I wrote the following grammar which should check for a conditional expression.
Examples below is what I want to achieve using this grammar:
test invalid
test = 1 valid
test = 1 and another_test>=0.2 valid
test = 1 kasd y = 1 invalid (two conditions MUST be separated by AND/OR)
a = 1 or (b=1 and c) invalid (there cannot be a lonely character like 'c'. It should always be a triplet. i.e, literal operator literal)
grammar expression;
expr
: literal_value
| expr ( '='|'<>'| '<' | '<=' | '>' | '>=' ) expr
| expr K_AND expr
| expr K_OR expr
| function_name '(' ( expr ( ',' expr )* | '*' )? ')'
| '(' expr ')'
;
literal_value
: NUMERIC_LITERAL
| STRING_LITERAL
| IDENTIFIER
;
keyword
: K_AND
| K_OR
;
name
: any_name
;
function_name
: any_name
;
database_name
: any_name
;
table_name
: any_name
;
column_name
: any_name
;
any_name
: IDENTIFIER
| keyword
| STRING_LITERAL
| '(' any_name ')'
;
K_AND : A N D;
K_OR : O R;
IDENTIFIER
: '"' (~'"' | '""')* '"'
| '`' (~'`' | '``')* '`'
| '[' ~']'* ']'
| [a-zA-Z_] [a-zA-Z_0-9]*
;
NUMERIC_LITERAL
: DIGIT+ ( '.' DIGIT* )? ( E [-+]? DIGIT+ )?
| '.' DIGIT+ ( E [-+]? DIGIT+ )?
;
STRING_LITERAL
: '\'' ( ~'\'' | '\'\'' )* '\''
;
fragment DIGIT : [0-9];
fragment A : [aA];
fragment B : [bB];
fragment C : [cC];
fragment D : [dD];
fragment E : [eE];
fragment F : [fF];
fragment G : [gG];
fragment H : [hH];
fragment I : [iI];
fragment J : [jJ];
fragment K : [kK];
fragment L : [lL];
fragment M : [mM];
fragment N : [nN];
fragment O : [oO];
fragment P : [pP];
fragment Q : [qQ];
fragment R : [rR];
fragment S : [sS];
fragment T : [tT];
fragment U : [uU];
fragment V : [vV];
fragment W : [wW];
fragment X : [xX];
fragment Y : [yY];
fragment Z : [zZ];
WS: [ \n\t\r]+ -> skip;
So my question is, how can I get the grammar to work for the examples mentioned above? Can we make certain words as mandatory between two triplets (literal operator literal)? In a sense I'm just trying to get a parser to validate the where clause condition but only simple condition and functions are permitted. I also want have a visitor that retrieves the values like function, parenthesis, any literal etc in Java, how to achieve that?
Yes and no.
You can change your grammar to only allow expressions that are comparisons and logical operations on the same:
expr
: term ( '='|'<>'| '<' | '<=' | '>' | '>=' ) term
| expr K_AND expr
| expr K_OR expr
| '(' expr ')'
;
term
: literal_value
| function_name '(' ( expr ( ',' expr )* | '*' )? ')'
;
The issue comes if you want to allow boolean variables or functions -- you need to classify the functions/vars in your lexer and have a different terminal for each, which is tricky and error prone.
Instead, it is generally better to NOT do this kind of checking in the parser -- have your parser be permissive and accept anything expression-like, and generate an expression tree for it. Then have a separate pass over the tree (called a type checker) that checks the types of the operands of operations and the arguments to functions.
This latter approach (with a separate type checker) generally ends up being much simpler, clearer, more flexible, and gives better error messages (rather than just 'syntax error').
Related
I am creating parser and lexer rules for Decaf programming language written in ANTLR4. I'm trying to parse a test file and keep getting an error, there must be something wrong in the grammar but i cant figure it out.
My test file looks like:
class Program {
int i[10];
}
The error is : line 2:8 mismatched input '10' expecting INT_LITERAL
And here is the full Decaf.g4 grammar file
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 ;
Whenever there are 2 or more lexer rules that match the same characters, the one defined first wins. In your case, these 2 rules both match 10:
DECIMAL_LITERAL : DIGIT(DIGIT)*;
INT_LITERAL : DECIMAL_LITERAL | HEX_LITERAL;
and since INT_LITERAL is defined after DECIMAL_LITERAL, the lexer will never create a INT_LITERAL token. If you now try to use it in a parser rule, you get an error message you posted.
