Question

我正在学习编程语言和编程语言的概念。我怎么翻译呢？

问题：

   <S> → (+ | -)[<C>]{<A>}

起初，我翻译成这样：

<S> -> epsilon
     | +<S>
     | -<S>
     | <C><S>
     | <A><S>

然而，它有一个问题，它可以重现C！

Answer 1

您的BNF版本不仅可以生成多个<C>，还可以生成多个+和- s（或者零，这也与原始语法不同）。< / p>

要解决这些问题，您需要引入其他非终端。具体来说，您可以使用仅匹配{<A>}的匹配和匹配[<C>]的匹配。您对<S>的定义可能是：

<S> -> + <OptionalC> <RepeatedA>
     | - <OptionalC> <RepeatedA>

Answer 2

Your EBNF original is not recursive in <S>, but your trial BNF is. If you want the BNF to generate the same language, you shouldn't do that. Also, your EBNF can't generate epsilon, so you shouldn't add that to your <S>-production, either.

As the other answer says, you need to introduce additional non-terminals:

<S> -> <plusminus> <optionalC> <repeatedA>
<plusminus> -> + | -
<optionalC> -> <C> | epsilon
<repeatedA> -> <A> <repeatedA> | epsilon

This is a straightforward translation of the EBNF elements in your original grammar. Even if you have additional requirements (such as epsilon-elimination), you should start with this kind of step.

Note that the <S> rule is not recursive. Also, although epsilon is used for some additional nonterminals, it is not part of the <S> or <plusminus> rules, so that, as in the EBNF original, <S> cannot produce it.

我正在努力从EBNF到BNF的翻译

2 个答案: