Macro expansion in Antlr4 - antlr4

We are developing a modeling tool in which the user can insert small pieces of code in various places. The user can add macros to this code, which we need to expand before parsing the final code. For example, when I parse the following code...
FOR i := 6 TO 1 DO
j := 6;
IF SM_MCP.State = oMcpApplication.UNDEFINED
THEN
Macro1(MsgNotInZeroPosition);
END_IF;
END_FOR;
... I just want to handle the macro in a visitor function. The result would be:
FOR i := 6 TO 1 DO
j := 6;
IF SM_MCP.State = oMcpApplication.UNDEFINED
THEN
<ClassName>.MessageQueue := MsgNotInZeroPosition;
END_IF;
END_FOR;
(This is just a made up example.) I have tried to create a simple grammar that parses the macros and simply copies everything around it. But, for some reason I cannot get it working properly.
I have tried several grammars, they all looked something like this:
grammar Macros;
source : (code macro )* code ;
macro : macro1 | macro2 | macro3 | macro4 ;
macro1 : Macro1 '(' parameters ')' ;
macro2 : Macro2 '(' parameters ')' ;
macro3 : Macro3 '(' parameters ')' ;
macro4 : Macro4 '(' parameters ')' ;
code : OTHER* ;
parameters : OTHER* ;
// Lexer rules
Macro1: 'Macro1';
Macro2: 'Macro2';
Macro3: 'Macro3';
Macro4: 'Macro4';
OTHER : . ;
I have tried non-greedy matching too, but most of the times everything ended up in one token or the parser crashed. Can anyone point me in the proper direction?

Your macro production shouldn't be using the macro tokens:
macro : Macro1 | Macro2 | Macro3 | Macro4 ;
but the macro productions:
macro : macro1 | macro2 | macro3 | macro4 ;
When I change that, your example input gets parsed as follows:

Related

ANTLR: how to debug a misidentified token

I am trying to implement a grammar in Antlr4 for a simple template engine. This engine consists of 3 different clauses:
IF ANSWERED ( variable )
END IF
Variable
Variable can be any upper or lowercase letter including white spaces. Both IF ANSWERED and END IF are always uppercase.
I have written the following grammar/lexer rules so far, but my problem is that IF ANSWERED keeps getting recognized as a Variable and not as 2 tokens IF and ANSWERED.
grammar program;
/**grammar */
command: (ifStart | ifEnd | VARIABLE ) EOF;
ifStart: IF ANSWERED '(' VARIABLE ')';
ifEnd: 'END IF';
/** lexer */
IF: 'IF';
ANSWERED: 'ANSWERED';
TEXT: (LOWERCASE | UPPERCASE | NUMBER) ;
VARIABLE: (TEXT | [ \t\r\n])+;
fragment LOWERCASE: [a-z];
fragment UPPERCASE: [A-Z];
fragment NUMBER: [0-9];
If I try to parse IF ANSWERED ( FirstName ) I get the following output:
[#0,0:10='IF ANSWERED',**<VARIABLE>**,1:0]
[#1,11:11='(',<'('>,1:11]
[#2,12:25='Execution date',<VARIABLE>,1:12]
[#3,26:26=')',<')'>,1:26]
[#4,27:26='<EOF>',<EOF>,1:27]
line 1:0 mismatched input 'IF ANSWERED' expecting 'IF'
I read that Antlr4 is greedy and tries to match the biggest possible token, but I fail to understand what is the correct approach, or how to think through the problem to find a solution.
Correct: ANTLR's lexer is greedy, and tries to consume as much as possible. That is why IF ANSWERED is tokenised as a TEXT token instead of 2 separate keywords. You'll need to change TEXT so that it does not match spaces.
Something like this could get you started:
parse
: command* EOF
;
command
: (ifStatement | variable)+
;
ifStatement
: IF ANSWERED '(' variable ')' command* END IF
;
variable
: TEXT
;
IF : 'IF';
END : 'END';
ANSWERED : 'ANSWERED';
TEXT : [a-zA-Z0-9]+;
SPACES : [ \t\r\n]+ -> skip;

