I have the little grammar below. node is the start production. When my input is (a:b) I get an error: line 1:1 extraneous input 'a' expecting {':', INAME}
Why is this?
EDIT - I forgot that the lexer and parser run as a separate phases. By the time the parser runs, the lexer has completed. When the lexer runs it has no knowledge of the parser rules. It has already made the TYPE/INAME decision choosing TYPE per #bart's reasoning below.
grammar g1;
TYPE: [A-Za-z_];
INAME: [A-Za-z_];
node: '(' namesAndTypes ')';
namesAndTypes:
INAME ':' TYPE
| ':' TYPE
| INAME
;
That is because the lexer will never produce an INAME token. The lexer works in the following was:
try to match as much characters as possible
when 2 or more lexer rules match the same characters, let the one defined first "win"
Because the input "a" and "b" both match the TYPE and INAME rules, the TYPE rule wins because it is defined first. It doesn't matter if the parser is trying to match an INAME rule, the lexer will not produce it. The lexer does not "listen" to he parser.
You could create some sort of ID rule, and then define type and iname parser rules instead:
ID: [A-Za-z_];
node
: '(' namesAndTypes ')'
;
namesAndTypes
: iname ':' type
| ':' type
| iname
;
type
: ID
;
iname
: ID
;
Related
My lexer (target language C++) contains a simple rule for parsing a string literal:
STRING: '"' ~'"'+ '"';
But based on the value returned by a function, I want my lexer to return either a STRING or an IDENT.
I've tried the following:
STRING_START: '"' -> mode(current_string_mode());
or
STRING_START: '"' -> mode(current_string_mode() == IDENT ? MODE_IDENT : MODE_STRING) ;
In either case, I get an error when trying to generate the lexer (error message says:'"' came as a complete surprise)
Alas, that is not possible.
If I look at the grammar of ANTLR itself, I see this:
lexerCommands
: RARROW lexerCommand (COMMA lexerCommand)*
;
lexerCommand
: lexerCommandName LPAREN lexerCommandExpr RPAREN
| lexerCommandName
;
lexerCommandName
: identifier
| MODE
;
lexerCommandExpr
: identifier
| INT
;
In short: the part between parenthesis (mode(...) or pushMode(...)) must be an identifier, or an integer literal. It cannot be an expression (what you're trying to do).
I want to match the following text:
test.define_shared_constant(:testConst, "12", false)
With this grammar it matches correctly:
grammar test;
statement: shared_constant_defioniton | method_call;
KEY: ':' ('a'..'z'|'A'..'Z'|'0'..'9'|'_'|'?'|'!'|'|'|'-'|'()')+;
expr: STRING;
STRING: '"' (~'"')* ('"' | NEWLINE) | '\'' (~'\'')* ('\'' | NEWLINE);
NEWLINE: '\r'? '\n' | '\r';
BOOLEAN: 'true' | 'false';
ID: ('a'..'z'|'A'..'Z'|'!') ('a'..'z'|'A'..'Z'|'0'..'9'|'_'|'!'|'?')*;
WS : [ \t\n\r]+ -> channel(HIDDEN);
DEF_SHARED_CONSTANT: 'define_shared_constant';
shared_constant_defioniton
: ID('.define_shared_constant' '(' KEY ',' expr ',' (BOOLEAN) ')')
;
method_call
: ID '.' ID? '('expr*(',' expr)*')'
;
With this grammar it does not match. It matches to method_call which is not even correct.
shared_constant_defioniton
: ID('.' DEF_SHARED_CONSTANT '(' KEY ',' expr ',' (BOOLEAN) ')')
;
It is interpreting 'define_shared_constant' as ID. So I have to specify that ID should not contain 'define_'. But how can I do that?
ID: ('a'..'z'|'A'..'Z'|'!') ('a'..'z'|'A'..'Z'|'0'..'9'|'_'|'!'|'?')*;
WS : [ \t\n\r]+ -> channel(HIDDEN);
DEF_SHARED_CONSTANT: 'define_shared_constant';
Here both ID and DEF_SHARED_CONSTANT could match the input define_shared_constant. In cases like this where multiple rules could match and would produce a match of the same length, the rule that's defined first wins. So defined_shared_constant is recognized as an ID token because ID is defined first.
