FLEX：有没有办法一次返回多个令牌

In flex, I want to return multiple tokens for one match of a regular expression. Is there a way to do this?

在flex中，我想为正则表达式的一个匹配返回多个标记。有没有办法做到这一点？

3 个解决方案

#1

The way I've been doing this is to create a queue of to-be-returned tokens, and at the beginning of yylex(), check for tokens and return them.

我这样做的方法是创建一个待返回令牌的队列，并在yylex（）的开头，检查令牌并返回它们。

#2

~~Do you mean all matches? Are you using regex functions or string functions? Use the global flag.~~

你的意思是所有的比赛？你在使用正则表达式函数或字符串函数吗？使用全局标志。

As for flex, I don't think you can do that. You test for a match with one pattern at a time so that's probably out of scope. Why'd you want that? As an optimization? Scoping issues?

至于flex，我认为你不能这样做。您一次测试一个模式的匹配，以便可能超出范围。你为什么要这样？作为优化？范围问题？

#3

-1

Usually, this is handled by a parser on top of the scanner which gives you much cleaner code. You can emulate that to some degree with states:

通常，这由扫描程序顶部的解析器处理，为您提供更清晰的代码。您可以在某种程度上模仿州：

%option noyywrap

%top {
#define TOKEN_LEFT_PAREN    4711
#define TOKEN_RIGHT_PAREN   4712
#define TOKEN_NUMBER        4713
}

%x PAREN_STATE
%%
"("         BEGIN(PAREN_STATE); return TOKEN_LEFT_PAREN;
<PAREN_STATE>{
   [0-9]+   return TOKEN_NUMBER;
   ")"      BEGIN(INITIAL); return TOKEN_RIGHT_PAREN;
   .|\n     /* maybe signal syntax error here */
}
%%
int main (int argc, char *argv [])
{
  int i;

  while ((i = yylex ()))
    printf ("%d\n", i);

  return 0;
}

but this will get very messy as soon as your grammar gets more complex.

但是一旦语法变得更复杂，这将变得非常混乱。

#1