Antlr production producing a bunch of single element arrays - antlr

My grammar is working, but I have a bunch of elements in the tree that are single element arrays, and I don't really understand why. I tried reading the information about visitors, but I'm pretty sure the "problem" is with the grammar and perhaps its verbosity. Does anything jump out here? Or perhaps I'm just visiting things incorrectly. In the example below I do not react to visitFnArgs or visitArgs, but just visitFunctionCall. Things like function arguments and statements seem to sometimes be wrapped in single element arrays.
grammar Txl;
root: program;
// High level language
program: stmt (NEWLINE stmt)* NEWLINE? EOF # Statement
;
stmt: require # Condition
| entry # CreateEntry
| assignment # Assign
;
require: REQUIRE valueExpression;
entry: (CREDIT | DEBIT) journal valueExpression (IF valueExpression)? (LPAREN 'id:' valueExpression RPAREN)?;
assignment: IDENT ASSIGN valueExpression;
journal: IDENT COLON IDENT;
valueExpression: expr # Expression;
expr: expr (MULT | DIV) expr # MulDiv
| expr (PLUS | MINUS) expr # AddSub
| expr MOD expr # Mod
| expr POW expr # Pow
| MINUS expr # Negative
| expr AND expr # And
| expr OR expr # Or
| NOT expr # Not
| expr EQ expr # Equality
| expr NEQ expr # Inequality
| expr (LTE | GTE) expr # CmpEqual
| expr (LT | GT) expr # Cmp
| expr QUESTION expr COLON expr # Ternary
| LPAREN expr RPAREN # Parens
| NUMBER # NumberLiteral
| IDENT LPAREN args RPAREN # FunctionCall
| IDENT # Identifier
| STRING_LITERAL # StringLiteral
;
fnArg: expr | journal;
args: (fnArg (',' fnArg)*)?;
// Reserved words
CREDIT: 'credit';
DEBIT: 'debit';
IF: 'if';
REQUIRE: 'require';
// Operators
MULT: '*';
DIV: '/';
MINUS: '-';
PLUS: '+';
POW: '^';
MOD: '%';
LPAREN: '(';
RPAREN: ')';
LBRACE: '[';
RBRACE: ']';
COMMA: ',';
EQ: '==';
NEQ: '!=';
GTE: '>=';
LTE: '<=';
GT: '>';
LT: '<';
ASSIGN: '=';
QUESTION: '?';
COLON: ':';
AND: 'and';
OR: 'or';
NOT: 'not';
HASH: '#';
NEWLINE : [\r\n];
WS: [ \t] + -> skip;
// Entities
NUMBER: ('0' .. '9') + ('.' ('0' .. '9') +)?;
IDENT: [a-zA-Z]+[0-9a-zA-Z]*;
EXTID: [a-zA-Z0-9-]+;
STRING_LITERAL : '"' (~('"' | '\\' | '\r' | '\n') | '\\' ('"' | '\\'))* '"';
This input:
require balance(assets:cash) + balance(assets:earnings) > AMT
Produces the following single element arrays:
SINGLE ELEMENT INSTRUCTION MathOperation (>)
SINGLE ELEMENT INSTRUCTION JournalReference { identifier: 'assets:cash' }
SINGLE ELEMENT INSTRUCTION JournalReference { identifier: 'assets:earnings' }
I wonder if partly my problem is I'm not visiting things properly. Here's my Math visitor:
visitMath(ctx) {
const visited = this.visitChildren(ctx);
return new MathOperation(
visited[0],
ctx.getChild(1).getText(),
visited[2],
);
}
But I assume the problem is in the thing that contains the math operation, which I think is visitRequire:
visitRequire(ctx) {
return new Condition(this.visitExpression(ctx.getChild(1)));
}
Or perhaps in visitValueExpression or visitCondition, which are not overridden in my visitor.

