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我有一个简单的递归规则:

i_stmt:
  | CHAIN LPAREN c=separated_nonempty_list(i_stmt, COMMA) RPAREN {Chain c}
  | ASSIGN LPAREN n=i_var COMMA e=i_expr RPAREN {Assign (n,e)}
  | CRETURN LPAREN i=i_expr RPAREN { Return i }
  ;

Menhir 编译成功,但 ocamlc 抱怨生成的代码:

File "parser.mly", line 38, characters 72-73:
Error: This expression has type unit list
       but an expression was expected of type Ast.istmt list
       Type unit is not compatible with type Ast.istmt 

类型定义:

type istmt =
  | Chain of (istmt list)
  | Assign of ivar*iexpr
  | Return of iexpr

我试图添加:

%type <Ast.istmt> i_stmt

但这也无济于事。我究竟做错了什么?

4

1 回答 1

4

menhir手册:

separated_nonempty_list(sep,X)用' Xs 分隔的'的非空序列sep

所以你应该写

| CHAIN LPAREN c=separated_nonempty_list(COMMA, i_stmt) RPAREN {Chain c}
于 2017-05-19T05:52:58.327 回答