To get the behaviour you want, you should move the definition of DEF_SHARED_CONSTANT before the definition of ID. If you don't define a named lexer rule for it at all and instead use 'define_shared_constant' directly in the parser rule, that also works because implicitly defined lexer rules act as if they had been defined at the beginning of the file.
This worked according to ANTLR specification. But running it as an IntelliJ Language plugin did not. I had use a predicated and the final solution looks like this:
ID: { getText().indexOf("define") == 0}? ('a'..'z'|'A'..'Z'|'!') ('a'..'z'|'A'..'Z'|'0'..'9'|'_'|'!'|'?')*;
I want to write a grammar using Antlr4 that will parse a some definition but I've been struggling to get Antlr to co-operate.
The definition has two kinds of lines, a type and a property. I can get my grammar to parse the type line correctly but it either ignores the property lines or fails to identify PROPERTY_TYPE depending on how I tweak my grammar.
Here is my grammar (attempt # 583):
grammar TypeDefGrammar;
start
: statement+ ;
statement
: type NEWLINE
| property NEWLINE
| NEWLINE ;
type
: TYPE_KEYWORD TYPE_NAME; // e.g. 'type MyType1'
property
: PROPERTY_NAME ':' PROPERTY_TYPE ; // e.g. 'someProperty1: int'
TYPE_KEYWORD
: 'type' ;
TYPE_NAME
: IDENTIFIER ;
PROPERTY_NAME
: IDENTIFIER ;
PROPERTY_TYPE
: IDENTIFIER ;
fragment IDENTIFIER
: (LETTER | '_') (LETTER | DIGIT | '_' )* ;
fragment LETTER
: [a-zA-Z] ;
fragment DIGIT
: [0-9] ;
NEWLINE
: '\r'? '\n' ;
WS
: [ \t] -> skip ;
Here is a sample input:
type SimpleType
intProp1: int
stringProp2 : String
(returns the type but ignores intProp1, stringProp2.)
What am I doing wrong?
Usually when a rule does not match the whole input, but does match a prefix of it, it will simply match that prefix and leave the rest of the input in the stream without producing an error. If you want your rule to always match the whole input, you can add EOF to the end of the rule. That way you'll get proper error messages when it can't match the entire input.
So let's change your start rule to start : statement+ EOF;. Now applying start to your input will lead to the following error messages:
line 3:0 extraneous input 'intProp1' expecting {, 'type', PROPERTY_NAME, NEWLINE}
line 4:0 extraneous input 'stringProp2' expecting {, 'type', PROPERTY_NAME, NEWLINE}
So apparently intProp1 and stringProp2 aren't recognized as PROPERTY_NAMEs. So let's look at which tokens are generated (you can do that using the -tokens option to grun or by just iterating over the token stream in your code):
[#0,0:3='type',<'type'>,1:0]
[#1,5:14='SimpleType',<TYPE_NAME>,1:5]
[#2,15:15='\n',<NEWLINE>,1:15]
[#3,16:16='\n',<NEWLINE>,2:0]
[#4,17:24='intProp1',<TYPE_NAME>,3:0]
[#5,25:25=':',<':'>,3:8]
[#6,27:29='int',<TYPE_NAME>,3:10]
[#7,30:30='\n',<NEWLINE>,3:13]
[#8,31:41='stringProp2',<TYPE_NAME>,4:0]
[#9,43:43=':',<':'>,4:12]
[#10,45:50='String',<TYPE_NAME>,4:14]
[#11,51:51='\n',<NEWLINE>,4:20]
[#12,52:51='<EOF>',<EOF>,5:0]
So all of the identifiers in the code are recognized as TYPE_NAMEs, not PROPERTY_NAMEs. In fact, it is not clear what should distinguish a TYPE_NAME from a PROPERTY_NAME, so now let's actually look at your grammar:
TYPE_NAME
: IDENTIFIER ;
PROPERTY_NAME
: IDENTIFIER ;
PROPERTY_TYPE
: IDENTIFIER ;
fragment IDENTIFIER
: (LETTER | '_') (LETTER | DIGIT | '_' )* ;
Here you have three lexer rules with exactly the same definition. That's a bad sign.
Whenever multiple lexer rules can match on the current input, ANTLR chooses the one that would produce the longest match, picking the one that comes first in the grammar in case of ties. This is known as the maximum munch rule.