Really short answer: There's nothing wrong with single element arrays. If there was only one instance of a thing that could exist multiple times, then it has to be an array (or List), and that list will have only the one item, if that's how many there are.
Antlr won't "unwrap" a single item to not be in an array. (That would only be valid in untyped languages or languages that allow Union types, and would be a pain to use as you'd always have to check whether you had a "thing" or a list of "thing"s)
Any time the "same type of thing" can exist more than once when matching a rule, ANTLR will make that available as an Array/List of that type.
Eample:
journal: IDENT COLON IDENT;
has 2 IDENT tokens, so it'll be made accessible via the context as a List of those types
(in Java, I'm not positive which language you're using).
public List<TerminalNode> IDENT() { return getTokens(TxlParser.IDENT); }
Two of your examples are of "JournalReference" so this would explain getting a list (if you use the ctx.IDENT() or the ctx.getChild(n) methods).
If I change the Journal rule to be:
journal: j1=IDENT COLON j2=IDENT;
I've given names to each IDENT so I get individual accessors for them (in addition to the IDENT() accessor that returns a list:
public static class JournalContext extends ParserRuleContext {
public Token j1;
public Token j2;
public TerminalNode COLON() { return getToken(TxlParser.COLON, 0); }
public List<TerminalNode> IDENT() { return getTokens(TxlParser.IDENT); }
With the labels you can use cox.j1 or cox.j2 to get individual tokens. (of course you'd name them as appropriate to your use case).
since the FunctionCall alternative of the expr rule uses the args rule
args: (fnArg (',' fnArg)*)?;
and that rule can have more than one fnArg, the it will necessarily be a list of fnArgs in the context:
public static class ArgsContext extends ParserRuleContext {
public List<FnArgContext> fnArg() {
return getRuleContexts(FnArgContext.class);
}
There's really not much you can do (or should want to do to not have that in a List, there can be one or more of them.
Since non of the code you present shows where you're writing your output, its a bit difficult to be more specific than that.
Your visitMath(cox) example is also a bit perplexing as math is not a rule in your grammar, so it would not exist in the Visitor interface.
I would suggest taking a closer look at the *Context classes that are generated for you. They'll provide utility methods that will be much easy to use and read in the future than getChild(n). getChild(n) is obscure, in that you'll have to refer back to the rule and diligently count rule members to determine which child to get, and it is also VERY brittle, in that n will change with any modification to your grammar. (Maintainers, or future you, will appreciate using the utility methods instead.)

