列表列表转换为修改后的单个列表 [英] List of lists into modified single list

问题描述

这些是从输入文件中读取的谓词

These are predicates which reads from input file

read_line(L,C) :-
    get_char(C),
    (isEOFEOL(C), L = [], !;
        read_line(LL,_),% atom_codes(C,[Cd]),
        [C|LL] = L).

%Tests if character is EOF or LF.
isEOFEOL(C) :-
    C == end_of_file;
    (char_code(C,Code), Code==10).

read_lines(Ls) :-
    read_line(L,C),
    ( C == end_of_file, Ls = [] ;
      read_lines(LLs), Ls = [L|LLs]
    ).

输入文件:

A B
C D
E F
G H

read_lines(L) 返回 L = [[A, ,B],[C, ,D],[E, ,F],[G, ,H]].我的目标是替换所有空格并将列表列表合并为单个列表.所以预期的输出应该是这样的:L = [A-B,C-D,E-F,G-H].到目前为止我得到的是修改 read_line 函数:

read_lines(L) returns L = [[A, ,B],[C, ,D],[E, ,F],[G, ,H]]. My goal is to replace all the spaces and merge list of lists into single list. So expected output should look like: L = [A-B,C-D,E-F,G-H]. What I got so far is modified read_line function:

read_line(L,C) :-
    get_char(C),
    ( (char_code(C,Code), Code == 32)
        -> C = '-'
        ; C = C),
    (isEOFEOL(C), L = [], !;
        read_line(LL,_),% atom_codes(C,[Cd]),
        [C|LL] = L).

当我使用它时，Prolog 说 Syntax error: Unexpected end of file.这有什么问题?

When I use it, Prolog says Syntax error: Unexpected end of file. What's wrong with that?

推荐答案

问题出在这段代码中:

( (char_code(C,Code), Code == 32)
    -> C = '-'
    ; C = C),

如果一个空格字符被读入变量 C，char_code 将 Code 绑定到 32，条件为真.然后Prolog 尝试将C 与'-' 统一起来，但是C 已经绑定到' '！这失败了，因此您的 read_line 调用失败并在标准输入流上留下一些未使用的输入.下次 Prolog 尝试从您那里读取输入时，它实际上会读取剩余的输入.

If a space character was read into variable C, char_code binds Code to 32, and the condition is true. Then Prolog tries to unify C with '-', but C is already bound to ' '! This fails, so your read_line call fails and leaves some unconsumed input on the standard input stream. The next time Prolog tries to read input from you, it actually reads that remaining input.

问题的根本原因是您似乎试图重新分配"变量 C.这在 Prolog 中是不可能的；一旦一个变量被绑定，它只能在回溯时解除绑定.

The underlying cause of the problem is that you seem to be trying to "reassign" the variable C. That is not possible in Prolog; once a variable is bound, it can only become unbound on backtracking.

使用 new 变量，如下所示:

Use a new variable, something like this:

( (char_code(C,Code), Code == 32)
    -> NewC = '-'
    ; NewC = C),

在适当的地方使用 NewC.

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