如何检查一行的第一个字符是否为“*"?在 ANTLR4 中? [英] How can I check if first character of a line is "*" in ANTLR4?
问题描述
我正在尝试为一种相对简单但特殊的语言编写解析器.
I am trying to write a parser for a relatively simple but idiosyncratic language.
简单地说,其中一条规则是注释行用星号表示仅,如果该星号是行的第一个字符.我该如何在 ANTLR4 中正式化这样的规则?我想过使用:
Simply put, one of the rules is that comment lines are denoted by an asterisk only if that asterisk is the first character of the line. How might I go about formalising such a rule in ANTLR4? I thought about using:
START_LINE_COMMENT: '\n*' .*? '\n' -> skip;
但我确信这不会连续处理多于一行的注释,因为末尾的换行符将作为 START_LINE_COMMENT
标记的一部分使用,这意味着任何后续的注释行将丢失所需的初始换行符,这将不起作用.有没有一种方法可以检查该行是否以 '*'
开头而无需使用先前的 '\n'
?
But I am certain this won't work with more than one line comment in a row, as the newline at the end will be consumed as part of the START_LINE_COMMENT
token, meaning any subsequent comment lines will be missing the required initial newline character, which won't work. Is there a way I can perhaps check if the line starts with a '*'
without needing to consume the prior '\n'
?
推荐答案
匹配注释行并不容易.当我每年写一个语法时,我不得不抓住 The Definitive ANTLR参考来刷新我的大脑.试试这个:
Matching a comment line is not easy. As I write one grammar per year, I had to grab to The Definitive ANTLR Reference to refresh my brain. Try this :
grammar Question;
/* Comment line having an * in column 1. */
question
: line+
;
line
// : ( ID | INT )+
: ( ID | INT | MULT )+
;
LINE_COMMENT
: '*' {getCharPositionInLine() == 1}? ~[\r\n]* -> channel(HIDDEN) ;
ID : [a-zA-Z]+ ;
INT : [0-9]+ ;
//WS : [ \t\r\n]+ -> channel(HIDDEN) ;
WS : [ \t\r\n]+ -> skip ;
MULT : '*' ;
编译执行:
$ echo $CLASSPATH
.:/usr/local/lib/antlr-4.6-complete.jar:
$ alias
alias a4='java -jar /usr/local/lib/antlr-4.6-complete.jar'
alias grun='java org.antlr.v4.gui.TestRig'
$ a4 Question.g4
$ javac Q*.java
$ grun Question question -tokens data.txt
[@0,0:3='line',<ID>,1:0]
[@1,5:5='1',<INT>,1:5]
[@2,9:12='line',<ID>,2:2]
[@3,14:14='2',<INT>,2:7]
[@4,16:26='* comment 1',<LINE_COMMENT>,channel=1,3:0]
[@5,32:35='line',<ID>,4:4]
[@6,37:37='4',<INT>,4:9]
[@7,39:48='*comment 2',<LINE_COMMENT>,channel=1,5:0]
[@8,51:78='* comment 3 after empty line',<LINE_COMMENT>,channel=1,7:0]
[@9,81:81='*',<'*'>,8:1]
[@10,83:85='not',<ID>,8:3]
[@11,87:87='a',<ID>,8:7]
[@12,89:95='comment',<ID>,8:9]
[@13,97:100='line',<ID>,9:0]
[@14,102:102='9',<INT>,9:5]
[@15,107:107='*',<'*'>,9:10]
[@16,109:110='no',<ID>,9:12]
[@17,112:118='comment',<ID>,9:15]
[@18,120:119='<EOF>',<EOF>,10:0]
具有以下 data.text 文件:
with the following data.text file :
line 1
line 2
* comment 1
line 4
*comment 2
* comment 3 after empty line
* not a comment
line 9 * no comment
请注意,如果解析器规则中的某处没有 MULT
标记或 '*'
,则标记中不会列出星号,但解析器会抱怨:
Note that without the MULT
token or '*'
somewhere in a parser rule, the asterisk is not listed in the tokens, but the parser complains :
line 8:1 token recognition error at: '*'
如果显示解析树
$ grun Question question -gui data.txt
您会看到整个文件都被一行规则吸收了.如果您需要识别线条,请像这样更改线条和空格规则:
you'll see that the whole file is absorbed by one line rule. If you need to recognize lines, change the line and white space rules like so :
line
: ( ID | INT | MULT )+ NL
| NL
;
//WS : [ \t\r\n]+ -> skip ;
NL : [\r\n] ;
WS : [ \t]+ -> skip ;
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