我的词法分析器中定义了一个CharacterString(例如'abcd'),我有以下标记定义:
CharacterString:
Apostrophe
(Alphanumeric)*
Apostrophe
;
是否可以忽略两个撇号,然后才能在词法分析器中获取没有它们的标记字符串(通过$ CharacterString.text->字符)?
我试过......
CharacterString:
Apostrophe { $channel = HIDDEN; }
(Alphanumeric)*
Apostrophe { $channel = HIDDEN; }
;
...没有成功......这个案例甚至不再匹配我的字符串了(例如'oiu'将在解析器中失败 - Missmatched Set Exception)。
谢谢:)
答案 0 :(得分:2)
内联代码{$channel=HIDDEN;}
会影响整个CharacterString
,因此您无法像尝试的那样执行此操作。
您需要添加一些自定义代码并自行删除引号。这是一个小C演示:
grammar T;
options {
language=C;
}
parse
: (t=. {printf(">\%s<\n", $t.text->chars);})+ EOF
;
CharacterString
: '\'' ~'\''* '\''
{
pANTLR3_STRING quoted = GETTEXT();
SETTEXT(quoted->subString(quoted, 1, quoted->len-1));
}
;
Any
: .
;
和一点测试功能:
#include "TLexer.h"
#include "TParser.h"
int main(int argc, char *argv[])
{
pANTLR3_UINT8 fName = (pANTLR3_UINT8)"input.txt";
pANTLR3_INPUT_STREAM input = antlr3AsciiFileStreamNew(fName);
if(input == NULL)
{
fprintf(stderr, "Failed to open file %s\n", (char *)fName);
exit(1);
}
pTLexer lexer = TLexerNew(input);
if(lexer == NULL)
{
fprintf(stderr, "Unable to create the lexer due to malloc() failure1\n");
exit(1);
}
pANTLR3_COMMON_TOKEN_STREAM tstream = antlr3CommonTokenStreamSourceNew(ANTLR3_SIZE_HINT, TOKENSOURCE(lexer));
if(tstream == NULL)
{
fprintf(stderr, "Out of memory trying to allocate token stream\n");
exit(1);
}
pTParser parser = TParserNew(tstream);
if(parser == NULL)
{
fprintf(stderr, "Out of memory trying to allocate parser\n");
exit(ANTLR3_ERR_NOMEM);
}
parser->parse(parser);
parser->free(parser); parser = NULL;
tstream->free(tstream); tstream = NULL;
lexer->free(lexer); lexer = NULL;
input->close(input); input = NULL;
return 0;
}
并且测试input.txt
文件包含:
'abc'
如果您现在1)生成词法分析器和解析器,2)编译所有.c
源文件,3)运行main
:
# 1
java -cp antlr-3.3.jar org.antlr.Tool T.g
# 2
gcc -Wall main.c TLexer.c TParser.c -l antlr3c -o main
# 3
./main
你会看到abc
(没有引号)正在打印到控制台。
答案 1 :(得分:1)
您可以通过词法分析器的RecognizerSharedState state
属性影响令牌构造:
CharacterString:
Apostrophe
CharSequence
Apostrophe
{ state.text = $CharSequence.text; }
;
fragment CharSequence:
Alphanumeric+
;