결정적인 ANTLR 4 참조 (페이지 106) :

fragment 접두사가 붙은 규칙은 다른 렉서 규칙에서만 호출 할 수 있습니다. 그들은 그 자체로 토큰이 아닙니다.

추상 개념 :

Case1 : (RULE1, RULE2, RULE3 엔티티 또는 그룹 정보가 필요한 경우)

rule0 : RULE1 | RULE2 | RULE3 ;
RULE1 : [A-C]+ ;
RULE2 : [DEF]+ ;
RULE3 : ('G'|'H'|'I')+ ;

Case2 : (RULE1, RULE2, RULE3을 신경 쓰지 않으면 RULE0에만 집중합니다.)

RULE0 : [A-C]+ | [DEF]+ | ('G'|'H'|'I')+ ;
// RULE0 is a terminal node. 
// You can't name it 'rule0', or you will get syntax errors:
// 'A-C' came as a complete surprise to me while matching alternative
// 'DEF' came as a complete surprise to me while matching alternative

Case3 : (Case2와 동일하므로 Case2보다 읽기 쉽습니다. )

RULE0 : RULE1 | RULE2 | RULE3 ;
fragment RULE1 : [A-C]+ ;
fragment RULE2 : [DEF]+ ;
fragment RULE3 : ('G'|'H'|'I')+ ;
// You can't name it 'rule0', or you will get warnings:
// warning(125): implicit definition of token RULE1 in parser
// warning(125): implicit definition of token RULE2 in parser
// warning(125): implicit definition of token RULE3 in parser
// and failed to capture rule0 content (?)

Case1과 Case2 / 3의 차이점은 무엇입니까?

1. 어휘 분석기 규칙은 동일합니다.
2. Case1의 각 RULE1 / 2 / 3은 Regex : (X)와 유사한 캡처 그룹입니다.
3. Case3의 각 RULE1 / 2 / 3은 Regex 🙁 ?: X)와 유사한 비 캡처 그룹입니다.
   

구체적인 예를 보겠습니다.

목표 : 식별 [ABC]+, [DEF]+, [GHI]+토큰

input.txt

ABBCCCDDDDEEEEE ABCDE
FFGGHHIIJJKK FGHIJK
ABCDEFGHIJKL

Main.py

import sys
from antlr4 import *
from AlphabetLexer import AlphabetLexer
from AlphabetParser import AlphabetParser
from AlphabetListener import AlphabetListener

class MyListener(AlphabetListener):
    # Exit a parse tree produced by AlphabetParser#content.
    def exitContent(self, ctx:AlphabetParser.ContentContext):
        pass

    # (For Case1 Only) enable it when testing Case1
    # Exit a parse tree produced by AlphabetParser#rule0.
    def exitRule0(self, ctx:AlphabetParser.Rule0Context):
        print(ctx.getText())
# end-of-class

def main():
    file_name = sys.argv[1]
    input = FileStream(file_name)
    lexer = AlphabetLexer(input)
    stream = CommonTokenStream(lexer)
    parser = AlphabetParser(stream)
    tree = parser.content()
    print(tree.toStringTree(recog=parser))

    listener = MyListener()
    walker = ParseTreeWalker()
    walker.walk(listener, tree)
# end-of-def

main()

사례 1 및 결과 :

Alphabet.g4 (Case1)

grammar Alphabet;

content : (rule0|ANYCHAR)* EOF;

rule0 : RULE1 | RULE2 | RULE3 ;
RULE1 : [A-C]+ ;
RULE2 : [DEF]+ ;
RULE3 : ('G'|'H'|'I')+ ;

ANYCHAR : . -> skip;

결과:

# Input data (for reference)
# ABBCCCDDDDEEEEE ABCDE
# FFGGHHIIJJKK FGHIJK
# ABCDEFGHIJKL

$ python3 Main.py input.txt
(content (rule0 ABBCCC) (rule0 DDDDEEEEE) (rule0 ABC) (rule0 DE) (rule0 FF) (rule0 GGHHII) (rule0 F) (rule0 GHI) (rule0 ABC) (rule0 DEF) (rule0 GHI) <EOF>)
ABBCCC
DDDDEEEEE
ABC
DE
FF
GGHHII
F
GHI
ABC
DEF
GHI

Case2 / 3 및 결과 :

