Question

如果你有这样的语法：

<assign> → <id> = <expr> 
<id> → A | B | C 
<expr> → <expr> + <term> 
         |  <term> 
<term> → <term> * <factor> 
         |  <factor> 
<factor> → ( <expr> ) 
          | <id>

然后句子A = B + C * A，你得到这个最左边的推导：

<assign> => <id> = <expr> 
         => A = <expr> 
         => A = <expr> + <term> 
         => A = <term> + <term> 
         => A = <factor> + <term> 
         => A = <id> + <term> 
         => A = B + <term> 
         => A = B + <term> * <factor> 
         => A = B + <factor> * <factor> 
         => A = B + <id> * <factor> 
         => A = B + C * <factor> 
         => A = B + C * <id> 
         => A = B + C * A

但是A = B +（C * A）怎么样？

Answer 1

A = B +（C * A）？

前五个步骤，与上述相同，然后......

     => A = B + <term>  
     => A = B + <factor>    
     => A = B + ( <expr>) 
     => A = B + ( <term> )  
     => A = B + ( <term> * <factor> ) 
     => A = B + ( <factor> * <factor> ) 
     => A = B + ( <id> * <id> ) 
     => A = B + (  C   *   A  )

Answer 2

( C * A )不需要paren，因为*具有更高的优先级。您可以在A = B * ( C + B )中找到一个案例。

您在最后两行中没有看到它，因为<factor>将变为<term> + <term>或<id>。在这种情况下，没有+所以它必须是<id>。

什么时候在EBNF中使用括号？

2 个答案: