解析规则无咖啡因文法antlr4

Question

我正在为用ANTLR4编写的Decaf编程语言创建解析器和lexer规则。我试图解析一个测试文件，但不断地得到一个错误，一定是语法有问题，但我无法理解。

我的测试文件如下所示：

class Program {
  int i[10];
}

错误是：第2行：8不匹配的输入“10”应为INT_LITERAL

下面是完整的decaf.g4语法文件

grammar Decaf;


/*
  LEXER RULES
  -----------
  Lexer rules define the basic syntax of individual words and symbols of a
  valid Decaf program. Lexer rules follow regular expression syntax.
  Complete the lexer rules following the Decaf Language Specification.
*/



CLASS : 'class';

INT : 'int';

RETURN : 'return';

VOID : 'void';

IF : 'if';

ELSE : 'else';

FOR : 'for';

BREAK : 'break';

CONTINUE : 'continue';

CALLOUT : 'callout';

TRUE : 'True' ;

FALSE : 'False' ;

BOOLEAN : 'boolean';

LCURLY : '{';

RCURLY : '}';

LBRACE : '(';

RBRACE : ')';


LSQUARE : '[';

RSQUARE : ']';
ADD : '+';

SUB : '-';

MUL : '*';

DIV : '/';

EQ : '=';

SEMI : ';';

COMMA : ',';

AND : '&&';

LESS : '<';

GREATER : '>';

LESSEQUAL : '<=' ;

GREATEREQUAL : '>=' ;

EQUALTO : '==' ;

NOTEQUAL : '!=' ;

EXCLAMATION : '!';



fragment CHAR : (' '..'!') | ('#'..'&') | ('('..'[') | (']'..'~') | ('\\'[']) | ('\\"') | ('\\') | ('\t') | ('\n');

CHAR_LITERAL : '\'' CHAR '\'';

//STRING_LITERAL : '"' CHAR+ '"' ;


HEXMARK : '0x';

fragment HEXA : [a-fA-F];

fragment HEXDIGIT : DIGIT | HEXA ;

HEX_LITERAL : HEXMARK HEXDIGIT+;


STRING : '"' (ESC|.)*? '"';

fragment ESC : '\\"' | '\\\\';




fragment DIGIT : [0-9];

DECIMAL_LITERAL : DIGIT(DIGIT)*;



COMMENT : '//' ~('\n')* '\n' -> skip;

WS : (' ' | '\n' | '\t' | '\r') + -> skip;

fragment ALPHA : [a-zA-Z] | '_';

fragment ALPHA_NUM : ALPHA | DIGIT;



ID : ALPHA ALPHA_NUM*;

INT_LITERAL : DECIMAL_LITERAL | HEX_LITERAL;

BOOL_LITERAL : TRUE | FALSE;

/*
  PARSER RULES
  ------------
  Parser rules are all lower case, and make use of lexer rules defined above
  and other parser rules defined below. Parser rules also follow regular
  expression syntax. Complete the parser rules following the Decaf Language
  Specification.
*/




program : CLASS ID LCURLY field_decl* method_decl* RCURLY EOF;

field_name : ID | ID LSQUARE INT_LITERAL RSQUARE;

field_decl : datatype field_name (COMMA field_name)* SEMI;

method_decl : (datatype | VOID) ID LBRACE ((datatype ID) (COMMA datatype ID)*)? RBRACE block;

block : LCURLY var_decl* statement* RCURLY;

var_decl : datatype ID (COMMA ID)* SEMI;


datatype : INT | BOOLEAN;

statement : location assign_op expr SEMI
        | method_call SEMI
        | IF LBRACE expr RBRACE block (ELSE block)?
        | FOR ID EQ expr COMMA expr block
        | RETURN (expr)? SEMI
        | BREAK SEMI
        | CONTINUE SEMI
        | block;
        
assign_op : EQ
          | ADD EQ
          | SUB EQ;
          
          
method_call : method_name LBRACE (expr (COMMA expr)*)? RBRACE
            | CALLOUT LBRACE STRING(COMMA callout_arg (COMMA callout_arg)*) RBRACE;


method_name : ID;

location : ID | ID LSQUARE expr RSQUARE;


expr : location
     | method_call
     | literal
     | expr bin_op expr
     | SUB expr
     | EXCLAMATION expr
     | LBRACE expr RBRACE;

 callout_arg : expr
            | STRING ;

bin_op : arith_op
      | rel_op
      | eq_op
      | cond_op;


arith_op : ADD | SUB | MUL | DIV | '%' ;

rel_op : LESS | GREATER | LESSEQUAL | GREATEREQUAL ;

eq_op : EQUALTO | NOTEQUAL ;

cond_op : AND | '||' ;

literal : INT_LITERAL | CHAR_LITERAL | BOOL_LITERAL ;

Answer 1

每当有2个或更多的lexer规则匹配相同的字符时，首先定义的规则将获胜。在您的示例中，这两个规则都符合10:

DECIMAL_LITERAL : DIGIT(DIGIT)*;

INT_LITERAL : DECIMAL_LITERAL | HEX_LITERAL;

由于int_literal是在decimal_literal之后定义的，所以lexer永远不会创建int_literal标记。如果您现在尝试在解析器规则中使用它，您将得到一条发布的错误消息。

解决方案：从lexer中删除int_literal并创建解析器规则：

int_literal : DECIMAL_LITERAL | HEX_LITERAL;