可可/ r：可删除因子

Question

我正在尝试在Coco / r中为C＃中的算术运算实现一种语言，该语言考虑了运算符的优先级。 我的ATG代码如下所示：

/* Coco/R lexer and parser specification for arithmetic expressions. */
/* 2006-09-14 */

/* Build with:
 *   Coco.exe -namespace Expressions Ex2.ATG
 */

using System.Collections.Generic;

COMPILER Expressions
  public int res;

/*--------------------------------------------------------------------------*/
CHARACTERS
  letter = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".
  digit = "0123456789".
  cr  = '\r'.
  lf  = '\n'.
  tab = '\t'.

TOKENS
  ident  = letter {letter | digit}.
  number = digit {digit}.
IGNORE cr + lf + tab

PRODUCTIONS
/*------------------------------------------------------------------------*/
Expr<out int n>   (. int n1, n2; .)
= Term<out n1>  (. n = n1; .)
{
  '+' Term<out n2>  (. n = n+n2; .)  
  | 
  '-' Term<out n2>  (. n = n-n2; .)    
  |
  Factor<out int n>
}
.
Factor<out int n>
=
{
  "==" Term<out n2> (. if(n1 == n2){ n = 1; } else { n = 2; } .)
  | 
  '<' Term<out n2>  (. if(n1 < n2) { n = 1; } else { n = 0; } .)
  | 
  '>' Term<out n2>  (. if(n1 > n2) { n = 1; } else { n = 0; } .)
  | 
  "!=" Term<out n2> (. if(n1 != n2){ n = 1; } else { n = 0; } .) 
  | 
  "<=" Term<out n2> (. if(n1 <= n2){ n = 1; } else { n = 0; } .)
  | 
  ">=" Term<out n2> (. if(n1 >= n2){ n = 1; } else { n = 0; } .)
  |
  "|" Term<out n2>  (. if(n1 != 0 | n2 != 0) { n = 1; } else { n = 0; } .)
  |
  "&" Term<out n2>  (. if(n1 != 0 & n2 != 0){ n = 1; } else { n = 0; } .)
}
.
Term<out int n> 
= number          (. n = Convert.ToInt32(t.val); .)
{
  '*' number  (. n = n*Convert.ToInt32(t.val); .) 
}
.

Expressions                        (. int n; .)
= Expr<out n>                (. res = n; .)

.


END Expressions.

除“ +”和“-”以外的其他运算符应具有较低的优先级。 此外，“＆”运算符的优先级应低于“ |”。

问题是，当我尝试测试代码时，出现以下错误：

Factor deletable
  LL1 warning in Expr: contents of [...] or {...} must not be deletable
  LL1 warning in Expr: "+" is start of several alternatives
  LL1 warning in Expr: "-" is start of several alternatives
  LL1 warning in Factor: "==" is start & successor of deletable structure
  LL1 warning in Factor: "<" is start & successor of deletable structure
  LL1 warning in Factor: ">" is start & successor of deletable structure
  LL1 warning in Factor: "!=" is start & successor of deletable structure
  LL1 warning in Factor: "<=" is start & successor of deletable structure
  LL1 warning in Factor: ">=" is start & successor of deletable structure
  LL1 warning in Factor: "|" is start & successor of deletable structure
  LL1 warning in Factor: "&" is start & successor of deletable structure

我真的是Coco / r和EBNF的新手。 我看了一下Coco \\ r的手册，但我并没有真正看到问题所在。 我想念什么？

先感谢您！

Answer 1

我认为，在Factor而不是

Factor<out int n>
=
{
  "==" Term<out n2> (. if(n1 == n2){ n = 1; } else { n = 2; } .)
  | 
...
}

你真的想要像

Factor<out int n>
=
  Term<out n1>
[
  "==" Term<out n2> (. if(n1 == n2){ n = 1; } else { n = 2; } .)
  | 
...
]

也就是说，您要无条件地要求Lead Term ，然后可以选择只跟一个关系。 否则，您将允许类似a < b > c == d语句。

Answer 2

Coco / R是LL（1）解析器。 这些错误基本上是在告诉您您编写的语法无法通过在前面看1个符号来解决。 您要么需要重构语法，要么提供冲突解决程序。 我会尝试重构语法，因为我认为您不需要在这里解决冲突。 有关更多详细信息，请参见“冲突”用户手册。