ANTLR3 c目标：当规则失败时，操作仍然执行，我可以'；我不明白

当规则失败时，操作仍然执行，为什么？ 这是我的测试：

grammar Test;
options{
    language = C;   
}
@parser::includes
{
    #include <iostream>
}
@lexer::includes
{
    #include <iostream>
}

rScript
    :   INT LR RR{std::cout << "Action Run...\n";};

INT :
    ('0'..'9')+;
LR  :   '{';
RR  :   '}';

语法测试；
选择权{
语言=C；
}
@解析器：：包含
{
#包括
}
@lexer:：includes
{
#包括
}
rScript
：INT LR RR{std:：cout rScript（psr）；
int parserrorcount=psr->pParser->rec->state->errorCount；
int-lexerrorcount=Lex->pLexer->rec->state->errorCount；
std:：cout[offtopic]为什么使用ANTLR3而不是ANTLR4？
int main()
{
    char teststr1[] = "111dd}";
    pANTLR3_INPUT_STREAM    input;
    input = antlr3StringStreamNew((pANTLR3_UINT8)teststr1, ANTLR3_ENC_8BIT, sizeof(teststr1) - 1, (pANTLR3_UINT8)"-memory-");

    pTestLexer Lex = TestLexerNew(input);
    pANTLR3_COMMON_TOKEN_STREAM tstream = antlr3CommonTokenStreamSourceNew(
        ANTLR3_SIZE_HINT, TOKENSOURCE(Lex));
    pTestParser psr = TestParserNew(tstream);
    psr->rScript(psr);

    int ParserErrorCount = psr->pParser->rec->state->errorCount;
    int LexerErrorCount = Lex->pLexer->rec->state->errorCount;
    std::cout << "Parser Error Count: " << ParserErrorCount << "\n";
    std::cout << "Lexer Error Count: " << LexerErrorCount << "\n";
    getchar();
    return 0;
}

-memory-(1) : lexer error 3 :
         at offset 4, near 'd' :
        dd}
-memory-(1) : lexer error 3 :
         at offset 5, near 'd' :
        d}
-memory-(1)  : error 10 : org.antlr.runtime.MissingTokenException, at offset 5
    near [Index: 0 (Start: 0-Stop: 0) ='<missing LR>', type<5> Line: 1 LinePos:5]
     : Missing LR
Action Run...
Parser Error Count: 1
Lexer Error Count: 2