it 语句的 LL(1) 语法是什么？

Question

I want to convert LALR(1) grammar to LL(1) grammar especially if statement.

I had used LALR(1) grammar for if statement

IF_Stmt -> Matched|Unmatched.
Matched -> if ( Expr_IF ) Matched else Matched | Other.
Unmatched -> if ( Expr_IF ) Matched
             |if ( Expr_IF ) Matched else Unmatched.

Here, I had realized that i have to remove Left factoring.

After that i couldn't solve epsilon production problem...

What is the LL(1) grammar for if-statement ?

Answer 1

IF_Stmt -> if ( Expr_IF ) Stmt Optional_Else_Stmt
Optional_Else_Stmt -> (empty-string)
                      |else Stmt.

Answer 2

Stmt -> if ( Expr ) Matched Optional_Else_Tail
      | Other

Matched -> if ( Expr ) Matched else Matched
         | Other

Optional_Else_Tail -> else Stmt
                    | ""