Alphabet.g4 (Case2)

grammar Alphabet;

content : (RULE0|ANYCHAR)* EOF;

RULE0 : [A-C]+ | [DEF]+ | ('G'|'H'|'I')+ ;

ANYCHAR : . -> skip;

Alphabet.g4 (Case3)

grammar Alphabet;

content : (RULE0|ANYCHAR)* EOF;

RULE0 : RULE1 | RULE2 | RULE3 ;
fragment RULE1 : [A-C]+ ;
fragment RULE2 : [DEF]+ ;
fragment RULE3 : ('G'|'H'|'I')+ ;

ANYCHAR : . -> skip;

결과:

# Input data (for reference)
# ABBCCCDDDDEEEEE ABCDE
# FFGGHHIIJJKK FGHIJK
# ABCDEFGHIJKL

$ python3 Main.py input.txt
(content ABBCCC DDDDEEEEE ABC DE FF GGHHII F GHI ABC DEF GHI <EOF>)

“그룹 캡처” 및 “그룹 캡처하지 않음” 부분을 보았습니까 ?

구체적인 example2를 봅시다.

목표 : 8 진수 / 10 진수 / 16 진수 식별

input.txt

0
123
 1~9999
 001~077
0xFF, 0x01, 0xabc123

Number.g4

grammar Number;

content
    : (number|ANY_CHAR)* EOF
    ;

number
    : DECIMAL_NUMBER
    | OCTAL_NUMBER
    | HEXADECIMAL_NUMBER
    ;

DECIMAL_NUMBER
    : [1-9][0-9]*
    | '0'
    ;

OCTAL_NUMBER
    : '0' '0'..'9'+
    ;

HEXADECIMAL_NUMBER
    : '0x'[0-9A-Fa-f]+
    ;

ANY_CHAR
    : .
    ;

Main.py

import sys
from antlr4 import *
from NumberLexer import NumberLexer
from NumberParser import NumberParser
from NumberListener import NumberListener

class Listener(NumberListener):
    # Exit a parse tree produced by NumberParser#Number.
    def exitNumber(self, ctx:NumberParser.NumberContext):
        print('%8s, dec: %-8s, oct: %-8s, hex: %-8s' % (ctx.getText(),
            ctx.DECIMAL_NUMBER(), ctx.OCTAL_NUMBER(), ctx.HEXADECIMAL_NUMBER()))
    # end-of-def
# end-of-class

def main():
    input = FileStream(sys.argv[1])
    lexer = NumberLexer(input)
    stream = CommonTokenStream(lexer)
    parser = NumberParser(stream)
    tree = parser.content()
    print(tree.toStringTree(recog=parser))

    listener = Listener()
    walker = ParseTreeWalker()
    walker.walk(listener, tree)
# end-of-def

main()

결과:

# Input data (for reference)
# 0
# 123
#  1~9999
#  001~077
# 0xFF, 0x01, 0xabc123

$ python3 Main.py input.txt
(content (number 0) \n (number 123) \n   (number 1) ~ (number 9999) \n   (number 001) ~ (number 077) \n (number 0xFF) ,   (number 0x01) ,   (number 0xabc123) \n <EOF>)
       0, dec: 0       , oct: None    , hex: None
     123, dec: 123     , oct: None    , hex: None
       1, dec: 1       , oct: None    , hex: None
    9999, dec: 9999    , oct: None    , hex: None
     001, dec: None    , oct: 001     , hex: None
     077, dec: None    , oct: 077     , hex: None
    0xFF, dec: None    , oct: None    , hex: 0xFF
    0x01, dec: None    , oct: None    , hex: 0x01
0xabc123, dec: None    , oct: None    , hex: 0xabc123

당신이에 수정 ‘조각’을 추가하는 경우 DECIMAL_NUMBER, OCTAL_NUMBER, HEXADECIMAL_NUMBER(그들은 더 이상 토큰이 아니기 때문에), 당신은 수 엔티티를 캡처 할 수 없습니다. 결과는 다음과 같습니다.

$ python3 Main.py input.txt
(content 0 \n 1 2 3 \n   1 ~ 9 9 9 9 \n   0 0 1 ~ 0 7 7 \n 0 x F F ,   0 x 0 1 ,   0 x a b c 1 2 3 \n <EOF>)

답변

grammar number;

number: INT;
DIGIT : '0'..'9';
INT   :  DIGIT+;