ANTLR4 ambiguity - how to solve

I would like to solve the following ambiguity:
grammar test;
WS : (' ' | '\t' | '\n' | '\r' | '\f')+ -> skip;
program
:
input* EOF;
input
: '%' statement
| inputText
;
inputText
: ~('%')+
;
statement
: Identifier '=' DecimalConstant ';'
;
DecimalConstant
: [0-9]+
;
Identifier
: Letter LetterOrDigit*
;
fragment
Letter
: [a-zA-Z$##_.]
;
fragment
LetterOrDigit
: [a-zA-Z0-9$##_.]
;
Sample input:
%a=5;
aa bbbb
As soon as I put a space after "aa" with values like "bbbb" an ambiguity is created.
In fact I want inputText to contain the full string "aa bbbb".
There is no ambiguity. The input aa bbbb will always be tokenised as 2 Identifier tokens. No matter what any parser rule is trying to match. The lexer operates independently from the parser.
Also, the rule:
inputText
: ~('%')+
;
does not match one or more characters other than '%'.
Inside parser rules, the ~ negates tokens, not characters. So ~'%' inside a parser rule will match any token, other than a '%' token. Inside the lexer, ~'%' matches any character other than '%'.
But creating a lexer rule like this:
InputText
: ~('%')+
;
will cause your example input to be tokenised as a single '%' token, followed by a large 2nd token that'd match this: a=5;\naa bbbb. This is how ANTLR's lexer works: match as much characters as possible (no matter what the parser rule is trying to match).
I found the solution:
grammar test;
WS : (' ' | '\t' | '\n' | '\r' | '\f')+ -> skip;
program
:
input EOF;
input
: inputText ('%' statement inputText)*
;
inputText
: ~('%')*
;
statement
: Identifier '=' DecimalConstant ';'
;
DecimalConstant
: [0-9]+
;
Identifier
: Letter LetterOrDigit*
;
fragment
Letter
: [a-zA-Z$##_.]
;
fragment
LetterOrDigit
: [a-zA-Z0-9$##_.]
;

Antlr4 parsing inconsistency

in a little test-parser I just wrote, I encountered a weird problem, which I don't quite understand.
Stripping it down to the smallest example showing the problem, let's start with the following grammar:
Testing.g4:
grammar Testing;
cscript // This is the construct I shortened
: (statement_list)* ;
statement_list
: statement ';' statement_list?
| block
;
statement
: assignment_statement
;
block : '{' statement_list? '}' ;
expression
: left=expression op=('*'|'/') right=expression # arithmeticExpression
| left=expression op=('+'|'-') right=expression # arithmeticExpression
| left=expression op=Comparison_operator right=expression # comparisonExpression
| ID # variableValueExpression
| constant # ignore // will be executed with the rule name
;
assignment_statement
: ID op=Assignment_operator expression
;
constant
: INT
| REAL;
Assignment_operator : ('=' | '+=' | '-=') ;
Comparison_operator : ('<' | '>' | '==' | '!=') ;
Comment : '//' .*? '\n' -> skip;
fragment NUM : [0-9];
INT : NUM+;
REAL
: NUM* '.' NUM+
| '.' NUM+
| INT
;
ID : [a-zA-Z_] [a-zA-Z_0-9]*;
WS : [ \t\r\n]+ -> skip;
Using the input
z = x + y;
everything is fine, we get a parse tree which goes from cscript to statement_list, statement, assignment_statement, id and expression. Great!
Now, if I add the possibility to declare variables, all goes down the drain:
This is the change to the grammar:
cscript
: (statement_list | variable_declaration ';')* ;
variable_declaration
: type ID ('=' expression)?
;
type
: 'int'
| 'real'
;
statement_list
: statement ';' statement_list?
| block
;
statement
: assignment_statement
;
// (continue as before)
All of a sudden, the same test-input gets wrongly dissected into two statement_lists, each continued to a statement with a "missing ';'" warning, the first going to an incomplete assignment_statement of "z =" and the second to an incomplete assignment_statement "x +".
My attempt to show the parse tree in text-form:
cscript
statement_list
statement
assignment_statement
'z'
'=' [marked as error]
[warning: missing ';']
statement_list
statement
assignment_statement
'x'
'+' [marked as error]
'y' [marked as error]
';'
Can anyone tell me what the problem is? (And how to fix it? ;-))
Edit on 2016-12-26, after Mike's comment:
After replacing all implicit lexer rules with explicit declarations, all of a sudden, the input "z = x + y" worked. (thumbs up)
The next thing I did was restoring more of the original example I had in mind, and adding a new input line
int x = 22;
to the input (which worked previously, but did not make it into the minimal example). Now, that line fails. This is the -token output of the test rig:
[#0,0:2='int',<4>,1:0]
[#1,4:4='x',<22>,1:4]
[#2,6:6='=',<1>,1:6]
[#3,8:9='22',<20>,1:8]
[#4,10:10=';',<12>,1:10]
[#5,13:13='z',<22>,2:0]
[#6,15:15='=',<1>,2:2]
[#7,17:17='x',<22>,2:4]
[#8,19:19='+',<18>,2:6]
[#9,21:21='y',<22>,2:8]
[#10,22:22=';',<12>,2:9]
[#11,25:24='<EOF>',<-1>,3:0]
line 1:6 mismatched input '=' expecting '='
As the problem seemed to be in the variable_declaration part, I even tried to split this into two parsing rules like this:
cscript
: (statement_list | variable_declaration_and_assignment SEMICOLON | variable_declaration SEMICOLON)* ;
variable_declaration_and_assignment
: type ID EQUAL expression
;
variable_declaration
: type ID
;
With the result:
line 1:6 no viable alternative at input 'intx='
Still stuck :-(
BTW: Splitting the "int x = 22;" into "int x;" and "x = 22;" works. sigh
Edit on 2016-12-26, after Mike's next comment:
Double-checked, and everything is lexer rules. Still, the mismatch between '=' and '=' (which I unfortunately cannot reconstruct anymore) gave me the idea to check the token types. The current status is:
(Shortened grammar)
cscript
: (statement_list | variable_declaration)* ;
...
variable_declaration
: type ID (EQUAL expression)? SEMICOLON
;
...
Assignment_operator : (EQUAL | PLUS_EQ | MINUS_EQ) ;
// among others
PLUS_EQ : '+=';
MINUS_EQ : '-=';
EQUAL: '=';
...
Shortened output:
[#0,0:2='int',<4>,1:0]
[#1,4:4='x',<22>,1:4]
[#2,6:6='=',<1>,1:6]
...
line 1:6 mismatched input '=' expecting ';'
Here, if I understand this correctly, the '=' is parsed to token type 1, which - according to the lexer.tokens output - is Assignment_Operator, while the expected EQUAL would be 13.
Might this be the problem?
Ok, seems the main take away here is: think about your definitions and how you define them. Create explicit lexer rules for your literals instead of defining them implicitly in the parser rules. Check the token values you get from the lexer if the parser gives you weird errors, because they must be correct in the first place or your parse has no chance to do its job.

