为什么我的 EOF / linebreak 非终结符总是有 shift/reduce 冲突？

Question

所以，在将解析语法放在一起方面，我还很初级。当我遇到 shift-reduce 冲突时，我需要帮助来剖析 Menhir 报告的冲突。

以这个小语法为例：

(* {2 Tokens } *)
%token EOF
%token COLON PIPE SEMICOLON
%token <string> COUNT
%token <string> IDENTIFIER

%start <AST.t> script
%start <AST.statement> statement

%%
(* {2 Rules } *)

script:
 | it = separated_list(break, statement); break?; EOF { { statements = it } }
 ;

statement:
 | COLON*; count = COUNT?; cmd = command { AST.make_statement ~count ~cmd }
 ;

command:
 | it = IDENTIFIER { it }
 ;

break:
 | SEMICOLON { }
 ;

%%

Menhir 的 --explain 标志生成了对由此产生的 shift/reduce 冲突的描述。不幸的是，我无法弄清它的正反面:

** Conflict (shift/reduce) in state 3.
** Token involved: SEMICOLON
** This state is reached from script after reading:

statement 

** The derivations that appear below have the following common factor:
** (The question mark symbol (?) represents the spot where the derivations begin to differ.)

script 
(?)

** In state 3, looking ahead at SEMICOLON, shifting is permitted
** because of the following sub-derivation:

loption(separated_nonempty_list(break,statement)) option(break) EOF 
separated_nonempty_list(break,statement) 
statement break separated_nonempty_list(break,statement) 
          . SEMICOLON 

** In state 3, looking ahead at SEMICOLON, reducing production
** separated_nonempty_list(break,statement) -> statement 
** is permitted because of the following sub-derivation:

loption(separated_nonempty_list(break,statement)) option(break) EOF // lookahead token appears because option(break) can begin with SEMICOLON
separated_nonempty_list(break,statement) // lookahead token is inherited
statement .

我花了一个晚上试图通过文档挖掘 shift/reduce 冲突实际上是，但我不得不承认我真的很难花时间理解我正在阅读的内容。有人可以给我一个关于 shift/reduce 冲突的简单（嗯，尽可能多的）解释吗？具体使用上面例子的上下文？

Answer 1

问题是，在查看分号时，解析器无法决定它是否应该期待 EOF 或列表的其余部分。原因是您使用 break 作为可选的终止符，而不是分隔符。

我建议您更改主要规则：

script:
 | it = optterm_list(break, statement); EOF { { statements = it } }
 ;

然后自己定义 optterm_list 组合子，像这样：

optterm_list(separator, X):
  | separator? {[]}
  | l=optterm_nonempty_list(separator, X) { l } 
optterm_nonempty_list(separator, X):
  | x = X separator? { [ x ] }
  | x = X
    separator
    xs = optterm_nonempty_list(separator, X)
     { x :: xs }

为什么我的 EOF / linebreak 非终结符总是有 shift/reduce 冲突？

Why do I always have a shift/reduce conflict with my EOF / linebreak nonterminals?

parsing

ocaml

shift-reduce-conflict

menhir