[英]Parsing DECAF grammar in ANTLR
I am creating a the parser for DECAF with Antlr grammar DECAF ; 我正在使用Antlr语法DECAF创建DECAF的解析器;
//********* LEXER ******************
LETTER: ('a'..'z'|'A'..'Z') ;
DIGIT : '0'..'9' ;
ID : LETTER( LETTER | DIGIT)* ;
NUM: DIGIT(DIGIT)* ;
COMMENTS: '//' ~('\r' | '\n' )* -> channel(HIDDEN);
WS : [ \t\r\n\f | ' '| '\r' | '\n' | '\t']+ ->channel(HIDDEN);
CHAR: (LETTER|DIGIT|' '| '!' | '"' | '#' | '$' | '%' | '&' | '\'' | '(' | ')' | '*' | '+'
| ',' | '-' | '.' | '/' | ':' | ';' | '<' | '=' | '>' | '?' | '@' | '[' | '\\' | ']' | '^' | '_' | '`'| '{' | '|' | '}' | '~'
'\t'| '\n' | '\"' | '\'');
// ********** PARSER *****************
program : 'class' 'Program' '{' (declaration)* '}' ;
declaration: structDeclaration| varDeclaration | methodDeclaration ;
varDeclaration: varType ID ';' | varType ID '[' NUM ']' ';' ;
structDeclaration : 'struct' ID '{' (varDeclaration)* '}' ;
varType: 'int' | 'char' | 'boolean' | 'struct' ID | structDeclaration | 'void' ;
methodDeclaration : methodType ID '(' (parameter (',' parameter)*)* ')' block ;
methodType : 'int' | 'char' | 'boolean' | 'void' ;
parameter : parameterType ID | parameterType ID '[' ']' ;
parameterType: 'int' | 'char' | 'boolean' ;
block : '{' (varDeclaration)* (statement)* '}' ;
statement : 'if' '(' expression ')' block ( 'else' block )?
| 'while' '(' expression ')' block
|'return' expressionA ';'
| methodCall ';'
| block
| location '=' expression
| (expression)? ';' ;
expressionA: expression | ;
location : (ID|ID '[' expression ']') ('.' location)? ;
expression : location | methodCall | literal | expression op expression | '-' expression | '!' expression | '('expression')' ;
methodCall : ID '(' arg1 ')' ;
arg1 : arg2 | ;
arg2 : (arg) (',' arg)* ;
arg : expression;
op: arith_op | rel_op | eq_op | cond_op ;
arith_op : '+' | '-' | '*' | '/' | '%' ;
rel_op : '<' | '>' | '<=' | '>=' ;
eq_op : '==' | '!=' ;
cond_op : '&&' | '||' ;
literal : int_literal | char_literal | bool_literal ;
int_literal : NUM ;
char_literal : '\'' CHAR '\'' ;
bool_literal : 'true' | 'false' ;
When I give it the input: 当我输入时:
class Program {
void main(){
return 3+5 ;
}
}
The parse tree is not building correctly since it is not recognizing the 3+5 as an expression. 解析树无法正确构建,因为它无法将3 + 5识别为表达式。 Is there anything wrong with my grammar that is causing the problem?
我的语法是否有引起问题的错误?
Lexer rules are matched from top to bottom. Lexer规则从上到下匹配。 When 2 or more lexer rules match the same amount of characters, the one defined first will win .
当2个或更多lexer规则匹配相同数量的字符时,首先定义的一个将获胜 。 Because of that, a single digit integer will get matched as a
DIGIT
instead of a NUM
. 因此,一位整数将作为
DIGIT
而不是NUM
匹配。
Try parsing the following instead: 尝试解析以下内容:
class Program {
void main(){
return 33 + 55 ;
}
}
which will be parsed just fine. 将被解析就好了。 This is because
33
and 55
are matched as NUM
s, because NUM
can now match 2 characters ( DIGIT
only 1, so NUM
wins ). 这是因为
33
和55
被匹配为NUM
,因为NUM
现在可以匹配2个字符( DIGIT
仅1个字符,所以NUM
wins )。
To fix it, make DIGIT
a fragment (and LETTER
as well): 要修复此问题,请使
DIGIT
成为片段(以及LETTER
):
fragment LETTER: ('a'..'z'|'A'..'Z') ;
fragment DIGIT : '0'..'9' ;
ID : LETTER( LETTER | DIGIT)* ;
NUM: DIGIT(DIGIT)* ;
Lexer fragments are only used internally by other lexer rules, and will never become tokens of their own. Lexer片段仅由其他lexer规则在内部使用,并且永远不会成为其自身的标记。
A couple of other things: your WS
rule matches way too much (it now also matches a |
and a '
), it should be: 还有两件事:您的
WS
规则匹配得太多(现在也匹配了|
和'
),应该是:
WS : [ \t\r\n\f]+ ->channel(HIDDEN);
and you shouldn't match a char literal in your parser: do it in the lexer: 而且您不应该在解析器中匹配char文字:在lexer中进行匹配:
CHAR : '\'' ( ~['\r\n\\] | '\\' ['\\] ) '\'';
If you don't, the following will not get parsed properly: 否则,将无法正确解析以下内容:
class Program {
void main(){
return '1';
}
}
because the 1
wil be tokenized as a NUM
and not as a CHAR
. 因为
1
将被标记为NUM
而不是CHAR
。
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