The solution: remove INT_LITERAL from your lexer and create a parser rule instead:
int_literal : DECIMAL_LITERAL | HEX_LITERAL;
and use int_literal in your parser rules instead.
This is freaking me out, I just can't find a solution to it. I have a grammar for search queries and would like to match any searchterm in a query composed out of printable letters except for special characters "(", ")". Strings enclosed in quotes are handled separately and work.
Here is a somewhat working grammar:
/* ANTLR Grammar for Minidb Query Language */
grammar Mdb;
start
: searchclause EOF
;
searchclause
: table expr
;
expr
: fieldsearch
| searchop fieldsearch
| unop expr
| expr relop expr
| lparen expr relop expr rparen
;
lparen
: '('
;
rparen
: ')'
;
unop
: NOT
;
relop
: AND
| OR
;
searchop
: NO
| EVERY
;
fieldsearch
: field EQ searchterm
;
field
: ID
;
table
: ID
;
searchterm
:
| STRING
| ID+
| DIGIT+
| DIGIT+ ID+
;
STRING
: '"' ~('\n'|'"')* ('"' )
;
AND
: 'and'
;
OR
: 'or'
;
NOT
: 'not'
;
NO
: 'no'
;
EVERY
: 'every'
;
EQ
: '='
;
fragment VALID_ID_START
: ('a' .. 'z') | ('A' .. 'Z') | '_'
;
fragment VALID_ID_CHAR
: VALID_ID_START | ('0' .. '9')
;
ID
: VALID_ID_START VALID_ID_CHAR*
;
DIGIT
: ('0' .. '9')
;
/*
NOT_SPECIAL
: ~(' ' | '\t' | '\n' | '\r' | '\'' | '"' | ';' | '.' | '=' | '(' | ')' )
; */
WS
: [ \r\n\t] + -> skip
;
The problem is that searchterm is too restricted. It should match any character that is in the commented out NOT_SPECIAL, i.e., valid queries would be:
Person Name=%
Person Address=^%Street%%%$^&*#^
But whenever I try to put NOT_SPECIAL in any way into the definition of searchterm it doesn't work. I have tried putting it literally into the rule, too (commenting out NOT_SPECIAL) and many others things, but it just doesn't work. In most of my attempts the grammar just complained about extraneous input after "=" and said it was expecting EOF. But I also cannot put EOF into NOT_SPECIAL.
Is there any way I can simply parse every text after "=" in rule fieldsearch until there is a whitespace or ")", "("?
N.B. The STRING rule works fine, but the user ought not be required to use quotes every time, because this is a command line tool and they'd need to be escaped.
Target language is Go.
You could solve that by introducing a lexical mode that you'll enter whenever you match an EQ token. Once in that lexical mode, you either match a (, ) or a whitespace (in which case you pop out of the lexical mode), or you keep matching your NOT_SPECIAL chars.
By using lexical modes, you must define your lexer- and parser rules in their own files. Be sure to use lexer grammar ... and parser grammar ... instead of the grammar ... you use in a combined .g4 file.
