我正在尝试为DSL实现预处理器,以代码/附加中的CPP示例为模型。但是,我没有使用令牌工厂。需要一个吗?调用emit(令牌)不会按预期将令牌注入令牌流。
这是词法分析器:
// string-delimited path
SPATH : '"' (~[\n\r])*? '"'
{
emit(); // inject the current token
// launch another lexer on the include file, get tokens,
// emit them all at once here
List<CommonToken> tokens = Preprocessor.include(getText());
if (null != tokens) {
for (CommonToken tok : tokens) {
emit(tok);
}
}
}
;
这是include方法:
@SuppressWarnings("unchecked")
public static List<CommonToken> include(String filename) {
List<CommonToken> tokens = null;
try (FileReader fr = openFile(filename.substring(1, filename.length() - 1));
BufferedReader br = new BufferedReader(fr)) {
ANTLRInputStream input = new ANTLRInputStream(br);
PreprocessorLexer lexer = new PreprocessorLexer(input);
tokens = (List<CommonToken>) lexer.getAllTokens();
} catch (IOException ioe) {
log.error("Can't load ~{}~", ioe.getLocalizedMessage());
}
return tokens;
}
答案 0 :(得分:3)
您需要覆盖Lexer.nextToken
才能提供此功能。在词法分析器中,保留Deque<Token>
尚未由nextToken
返回的nextToken
个注入的令牌。当队列为空时,private final Deque<Token> pendingTokens = new ArrayDeque<>();
@Override
public Token nextToken() {
Token pending = pendingTokens.pollFirst();
if (pending != null) {
return pending;
}
Token next = super.nextToken();
pending = pendingTokens.pollFirst();
if (pending != null) {
pendingTokens.addLast(next);
return pending;
}
return next;
}
的实现应根据超类实现返回下一个标记。
这是一些示例代码。我没有尝试编译或运行它,所以它可能不完美。
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