Related

Antlr4 mismatched input '<' expecting '<' with (seemingly) no lexer ambiguity

I cannot seem to figure out what antlr is doing here in this grammar. I have a grammar that should match an input like:
i,j : bool;
setvar : set<bool>;
i > 5;
j < 10;
But I keep getting an error telling me that "line 3:13 mismatched input '<' expecting '<'". This tells me there is some ambiguity in the lexer, but I only use '<' in a single token.
Here is the grammar:
//// Parser Rules
grammar MLTL1;
start: block*;
block: var_list ';'
| expr ';'
;
var_list: IDENTIFIER (',' IDENTIFIER)* ':' type ;
type: BASE_TYPE
| KW_SET REL_LT BASE_TYPE REL_GT
;
expr: expr REL_OP expr
| '(' expr ')'
| IDENTIFIER
| INT
;
//// Lexical Spec
// Types
BASE_TYPE: 'bool'
| 'int'
| 'float'
;
// Keywords
KW_SET: 'set' ;
// Op groups for precedence
REL_OP: REL_EQ | REL_NEQ | REL_GT | REL_LT
| REL_GTE | REL_LTE ;
// Relational ops
REL_EQ: '==' ;
REL_NEQ: '!=' ;
REL_GT: '>' ;
REL_LT: '<' ;
REL_GTE: '>=' ;
REL_LTE: '<=' ;
IDENTIFIER
: LETTER (LETTER | DIGIT)*
;
INT
: SIGN? NONZERODIGIT DIGIT*
| '0'
;
fragment
SIGN
: [+-]
;
fragment
DIGIT
: [0-9]
;
fragment
NONZERODIGIT
: [1-9]
;
fragment
LETTER
: [a-zA-Z_]
;
COMMENT : '#' ~[\r\n]* -> skip;
WS : [ \t\r\n]+ -> channel(HIDDEN);
I tested the grammar to see what tokens it is generating for the test input above using this python:
from antlr4 import InputStream, CommonTokenStream
import MLTL1Lexer
import MLTL1Parser
input="""
i,j : bool;
setvar: set<bool>;
i > 5;
j < 10;
"""
lexer = MLTL1Lexer.MLTL1Lexer(InputStream(input))
stream = CommonTokenStream(lexer)
stream.fill()
tokens = stream.getTokens(0,100)
for t in tokens:
print(str(t.type) + " " + t.text)
parser = MLTL1Parser.MLTL1Parser(stream)
parse_tree = parser.start()
print(parse_tree.toStringTree(recog=parser))
And noticed that both '>' and '<' were assigned the same token value despite being two different tokens. Am I missing something here?
(There may be more than just these two instances, but...)
Change REL_OP and BASE_TYPE to parser rules (i.e. make them lowercase.
As you've used them, you're turning many of your intended Lexer rules, effectively into fragments.
I't important to understand that tokens are the "atoms" you have in your grammar, when you combine several of them into another Lexer rule, you just make that the token type.
(If you used grun to dump the tokens you would have seen them identified as REL_OP tokens.
With the changes below, your sample input works just fine.
grammar MLTL1
;
start: block*;
block: var_list ';' | expr ';';
var_list: IDENTIFIER (',' IDENTIFIER)* ':' type;
type: baseType | KW_SET REL_LT baseType REL_GT;
expr: expr rel_op expr | '(' expr ')' | IDENTIFIER | INT;
//// Lexical Spec
// Types
baseType: 'bool' | 'int' | 'float';
// Keywords
KW_SET: 'set';
// Op groups for precedence
rel_op: REL_EQ | REL_NEQ | REL_GT | REL_LT | REL_GTE | REL_LTE;
// Relational ops
REL_EQ: '==';
REL_NEQ: '!=';
REL_GT: '>';
REL_LT: '<';
REL_GTE: '>=';
REL_LTE: '<=';
IDENTIFIER: LETTER (LETTER | DIGIT)*;
INT: SIGN? NONZERODIGIT DIGIT* | '0';
fragment SIGN: [+-];
fragment DIGIT: [0-9];
fragment NONZERODIGIT: [1-9];
fragment LETTER: [a-zA-Z_];
COMMENT: '#' ~[\r\n]* -> skip;
WS: [ \t\r\n]+ -> channel(HIDDEN);

How do I convert this Antlr3 AST to Antlr4?

