这里提出了类似的问题，但我没有设法完全理解答案(也不是问题)，因为它立即处理了我在 Prolog 中还没有遇到过的 dynamic 事情.我一直在寻找一个更一般、更高级"的答案，我通过提出这个问题找到了这个答案.

解决方案

我在程序中定义为List1 = [a,b,c,d].

这不是它的作用.它所做的是定义一个谓词 =/2:

<块引用>

2 ?- write_canonical( (List1 = [a,b,c,d]) ).=(_,[a,b,c,d])

^{(您在那里看到的 ?- 或 2 ?-，是Prolog 系统的交互式提示；在我的情况下为 SWI Prolog.无论在那一行之后发生什么，都是我输入的内容；然后在下一行我们看到系统的响应).}

当然，这践踏了 = 作为统一谓词的现有内置定义.因此，错误恰恰说明了这一点.是的，它很重要.

要在 Prolog 中定义"一个列表，我们可以定义一个谓词

8 ?- [用户].p([1,2,3,4]).

这样我们就可以查询

9 ?- p(List1).列表 1 = [1, 2, 3, 4].

并进一步使用 List1,

10 ?- p(List1), member(A,List1).列表 1 = [1, 2, 3, 4],一 = 1 ;列表 1 = [1, 2, 3, 4],A = 2 ;列表 1 = [1, 2, 3, 4],一 = 3 ;列表 1 = [1, 2, 3, 4],A = 4.

我们也可以直接将列表指定为查询的子目标，

11 ?- List1 = [1,2,3,4], member(A,List1).列表 1 = [1, 2, 3, 4],一 = 1 ;列表 1 = [1, 2, 3, 4],A = 2 ;列表 1 = [1, 2, 3, 4],一 = 3 ;列表 1 = [1, 2, 3, 4],A = 4.

使使用谓词=/2，而不是重新定义，这是禁止的.

<小时>

以上回答了您的1).至于2)，你并没有告诉我们全部真相.你似乎做了什么，是先进行查询

12 ?- List1 = [a,b,c,d].列表 1 = [a, b, c, d].

这很好，很花哨；然后然后进行另一个查询，

13 ?- member(a,List1).列表 1 = [a|_G2181] ;List1 = [_G2180, a|_G2184] ;List1 = [_G2180, _G2183, a|_G2187];List1 = [_G2180, _G2183, _G2186, a|_G2190] ;List1 = [_G2180, _G2183, _G2186, _G2189, a|_G2193] .

Prolog 提示不是 REPL.我们不对它下定义.我们进行查询.

During some playing around with different list predicates in SWI-Prolog (SWISH), I was trying to check if an atom a was part of the list List1 which I defined in the program as List1 = [a,b,c,d].

I formulated my query as member(a, List1)., expecting something along the lines of a simple 'yes' (just as it shows in this youtube video at 59:25), but instead I got a warning

Singleton variables: [List1]

and an error

No permission to modify static procedure `(=)/2'

From what I understand from looking this up online, the warning is not that important here. I do not understand, however, why I get an error message while a is clearly a member of List1.

I tried this in two different ways:

1) By adding List1 = [a,b,c,d]. to the program and querying with member(a,List1). (which resulted in the error above);

2) By passing List1 = [a,b,c,d] directly to the interpreter and then using the same query ( member(a,List1). ), which resulted in an endless amount of results where a shifted positions in the Head of the list, like so:

List1 = [a|_1186]
List1 = [_1062, a|_1070]
List1 = [_1062, _1068, a|_1076]
List1 = [_1062, _1068, _1074, a|_1082]
List1 = [_1062, _1068, _1074, _1080, a|_1088]

Is this something about the specific Prolog version I am using, or am I missing something very simple?

EDIT

I was aware that a similar question was posed here , but I did not manage to fully understand the answer (nor the question) as it was immediately going about things as dynamic which I have not yet encountered in Prolog. I was looking for a more general, more 'high-level' answer which I have found by posing this question.

解决方案

I defined in the program as List1 = [a,b,c,d].

This is not what it does. What it does is define a predicate =/2:

2 ?- write_canonical( (List1 = [a,b,c,d]) ).
=(_,[a,b,c,d])

^{(The ?-, or 2 ?- that you see there, is the interactive prompt of a Prolog system; SWI Prolog in my case. Whatever goes on that line after it is what I have typed; and then on the next line we see the system's response).}

Of course this tramples over the already existing built-in definition for = as the unification predicate. And hence the error which says precisely that. And yes, it is important.

To "define" a list in Prolog, we can define a predicate

8 ?- [user].
p([1,2,3,4]).

such that we can then query

9 ?- p(List1).
List1 = [1, 2, 3, 4].

and work further with List1,

10 ?- p(List1), member(A,List1).
List1 = [1, 2, 3, 4],
A = 1 ;
List1 = [1, 2, 3, 4],
A = 2 ;
List1 = [1, 2, 3, 4],
A = 3 ;
List1 = [1, 2, 3, 4],
A = 4.

We could also just directly specify the list as a sub-goal of our query,

11 ?- List1 = [1,2,3,4], member(A,List1).
List1 = [1, 2, 3, 4],
A = 1 ;
List1 = [1, 2, 3, 4],
A = 2 ;
List1 = [1, 2, 3, 4],
A = 3 ;
List1 = [1, 2, 3, 4],
A = 4.

making use of the predicate =/2, as opposed to redefining it, which is forbidden.

---

The above answers your 1). As for 2), you aren't telling us the whole truth. What you appear to have done, was to first make a query

12 ?- List1 = [a,b,c,d].
List1 = [a, b, c, d].

which is fine and dandy; and then make another query,

13 ?- member(a,List1).
List1 = [a|_G2181] ;
List1 = [_G2180, a|_G2184] ;
List1 = [_G2180, _G2183, a|_G2187] ;
List1 = [_G2180, _G2183, _G2186, a|_G2190] ;
List1 = [_G2180, _G2183, _G2186, _G2189, a|_G2193] .

Prolog prompt is not a REPL. We don't make definitions at it. We make queries.

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