我有一个手写的LL1解析器。我的AST并没有尽可能简化。语句部分如下所示:
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]
如您所见,我仍然保留许多不必要的信息。我想像这样建立自己的声明:
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
这样做我有点迷茫,因为我真的是用书本来构建我的LL1解析器的(我相信不会期望很长的语法):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令牌表, ast。
我认为,为了像我的目标语句AST中那样构建Block类型,我需要在递归parseStmt方法中构建一个语句列表。我已将解析器代码简化为仅调用parseStmtList的解析器方法以及它们调用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
但是,我遇到了错误:
Error: This expression has type 'a -> token_list * Ast.stmtlist
but an expression was expected of type 'b * 'c
对于let (tokenlist_stmt_list, stmt_list) = tokenlist_stmt new_lst |> parseStmtList in
中的行parseStmtList
答案 0 :(得分:1)
tokenlist_stmt new_lst |> parseStmtList
这里,您要将tokenlist_stmt
应用于自变量new_lst
,然后将parseStmtList
应用于结果。但是tokenlist_stmt
实际上不是函数,因此这是类型错误。
大概您打算用parseStmtList
和tokenlist_stmt
作为两个参数来调用new_lst
。语法很简单:
parseStmtList tokenlist_stmt new_lst
另外,lst::stmt
也是类型错误,原因有两个:
::
将列表作为其右侧操作数,而不是左侧,因此应为stmt::lst
lst
实际上不是列表,而是Ast.Block
,因为这是parseStmtList
返回的内容。修复所有问题后,您会注意到该列表将以错误的方式出现(大概是因为这是您首先尝试lst::stmt
的原因,但是您无法在列表末尾添加内容)像这样的列表)。在使用累加器构建列表时,这是一个常见问题。解决方案是要么在构建完列表后就反转列表,要么首先不使用累加器。
需要指出的一件事是,在使用Ast.stmtlist
时,所有这些问题也将适用。也就是说,如果您的代码如下所示:
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))
然后您将得到完全相同的错误。这使我认为,您已更改了不必要的代码方式。由于您的旧代码可能正常工作,因此我假设它看起来像这样:
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,您拥有:
let (tokenlist_stmtlist, stmtlist) = parseStmtList tokenlist_leftbrace in
begin
match tokenlist_expr.head with
| Lexer.RightBrace -> (next tokenlist_stmtlist, Ast.Block(stmtlist))
现在,除去Ast.stmtlist
之后,您需要更改的是实际使用其构造函数的部分,并用列表构造函数(::
和[]
替换那些部分。因此,parseStmt
中的代码将完全保持不变,parseStmtList
中的唯一更改应该是替换行
| Lexer.RightBrace -> (tokenlist, Ast.StmtlistNil )
使用
| Lexer.RightBrace -> (tokenlist, [] )
和行
(tokenlist_stmt_list, Ast.StmtList (stmt, stmt_lst))
使用
(tokenlist_stmt_list, stmt :: stmt_lst)
如果您的旧代码看起来与上面的代码有所不同,则可能必须更改不同的行,但想法仍然相同:将Ast.StmtList
替换为::
和Ast.StmtListNil
与[]
。
就是这样。这就是所有必要的更改。您太复杂了。