是否有可能获得一个默认的令牌不覆盖ANTLR3的已经存在的令牌？ [英] Is it possible to get a default token which doesn't overrides the already present tokens in ANTLR3?

问题描述

我有其中我已经写了词法令牌作为一项要求：

I have a requirement wherein I have written the Lexer token as:

IF_LEXER_TOKEN: 'IF' (.)* 'END_IF'
ANY :(options {greedy=true;}: .)* ;

但是，如果输入被给定为：

But if the input is given as:

IF a>b then a=b END_IF
IF c>d then c=d

在这种情况下，预期的行为是它应该使用标记 IF_LEXER_TOKEN 的第一行和第二行中的任何标记，而是其考虑两条线的任何标记。请帮助。注：由于一些限制，我不能创建对于上述方案解析器规则

In this case the expected behavior is that it should use the token IF_LEXER_TOKEN for first line and ANY token for second line, but instead its considering the ANY token for both lines. Kindly help. Note:Due to some constraints I can't create a parser rule for the above scenario.

推荐答案

我的答案是：不要'吨做你想做什么。切勿使用 * 或 + 在词法规则的末尾：。它消耗的一切，直到 EOF 。

My answer would be: don't do what you're trying to do. Never use .* or .+ at the end of a lexer rule: it consumes everything until EOF.

我不知道你想要做的事情，但在我看来，你正在试图获得回答问题是 ¹ ，而你应该被解释问题 X ¹

I don't know exactly what you're trying to do, but it seems to me you're trying to get an answer to problem Y¹, while you ought to be explaining problem X¹.

¹ 的 http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem

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