I'm trying to convert my existing Antlr3 project to Antlr4 to get more functionality. I have this grammar that wouldn't compile with Antlr4.9
expr
: term ( OR^ term )* ;
and
factor
: ava | NOT^ factor | (LPAREN! expr RPAREN!) ;
Mostly because Antlr4 doesn't support ^ and ! anymore. From the documentation it seems like those are
AST root operator. When generating abstract syntax trees (ASTs), token
references suffixed with the "^" root operator force AST nodes to be
created and added as the root of the current tree. This symbol is only
effective when the buildAST option is set. More information about ASTs
is also available.
AST exclude operator. When generating abstract syntax trees, token
references suffixed with the "!" exclude operator are not included in
the AST constructed for that rule. Rule references can also be
suffixed with the exclude operator, which implies that, while the tree
for the referenced rule is constructed, it is not linked into the tree
for the referencing rule. This symbol is only effective when the
buildAST option is set. More information about ASTs is also available.
If I took those out it would compile but I'm not sure what do those mean and how would Antlr4 supports it.
LPAREN and RPAREN is tokens
tokens {
EQUALS = '=';
LPAREN = '(';
RPAREN = ')';
}
which Antlr4 kindly provides the way to convert that in the error messages but not ^ and !. The grammar is for parsing boolean expression for example (a=b AND b=c)
I think this is the rule
targetingexpr returns [boolean value]
: expr { $value = $expr.value; } ;
expr returns [boolean value]
: ^(NOT a=expr) { $value = !a; }
| ^(AND a=expr b=expr) { $value = a && b; }
| ^(OR a=expr b=expr) { $value = a || b; }
| ^(EQUALS A=ALPHANUM B=ALPHANUM) { $value = targetingContext.contains($A.text,$B.text); }
;
The v3 grammar:
...
tokens {
EQUALS = '=';
LPAREN = '(';
RPAREN = ')';
}
...
expr
: term ( OR^ term )* ;
factor
: ava | NOT^ factor | (LPAREN! expr RPAREN!) ;
in v4 would look like this:
...
expr
: term ( OR term )* ;
factor
: ava | NOT factor | (LPAREN expr RPAREN) ;
EQUALS : '=';
LPAREN : '(';
RPAREN : ')';
So, just remove the inline ^ and ! operators (tree rewriting is no longer available in ANTLR4), and move the literal tokens in the tokens { ... } sections into own lexer rules.
I think this is the rule
targetingexpr returns [boolean value]
: expr { $value = $expr.value; } ;
expr returns [boolean value]
: ^(NOT a=expr) { $value = !a; }
| ^(AND a=expr b=expr) { $value = a && b; }
| ^(OR a=expr b=expr) { $value = a || b; }
| ^(EQUALS A=ALPHANUM B=ALPHANUM) { $value = targetingContext.contains($A.text,$B.text); }
;
What you posted there is part of a tree grammar for which there is no equivalent. In ANTLR4 you'd use a visitor to evaluate your expressions instead of inside a tree grammar.

String interpolation: is it possible without adding members?