A quick demo:
lexer grammar MdbLexer;
STRING
: '"' ~[\r\n"]* '"'
;
OPAR
: '('
;
CPAR
: ')'
;
AND
: 'and'
;
OR
: 'or'
;
NOT
: 'not'
;
NO
: 'no'
;
EVERY
: 'every'
;
EQ
: '=' -> pushMode(NOT_SPECIAL_MODE)
;
ID
: VALID_ID_START VALID_ID_CHAR*
;
DIGIT
: [0-9]
;
WS
: [ \r\n\t]+ -> skip
;
fragment VALID_ID_START
: [a-zA-Z_]
;
fragment VALID_ID_CHAR
: [a-zA-Z_0-9]
;
mode NOT_SPECIAL_MODE;
OPAR2
: '(' -> type(OPAR), popMode
;
CPAR2
: ')' -> type(CPAR), popMode
;
WS2
: [ \t\r\n] -> skip, popMode
;
NOT_SPECIAL
: ~[ \t\r\n()]+
;
Your parser grammar would start like this:
parser grammar MdbParser;
options {
tokenVocab=MdbLexer;
}
start
: searchclause EOF
;
// your other parser rules
My Go is a bit rusty, but a small Java test:
String source = "Person Address=^%Street%%%$^&*#^()";
MdbLexer lexer = new MdbLexer(CharStreams.fromString(source));
CommonTokenStream tokens = new CommonTokenStream(lexer);
tokens.fill();
for (Token t : tokens.getTokens()) {
System.out.printf("%-15s %s\n", MdbLexer.VOCABULARY.getSymbolicName(t.getType()), t.getText());
}
print the following:
ID Person
ID Address
EQ =
NOT_SPECIAL ^%Street%%%$^&*#^
OPAR (
CPAR )
EOF <EOF>
I developed this small grammar here i have an issue with:
grammar test;
term : above_term | below_term;
above_term :
<assoc=right> 'forall' binders ',' forall_term
| <assoc=right> above_term '->' above_term
| <assoc=right> above_term '->' below_term
| <assoc=right> below_term '->' above_term
| <assoc=right> below_term '->' below_term
;
below_term :
<assoc = right> below_term arg (arg)*
| '#' qualid (term)*
| below_term '%' IDENT
| qualid
| sort
| '(' term ')'
;
forall_term : term;
arg : term| '(' IDENT ':=' term ')';
binders : binder (binder)*;
binder : name |<assoc=right>name (name)* ':' term | '(' name (name)* ':' term ')' |<assoc=right> name (':' term)? ':=' term;
name : IDENT | '_';
qualid : IDENT | qualid ACCESS_IDENT;
sort : 'Prop' | 'Set' | 'Type' ;
/**************************************
* LEXER RULES
**************************************/
/*
* STRINGS
*/
STRING : '"' (~["])* '"';
/*
* IDENTIFIER AND ACCESS IDENTIFIER
*/
ACCESS_IDENT : '.' IDENT;
IDENT : FIRST_LETTER (SUBSEQUENT_LETTER)*;
fragment FIRST_LETTER : [a-z] | [A-Z] | '_' | UNICODE_LETTER;
fragment SUBSEQUENT_LETTER : [a-z] | [A-Z] | DIGIT | '_' | '"' | UNICODE_LETTER | UNICODE_ID_PART;
fragment UNICODE_LETTER : '\\' 'u' HEX HEX HEX HEX;
fragment UNICODE_ID_PART : '\\' 'u' HEX HEX HEX HEX;
fragment HEX : [0-9a-fA-F];
/*
* NATURAL NUMBERS AND INTEGERS
*/
NUM : DIGIT (DIGIT)*;
INTEGER : ('-')? NUM;
fragment DIGIT : [0-9];
WS : [ \n\t\r] -> skip;
You can copy this grammar and test it with antlr if you want, it will work. Now for my question:
Let's consider an expression like this: a b -> c d -> forall n:nat, c.
Now according to my grammar the ("->") rule (right after forall rule) has the highest precedence.
As for this I want this term to be parsed so that both ("->") rules are on top of the parse tree. like this: (Please note, that this is an abstract view, i know that there are many above and below terms between the leafs)
However sadly it doesn't get parsed this way but this way:
Howcome the parser doesn't see the (->) rules both on top of the parse tree? Is this a precedence issue?
By changing term to below_term in the (arg) rule we can fix the problem arg : below_term| '(' IDENT ':=' term ')'; .
Lets take this expression as an example: a b c.
Once the parser sees, that the pattern a b matches this rule: below_term arg (arg)* he puts a as a below_term and trys to match b with the arg rule. However since arg points to the below_term rule now, no above_term is alowed except when it is braced. This solved my problem.
The term a b -> a b c -> forall n:nat, n now gets parsed this way:
I'm writing an ANTLR lexer/parser for context free grammar.
