我有一个手写的LL1解析器。我的AST并没有尽可能简化。语句部分如下:
type stmt_opt = StmtExpression of assignment | OptNil
[@@deriving show]
(*stmt_list -> stmt stmt_list | ε *)
type stmtlist =
| StmtList of stmt * stmtlist
| StmtlistNil
[@@deriving show]
and stmt =
| Assignment of assignment
| Return of stmt_opt
| Parentheses of stmtlist
| If of assignment * stmt
| For of assignment * assignment * assignment * stmt
| While of assignment * stmt
(*“lparen” formals_opt “rparen” “LBRACE” vdecl_list stmt_list “RBRACE”*)
[@@deriving show]
正如你所看到的,我仍然掌握着很多不必要的信息。我想把我的声明建立成这样:
type stmt =
Block of stmt list
| Expr of expr
| Return of expr
| If of expr * stmt * stmt
| For of expr * expr * expr * stmt
| While of expr * stmt
我有点不知所措,因为我真的是根据书本构建了我的LL1解析器(我相信这些书预计不会有很长的语法(:每个非终结符都有一个解析方法,每个解析方法都返回一个标记登记和一个ast。
我认为,为了构建像我的目标语句AST中那样的Block类型,我需要在递归parseStmt方法中构建一个语句列表。我已经将我的解析器代码简化为只调用parseStmtList的解析器方法和它们调用parseStmtList 的特定实例
(*stmt_list = stmt stmt_list | epsilon*)
let rec parseStmtList tokenlist lst =
match tokenlist.head with
| Lexer.RightBrace -> (tokenlist, Ast.StmtlistNil )
| _ -> let (tokenlist_stmt, stmt) = parseStmt tokenlist in
let new_lst = lst::stmt in
let (tokenlist_stmt_list, stmt_list) = tokenlist_stmt new_lst |> parseStmtList in
(tokenlist_stmt_list, Ast.Block(stmt_lst))
(*stmt -> assignment SEMI
| RETURN stmt_opt SEMI
| LBRACE stmt_list RBRACE
| IF LPAREN assignment RPAREN stmt
| FOR LPAREN assignment SEMI assignment SEMI assignment RPAREN stmt
| WHILE LPAREN assignment RPAREN stmt
*)
and parseStmt tokenlist =
begin
match tokenlist.head with
| Lexer.ID identifier -> let (tokenlist_assignment, assignment) = parseAssignment tokenlist in
begin
match tokenlist_assignment.head with
| Lexer.Semicolon -> (next tokenlist_assignment, Ast.Assignment(assignment))
| _-> let err_msg = __LOC__ ^ "Syntax Error semicolon expected but received" ^ show_token_list tokenlist in
raise (Syntax_error err_msg)
end
| Lexer.LeftBrace -> let tokenlist_leftbrace = next tokenlist in
let (tokenlist_expr, expr) = parseStmtList tokenlist_leftbrace [] in
begin
match tokenlist_expr.head with
| Lexer.RightBrace -> (next tokenlist_expr, Ast.Parentheses(expr))
| _-> let err_msg = __LOC__ ^ "Syntax Error right brace expected but received" ^ show_token_list tokenlist in
raise (Syntax_error err_msg)
end
| _-> let err_msg = __LOC__ ^ "Syntax Error left brace expected but received" ^ show_token_list tokenlist in
raise (Syntax_error err_msg)
end
然而,我得到了错误:
Error: This expression has type 'a -> token_list * Ast.stmtlist
but an expression was expected of type 'b * 'c
对于parseStmtList
中的线路let (tokenlist_stmt_list, stmt_list) = tokenlist_stmt new_lst |> parseStmtList in
tokenlist_stmt new_lst |> parseStmtList
这里,将tokenlist_stmt
应用于参数new_lst
,然后将parseStmtList
应用于结果。但是tokenlist_stmt
实际上并不是一个函数,所以这是一个类型错误。
假设您的意图是用tokenlist_stmt
和new_lst
作为其两个自变量来调用parseStmtList
。语法很简单:
parseStmtList tokenlist_stmt new_lst
此外,lst::stmt
也是一个类型错误,原因有两个:
::
将列表作为其右操作数,而不是左操作数,因此它应该是stmt::lst
lst
实际上不是一个列表,它是一个Ast.Block
,因为这是parseStmtList
返回的内容
一旦你修复了所有这些,你会注意到列表将是错误的(大概这就是你最初尝试lst::stmt
的原因,但你不能像那样附加到列表的末尾(。当使用累加器构建列表时,这是一个常见的问题。解决方案是,一旦构建完成,就反转列表,或者一开始就不使用累加器。
需要指出的一件事是,当使用Ast.stmtlist
时,所有这些问题都会适用。也就是说,如果你的代码看起来像这样:
let new_lst = Ast.StmtList(lst, stmt) in
let (tokenlist_stmt_list, stmt_list) = tokenlist_stmt new_lst |> parseStmtList in
(tokenlist_stmt_list, Ast.Block(stmt_lst))
然后你会得到完全相同的错误。这让我觉得,你已经改变了比你需要的更多的代码。由于你的旧代码可能有效,我认为它看起来像这样:
let rec parseStmtList tokenlist =
match tokenlist.head with
| Lexer.RightBrace -> (tokenlist, Ast.StmtlistNil )
| _ -> let (tokenlist_stmt, stmt) = parseStmt tokenlist in
let (tokenlist_stmt_list, stmt_list) = parseStmtList tokenlist_stmt in
(tokenlist_stmt_list, Ast.StmtList (stmt, stmt_lst))
然后在parseStmt
中,您有:
let (tokenlist_stmtlist, stmtlist) = parseStmtList tokenlist_leftbrace in
begin
match tokenlist_expr.head with
| Lexer.RightBrace -> (next tokenlist_stmtlist, Ast.Block(stmtlist))
现在,在删除Ast.stmtlist
之后,您只需要更改实际使用其构造函数的部分,并将这些部分替换为列表构造函数(::
和[]
(。因此,parseStmt
中的代码将完全保持不变,parseStmtList
中唯一的变化应该是替换行
| Lexer.RightBrace -> (tokenlist, Ast.StmtlistNil )
带有
| Lexer.RightBrace -> (tokenlist, [] )
和线路
(tokenlist_stmt_list, Ast.StmtList (stmt, stmt_lst))
带有
(tokenlist_stmt_list, stmt :: stmt_lst)
如果您的旧代码看起来与我上面提出的不同,您可能需要更改不同的行,但想法保持不变:用::
替换Ast.StmtList
,用[]
替换Ast.StmtListNil
。
就这样。这就是所有必要的改变。你太复杂了。