If you have multiple rules with the same definition, that means those rules will always match on the same input and they will always produce matches of the same length. So by the maximum much rule, the first definition (TYPE_NAME) will always be used and the other ones might as well not exist.
The problem basically boils down to the fact that there's nothing that lexically distinguishes the different types of names, so there's no basis on which the lexer could decide which type of name a given identifier represents. That tells us that the names should not be lexer rules. Instead IDENTIFIER should be a lexer rule and the FOO_NAMEs should either be (somewhat unnecessary) parser rules or removed altogether (you can just use IDENTIFIER wherever you're currently using FOO_NAME).
grammar TestGrammar;
AND : 'AND' ;
OR : 'OR'|',' ;
NOT : 'NOT' ;
LPAREN : '(' ;
RPAREN : ')' ;
DQUOTE : '"' ;
WORD : [a-z0-9._#+=]+(' '[a-z0-9._#+=]+)* ;
WS : [ \t\r\n]+ -> skip ;
quotedword : DQUOTE WORD DQUOTE;
expression
: LPAREN expression+ RPAREN
| expression (AND expression)+
| expression (OR expression)+
| expression (NOT expression)+
| NOT expression+
| quotedword
| WORD;
I've managed to implement the above grammar for antlr4.
I've got a long way to go but for now my question is,
how can I make WORD generic? Basically I want this [a-z0-9._#+=] to be anything except the operators (AND, OR, NOT, LPAREN, RPAREN, DQUOTE, SPACE).
The lexer will use the first rule that can match the given input. Only if that rule can't match it, it will try the next one.
Therefore you can make your WORD rule generic by using this grammar:
AND : 'AND' ;
OR : 'OR'|',' ;
NOT : 'NOT' ;
LPAREN : '(' ;
RPAREN : ')' ;
DQUOTE : '"' ;
WS : [ \t\r\n]+ -> skip ;
WORD: .+? ;
Make sure to use the non-greedy operator ? in this case becaue otherwise once invoked the WORD rule will consume all following input.
As WORD is specified last, input will only be tried to be consumed by it if all previous lexer rules (all that have been defined above in the source code) have failed.
EDIT: If you don't want your WORD rule to match any input then you just have to modify the rule I provided. But the essence of my answer is that in the lexer you don't have to worry about two rules potentially matching the same input as long as you got the order in the source code right.
Try something like this grammar:
grammar TestGrammar;
...
WORD : Letter+;
QUOTEDWORD : '"' (~["\\\r\n])* '"' // disallow quotes, backslashes and crlf in literals
WS : [ \t\r\n]+ -> skip ;
fragment Letter :
[a-zA-Z$_] // these are the "java letters" below 0x7F
| ~[\u0000-\u007F\uD800-\uDBFF] // covers all characters above 0x7F which are not a surrogate
| [\uD800-\uDBFF] [\uDC00-\uDFFF] // covers UTF-16 surrogate pairs encodings for U+10000 to U+10FFFF
;
expression:
...
| QUOTEDWORD
| WORD+;
Maybe you want to use escape sequences in QUOTEDWORD, then look in this example how to do this.
This grammar allows you:
to have quoted words interpreted as string literals (preserving all spaces within)
to have multiple words separated by whitespace (which is ignored)
How do I build a token in lexer that can handle recursion inside as this string:
${*anything*${*anything*}*anything*}
?
Yes, you can use recursion inside lexer rules.
Take the following example:
${a ${b} ${c ${ddd} c} a}
which will be parsed correctly by the following grammar:
parse
: DollarVar
;
DollarVar
: '${' (DollarVar | EscapeSequence | ~Special)+ '}'
;
fragment
Special
: '\\' | '$' | '{' | '}'
;
fragment
EscapeSequence
: '\\' Special
;
as the interpreter inside ANTLRWorks shows:
alt text http://img185.imageshack.us/img185/5471/recq.png
ANTLR's lexers do support recursion, as #BartK adeptly points out in his post, but you will only see a single token within the parser. If you need to interpret the various pieces within that token, you'll probably want to handle it within the parser.
IMO, you'd be better off doing something in the parser:
variable: DOLLAR LBRACE id variable id RBRACE;
By doing something like the above, you'll see all the necessary pieces and can build an AST or otherwise handle accordingly.