[antlr-interest] Syntaxerror not found if first symbol is wrong

[Oliver B. Fischer]

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Hi Ben,

I am reading the input from ANTLRStringStream, so I don't get an EOF.
How to fix this?

Bye

Oliver

Benjamin Niemann schrieb:
| Hi Oliver,
|
| If the very first token is an ID, the code will throw an
| EarlyExitException - doesn't it? (alt1 sticks to its default value 2,
| which is handled by the default case in the switch, which in turn will
| raise the exception, because cnt1 is 0)
|
| If you have a sequence INT ID ';' ID, then the loop will exit after
| the semicolon, and the parser will not complain about the dangling ID.
| If that's what the problem is, then you can solve it by using
|
| program : (a b ';')+  EOF;
|
| -Ben
|
| On Sun, Aug 10, 2008 at 6:11 PM, Oliver B. Fischer
<o.b.fischer at gmx.de> wrote:
| Hello,
|
| my grammar is not able to recognize an syntax error if already the first
| ~ symbol found is not an expected one:
|
| My grammar looks like this:
|
| INT : 'int' ;
|
| ID : ('a'..'z')+ ;
|
| WS : (' '|'\t'|'\n'|'\r')+ { $channel=HIDDEN; };
|
| program : (a b ';')+ ;
|
| a : INT ;
|
| b : ID ;
|
| So, any valid input must start with 'int', but if the first symbol found
| by the lexer is an ID, the generated parser does not recognize the error.
|
| ANTLR generates the following code:
| public final void program() throws RecognitionException {
| ~  try {
| ~    {
| ~      int cnt1 = 0;
| ~      loop1:
| ~      do {
| ~        int alt1 = 2;
| ~        int LA1_0 = input.LA(1);
|
| ~        if ((LA1_0 == INT)) {
| ~          alt1 = 1;
| ~        }
|
|
| ~        switch (alt1) {
| ~          case 1: {
| ~            pushFollow(FOLLOW_a_in_program154);
| ~            a();
| ~            _fsp--;
|
| ~            pushFollow(FOLLOW_b_in_program156);
| ~            b();
| ~            _fsp--;
|
| ~            match(input, SEM, FOLLOW_SEM_in_program158);
|
| ~          }
| ~          break;
|
| ~          default:
| ~            if (cnt1 >= 1) break loop1;
| ~            EarlyExitException eee =
| ~                    new EarlyExitException(1, input);
| ~            throw eee;
| ~        }
| ~        cnt1++;
| ~      } while (true);
|
|
| ~    }
|
| ~  }
|
| ~  catch (RecognitionException e) {
| ~    throw e;
| ~  }
| ~  finally {
| ~  }
| ~  return;
| }
|
| So, the ID token falls through the switch-statement. How can I avoid this?
|
| Thank you for your help!
|
| Bye
|
| Oliver
|
|
|
|>

- --
Oliver B. Fischer, Schönhauser Allee 64, 10437 Berlin
Tel. +49 30 44793251, Mobil: +49 178 7903538
Mail: o.b.fischer at gmx.de Blog: http://www.sw-blog.net
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