This is what I have now:
statement
: assignment_statement
;
assignment_statement
: IDENTIFIER '=' expression ';'
;
term
: IDENT
| '(' expression ')'
| INTEGER
| STRING_LITERAL
| CHAR_LITERAL
| IDENT '(' actualParameters ')'
;
negation
: 'not'* term
;
unary
: ('+' | '-')* negation
;
mult
: unary (('*' | '/' | 'mod') unary)*
;
add
: mult (('+' | '-') mult)*
;
relation
: add (('=' | '/=' | '<' | '<=' | '>=' | '>') add)*
;
expression
: relation (('and' | 'or') relation)*
;
IDENTIFIER : LETTER (LETTER | DIGIT)*;
fragment DIGIT : '0'..'9';
fragment LETTER : ('a'..'z' | 'A'..'Z');
So my assignment statement is identified by the form
IDENTIFIER = expression;
However, assignment statement should also take into account cases when the right hand side is a function call (the return value of the statement). For example,
items = getItems();
What grammar rule should I add for this? I thought of adding a function call to the "expression" rule, but I wasn't sure if function call should be regarded as expression..
Thanks
This grammar looks fine to me. I am assuming that IDENT and IDENTIFIER are the same and that you have additional productions for the remaining terminals.
This production seems to define a function call.
| IDENT '(' actualParameters ')'
You need a production for the actual parameters, something like this.
actualParameters : nothing | expression ( ',' expression )*
I have a grammar and everything works fine until this portion:
lexp
: factor ( ('+' | '-') factor)*
;
factor :('-')? IDENT;
This of course introduces an ambiguity. For example a-a can be matched by either Factor - Factor or Factor -> - IDENT
I get the following warning stating this:
[18:49:39] warning(200): withoutWarningButIncomplete.g:57:31:
Decision can match input such as "'-' {IDENT, '-'}" using multiple alternatives: 1, 2
How can I resolve this ambiguity? I just don't see a way around it. Is there some kind of option that I can use?
Here is the full grammar:
program
: includes decls (procedure)*
;
/* Check if correct! */
includes
: ('#include' STRING)*
;
decls
: (typedident ';')*
;
typedident
: ('int' | 'char') IDENT
;
procedure
: ('int' | 'char') IDENT '(' args ')' body
;
args
: typedident (',' typedident )* /* Check if correct! */
| /* epsilon */
;
body
: '{' decls stmtlist '}'
;
stmtlist
: (stmt)*;
stmt
: '{' stmtlist '}'
| 'read' '(' IDENT ')' ';'
| 'output' '(' IDENT ')' ';'
| 'print' '(' STRING ')' ';'
| 'return' (lexp)* ';'
| 'readc' '(' IDENT ')' ';'
| 'outputc' '(' IDENT ')' ';'
| IDENT '(' (IDENT ( ',' IDENT )*)? ')' ';'
| IDENT '=' lexp ';';
lexp
: term (( '+' | '-' ) term) * /*Add in | '-' to reveal the warning! !*/
;
term
: factor (('*' | '/' | '%') factor )*
;
factor : '(' lexp ')'
| ('-')? IDENT
| NUMBER;
fragment DIGIT
: ('0' .. '9')
;
IDENT : ('A' .. 'Z' | 'a' .. 'z') (( 'A' .. 'Z' | 'a' .. 'z' | '0' .. '9' | '_'))* ;
NUMBER
: ( ('-')? DIGIT+)
;
CHARACTER
: '\'' ('a' .. 'z' | 'A' .. 'Z' | '0' .. '9' | '\\n' | '\\t' | '\\\\' | '\\' | 'EOF' |'.' | ',' |':' ) '\'' /* IS THIS COMPLETE? */
;
As mentioned in the comments: these rules are not ambiguous:
lexp
: factor (('+' | '-') factor)*
;
factor : ('-')? IDENT;
This is the cause of the ambiguity:
'return' (lexp)* ';'
which can parse the input a-b in two different ways:
a-b as a single binary expression
a as a single expression, and -b as an unary expression
You will need to change your grammar. Perhaps add a comma in multiple return values? Something like this:
'return' (lexp (',' lexp)*)? ';'
which will match:
return;
return a;
return a, -b;
return a-b, c+d+e, f;
...