I would like to create an Antlr parser for custom language and decided to pick a simple calculator as an example. In my new grammar it should be possible to define a string, like this:
s = "Hello, I am a string"
and handle string interpolation.
Text in double quotes enclosed in persent should be treated as interpolated, e.g.
s = "Hello, did you know that %2 + 2% is 4?"
Double percent sign should not be processed, e.g.
s = "He wants 50%% of this deal."
But at the same time my calculator should support modulus operation:
x = 5 % 2
So far, I was able to craft a Lexer/Grammar, which could switch mode and parse simple strings, here they are:
lexer grammar CalcLexer;
EQ: '=';
PLUS: '+';
MINUS: '-';
MULT: '*';
DIV: '/';
LPAREN : '(' ;
RPAREN : ')' ;
SINGLE_PERCENT_POP: '%' -> popMode;
ID : [a-zA-Z]+ ;
INT : [0-9]+ ;
OPEN_DOUBLE_QUOTE: '"' -> pushMode(STRING_MODE);
NEWLINE:'\r'? '\n' ;
WS : [ \t]+ -> skip;
mode STRING_MODE;
DOUBLE_PERCENT: '%%';
SINGLE_PERCENT: '%' -> pushMode(DEFAULT_MODE);
TEXT: ~('%'|'\n'|'"')+;
CLOSE_DOUBLE_QUOTE: '"' -> popMode;
and
parser grammar CalcGrammar;
options { tokenVocab=CalcLexer; } // use tokens from CalcLexer.g4
prog: stat+ ;
stat: expr NEWLINE
| ID EQ (expr|text) NEWLINE
| NEWLINE
;
text: OPEN_DOUBLE_QUOTE content* CLOSE_DOUBLE_QUOTE;
content: DOUBLE_PERCENT | TEXT | SINGLE_PERCENT expr SINGLE_PERCENT_POP;
expr: expr (MULT|DIV) expr
| expr (PLUS|MINUS) expr
| INT
| ID
| LPAREN expr RPAREN
;
But only thing doesn't work and I'm not sure if it ever possible to implement without custom code (members) is modulus operation:
x = 5 % 2
There is no way I can ask Anltr to check for previous mode and safely pop mode.
But I hope my understanding is wrong and there is some way to treat % sign as operator in default mode?
I have found several sources for inspiration, probably they would help you as well:
Parsing string interpolation in ANTLR
ANTLR String interpolation
Parsing String Interpolations with ANTLR4
String interpolation and lexer modes
Murphy's law for StackOverflow: you will find an answer to your own question after several minutes you post detailed question to SO.
Instead of switching to DEFAULT_MODE, I should create separate one - STRING_INTERPOLATION. This way I have to define separate tokens for this mode, which will let use % sign in normal mode (and prohibit in interpolated).
Here is Lexer and Grammar which works for me:
lexer grammar CalcLexer;
EQ: '=';
PLUS: '+';
MINUS: '-';
MULT: '*';
DIV: '/';
MOD: '%';
LPAREN : '(' ;
RPAREN : ')' ;
ID : F_ID;
INT : F_INT;
fragment F_ID: [a-zA-Z]+ ;
fragment F_INT: [0-9]+ ;
OPEN_DOUBLE_QUOTE: '"' -> pushMode(STRING_MODE);
NEWLINE:'\r'? '\n' ;
WS : [ \t]+ -> skip;
mode STRING_MODE;
DOUBLE_PERCENT: '%%';
SINGLE_PERCENT: '%' -> pushMode(STRING_INTERPOLATION);
TEXT: ~('%'|'\n'|'"')+;
CLOSE_DOUBLE_QUOTE: '"' -> popMode;
mode STRING_INTERPOLATION;
SINGLE_PERCENT_POP: '%' -> popMode;
I_PLUS: PLUS -> type(PLUS);
I_MINUS: MINUS -> type(MINUS);
I_MULT: MULT -> type(MULT);
I_DIV: DIV -> type(DIV);
I_MOD: MOD -> type(MOD);
I_LPAREN: LPAREN -> type(LPAREN);
I_RPAREN: RPAREN -> type(RPAREN);
I_ID : F_ID -> type(ID);
I_INT : F_INT -> type(INT);
WS1 : [ \t]+ -> skip;
and
parser grammar CalcGrammar;
options { tokenVocab=CalcLexer; } // use tokens from CalcLexer.g4
prog: stat+ ;
stat: expr NEWLINE
| ID EQ (expr|text) NEWLINE
| NEWLINE
;
text: OPEN_DOUBLE_QUOTE content* CLOSE_DOUBLE_QUOTE;
content: DOUBLE_PERCENT | TEXT | SINGLE_PERCENT expr SINGLE_PERCENT_POP;
expr: expr (MULT|DIV|MOD) expr
| expr (PLUS|MINUS) expr
| INT
| ID
| LPAREN expr RPAREN
;
I hope this would help someone. Probably, future me.

How to handle the precedence of assignment operator in a PHP parser?

