如何简化JavaScript / ECMAScript数组文字的生成？ [英] How to simplify JavaScript/ECMAScript array literal production?

问题描述

我目前正在实现JavaScript / ECMAScript 5.1 使用JavaCC解析器，并且 ArrayLiteral 生产。

  ArrayLiteral：
 [Elision_opt] 
 [ElementList] 
 [ElementList， Elision_opt] 
 
 ElementList：
 Elision_opt AssignmentExpression 
 ElementList，Elision_opt AssignmentExpression 
 
 Elision：
，
 Elision，
  

我有三个问题，我将一个接一个地问他们。

---

我试图简化/重写上面描述的 ArrayLiteral 生产，最后到达以下生产（伪语法）：

  ArrayLiteral：
 [（， | AssignmentExpression，）* Assi gnmentExpression？ ] 
  

我的第一个问题：这个重写正确吗？

另外两个折弯：

• 用于JavaScript / ECMAScript数组文字生产的LOOKAHEADs


• 如何为广告系列实施否定的LOOKAHEAD检查是JavaCC中的令牌吗？

解决方案

是的，可以正确捕获所呈现的语法。

但是，更好的重写方法是：

  [ AssignmentExpression？ （， AssignmentExpression？）*] 
  

因为在OP中的重写不是LL（ 1）-您必须先阅读完整的 AssignmentExpression 才能区分各种可能性，而通过阅读第一个令牌，您就可以找出使用哪种选择。 / p>

I currently implementing a JavaScript/ECMAScript 5.1 parser with JavaCC and have problems with the ArrayLiteral production.

ArrayLiteral :
    [ Elision_opt ]
    [ ElementList ]
    [ ElementList , Elision_opt ]

ElementList :
    Elision_opt AssignmentExpression
    ElementList , Elision_opt AssignmentExpression

Elision :
    ,
    Elision ,

I have three questions, I'll ask them one by one.

---

I have tried to simplify/to rewrite the ArrayLiteral production depicted above and finally arrived to the following production (pseudo-grammar):

ArrayLiteral:
    "[" ("," | AssignmentExpression ",") * AssignmentExpression ? "]"

My first question: is this rewriting correct?

Two other quetsions:

• LOOKAHEADs for the JavaScript/ECMAScript array literal production
• How to implement a negative LOOKAHEAD check for a token in JavaCC?

解决方案

Yes, that correctly captures the grammar presented.

However, a better rewrite would be:

"[" AssignmentExpression ? ( "," AssignmentExpression ? ) * "]"

because the rewriting in the OP is not LL(1) -- you cannot distinguish the possibilities without reading the entire AssignmentExpression -- whereas with this one you can figure out which alternative to use simply by looking at the first token.

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