ANTLR4解析时互左递归错误 [英] ANTLR4 mutually left-recursive error when parsing

问题描述

我有这个 ANTLR 4 语法:

I have this ANTLR 4 grammar:

constantFixedExpresion : term (('+'|'-') term)+;

term : factor (('*'|'//'|'REM')factor)+;

factor : ('+'|'-')*
           ( wholeNumberConstant
           | constantFixedExpresion
           | 'TOFIXED' (stringConstant | bitCodeConstant)      
           | identifier)
         ('FIT'constantFixedExpresion)*;

我收到以下错误:

error(119): LanguageA.g4::: 以下规则集是相互左递归的 [constantFixedExpresion, factor, term]

error(119): LanguageA.g4::: The following sets of rules are mutually left-recursive [constantFixedExpresion, factor, term]

我尝试了很多方法，但无法解决.是什么问题，我该如何解决?

I tried so many ways but can't fix it. What is the problem and how can I solve it?

推荐答案

Antlr 是一个 LL(*) 解析器，它在很多方面都比 LL(k) 解析器，但仍有许多缺点.其中之一是它无法处理左递归(实际上，版本 4 可以在同一规则内处理左递归).错误是说你有一个语法的左递归，这是 LL 解析器的祸根.

Antlr is a LL(*) parser, which is in many ways "better" than a LL(k) parser, but still has many of its disavantages. One of these being the fact it can't deal with left-recursion (in fact, the version 4 can deal with left-recursion within the same rule). What the error is saying is that you have a left-recursion of a grammar, a bane for the LL parsers.

这是由语法中的这种结构引起的:

This is caused by this construction in your grammar:

constantFixedExpression: term ...;
term: factor ...;
factor: ('+' | '-')* (constantFixedExpression | ...) ...;

由于*操作符表示0或更多，我可以用0实例化它，所以解析器会这样做:尝试constantFixedExpression，所以它需要尝试term，所以它需要尝试 factor，所以它需要尝试 constantFixedEXpression，所以它 [...]无限循环.

Since the * operator means 0 or more, I can instantiate it with 0, so the parser will do this: "try constantFixedExpression, so it needs to try term, so it needs to try factor, so it needs to try constantFixedEXpression, so it [...]" and you've got yourself an infinite loop.

幸运的是，上下文无关的形式语法有一个等效的转换来消除左递归！它可以一般地表示为:

Fortunately, context-free formal grammars have an equivalent transformation for removing left-recursion! It can be expressed generically by:

A -> Aa | b
-- becomes --
A -> bR
R -> aR | ε

或者用 Antlr 表示法:

Or in Antlr notation:

A: Aa | b;
// becomes
A: bR;
R: (aR)?;

有关此过程的更多信息可以在自动机/语法书籍或维基百科中找到.

More information about this process can be found in automaton/grammars books or in the Wikipedia.

我将通过重构来纠正您的语法，以消除左递归作为您的工作.不过，我想再谈一点:Antlr 4 可以做左递归！正如我提到的，版本 4 可以在相同的规则内处理左递归.在 Antlr4 中，除了直接在解析中指示运算符的优先级和关联性之外，还有其他方法可以指示运算符的优先级和关联性.让我们看看它是如何工作的:

I'll leave correcting your grammar with the refactoration to remove left-recursion as your work. However, I want to touch in another point: Antlr 4 can do left-recursion! As I mentioned, the version 4 can deal with left-recursion within the same rule. There are ways to indicate precedence and associativity of operators other than directly in parsing, as you're doing, in Antlr4. Let's see how it works:

expr: NUMBER
      |<assoc=right> expr '^' expr
      | expr '*' expr
      | expr '/' expr
      | expr '+' expr
      | expr '-' expr;

这是一个基本的计算器语法示例.顶部的运算符优先级最高，底部的运算符优先级较低.这意味着 2+2*3 将被解析为 2+(2*3) 而不是 (2+2)*3.<assoc=right> 结构意味着运算符处于右结合，所以 1^2^3 将被解析为 1^(2^3) 而不是 (1^2)^3.

This is an example of a basic calculator grammar. The operators at the top are those with highest precedence, and those at the bottom are of lower precedence. This means 2+2*3 will be parsed as 2+(2*3) rather than (2+2)*3. The <assoc=right> construction means the operator in right-associative, so 1^2^3 will be parsed as 1^(2^3) rather than (1^2)^3.

如您所见，使用左递归指定运算符要容易得多，因此 Antlr 4 在这些时刻有很大帮助！我建议您重新编写语法以利用此功能.

As you can see, it is much easier to specify operators with left-recursion, so Antlr 4 is of big help in these moments! I recommend re-writing your grammar to make use of this feature.

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