I wrote a PHP5 parser in ANTLR 3.4, which is almost ready, but I can not handle one of the tricky feature of PHP. My problem is with the precedence of assignment operator. As the PHP manual says the precedence of assignment is almost at the end of the list. Only and, xor, or and , are after it in the list.
But there is a note on this the manual page which says:
Although = has a lower precedence than most other operators, PHP will
still allow expressions similar to the following: if (!$a = foo()), in
which case the return value of foo() is put into $a.
The small example in the note isn't a problem for my parser, I can handle this as a special case in the assigment rule.
But there are more complex codes eg:
if ($a && $b = func()) {}
My parser fails here, because it recognizes first $a && $b and can not deal with the rest of the conditioin. This is because the && has higher precedence, than =.
If I put brackets around the right side of &&:
if ($a && ($b = func())) {}
In this way the parser recognizes the structure well.
The operators are built in the way that the ANTLR book recommends: there are the base exressions at the first step and each level of operators are coming after each other.
Is there any way to handle this precedence jumping?
Don't look at it as an assignment, but let's name it an assignment expression. Put this assignment expression "below" the unary expressions (so they have a higher precedence than the unary ones):
grammar T;
options {
output=AST;
}
tokens {
BLOCK;
FUNC_CALL;
EXPR_LIST;
}
parse
: stat* EOF!
;
stat
: assignment ';'!
| if_stat
;
assignment
: Var '='^ expr
;
if_stat
: If '(' expr ')' block -> ^(If expr block)
;
block
: '{' stat* '}' -> ^(BLOCK stat*)
;
expr
: or_expr
;
or_expr
: and_expr ('||'^ and_expr)*
;
and_expr
: unary_expr ('&&'^ unary_expr)*
;
unary_expr
: '!'^ assign_expr
| '-'^ assign_expr
| assign_expr
;
assign_expr
: Var ('='^ atom)*
| atom
;
atom
: Num
| func_call
;
func_call
: Id '(' expr_list ')' -> ^(FUNC_CALL Id expr_list)
;
expr_list
: (expr (',' expr)*)? -> ^(EXPR_LIST expr*)
;
If : 'if';
Num : '0'..'9'+;
Var : '$' Id;
Id : ('a'..'z')+;
Space : (' ' | '\t' | '\r' | '\n')+ {skip();};
If you'd now parse the source:
if (!$a = foo()) { $a = 1 && 2; }
if ($a && $b = func()) { $b = 2 && 3; }
if ($a = baz() && $b) { $c = 3 && 4; }
the following AST would get constructed:

ANTLR expression interpreter

I have created the following grammar: I would like some idea how to build an interpreter that returns a tree in java, which I can later use for printing in the screen, Im bit stack on how to start on it.
grammar myDSL;
options {
language = Java;
}
#header {
package DSL;
}
#lexer::header {
package DSL;
}
program
: IDENT '={' components* '}'
;
components
: IDENT '=('(shape)(shape|connectors)* ')'
;
shape
: 'Box' '(' (INTEGER ','?)* ')'
| 'Cylinder' '(' (INTEGER ','?)* ')'
| 'Sphere' '(' (INTEGER ','?)* ')'
;
connectors
: type '(' (INTEGER ','?)* ')'
;
type
: 'MG'
| 'EL'
;
IDENT: ('a'..'z' | 'A'..'Z')('a'..'z' | 'A'..'Z' | '0'..'0')*;
INTEGER: '0'..'9'+;
// This if for the empty spaces between tokens and avoids them in the parser
WS: (' ' | '\t' | '\n' | '\r' | '\f')+ {$channel=HIDDEN;};
COMMENT: '//' .* ('\n' | '\r') {$channel=HIDDEN;};
A couple of remarks:
There's no need to set the language for Java, which is the default target language. So you can remove this:
options {
language = Java;
}
Your IDENT contains an error:
IDENT: ('a'..'z' | 'A'..'Z')('a'..'z' | 'A'..'Z' | '0'..'0')*;
the '0'..'0') should most probably be '0'..'9').
The sub rule (INTEGER ','?)* also matches source like 1 2 3 4 (no comma's at all!). Perhaps you meant to do: (INTEGER (',' INTEGER)*)?
Now, as to your question: how to let ANTLR construct a proper AST? This can be done by adding output = AST; in your options block:
options {
//language = Java;
output = AST;
}
And then either adding the "tree operators" ^ and ! in your parser rules, or by using tree rewrite rules: rule: a b c -> ^(c b a).
The "tree operator" ^ is used to define the root of the (sub) tree and ! is used to exclude a token from the (sub) tree.
Rewrite rules have ^( /* tokens here */ ) where the first token (right after ^() is the root of the (sub) tree, and all following tokens are child nodes of the root.
An example might be in order. Let's take your first rule:
program
: IDENT '={' components* '}'
;
and you want to let IDENT be the root, components* the children and you want to exclude ={ and } from the tree. You can do that by doing:
program
: IDENT^ '={'! components* '}'!
;
or by doing:
program
: IDENT '={' components* '}' -> ^(IDENT components*)
;