Is there a way to resolve this ambiguity without using predicates?

Grammar:
grammar Test;
file: (procDef | statement)* EOF;
procDef: 'procedure' ID NL statement+ ;
statement: 'statement'? NL;
WS: (' ' | '\t') -> skip;
NL: ('\r\n' | '\r' | '\n');
ID: [a-zA-Z0-9]+;
Test data:
statement
procedure Proc1
statement
statement
The parser does what I want (i.e. statement+ is greedy), but it reports an ambiguity because it doesn't know whether the last statement belongs to procDef or file (as I understand it).
As predicates are language dependent I'd prefer not to use one.
The procedure is supposed to end when a statement that can't belong to it, such as 'procedure', occurs.
I also would prefer to have the statements bound to the procedure to avoid having to rearrange the structure later.
Edit
It seems I should expand my test data a bit (but I will leave the original as it is small and shows the ambiguity I want to solve).
I want to be able to handle situations like this:
statement
procedure Proc1
statement
statement
procedure Proc2
statement
statement
procedure Proc2a
statement
statement
global
statement
procedure Proc3
statement
statement
(The indentation is not significant.) I can do it without predicates with something like
file: (
commonStatement
| globalStatement
)* EOF;
procDef: 'procedure' ID NL commonStatement+ ;
commonStatement: 'statement'? NL;
globalStatement: 'global' NL | procDef (globalStatement | EOF);
but then the tree becomes deeper with each consecutive procDef, and that feels very undesirable.
Then a solution with predicates is actually preferable.
#parser::members { boolean inProc; }
file: (
{!inProc}? commonStatement
| globalStatement
)* EOF;
procDef: 'procedure' ID {inProc = true;} NL commonStatement+ ;
commonStatement: 'statement'? NL;
globalStatement: ('global' NL {inProc = false;} | procDef) ;
The situation is actually worse than this, as globally acessible commonStatements can occur without an intervening globalStatement (accessible through gotos), but there is no way a parser can distinguish between that and statements belonging to the procedure, so my plan was to just discourage such use (and I don't think it's common). In fact, it is perfectly legal to jump into procedure code as well ...
It may turn out that in the end I will have to examine runtime paths anyway (scope is very much determined at runtime), and the grammar might end up something like
file: (
commonStatement
| globalStatement
| procDef
)* EOF;
procDef: 'procedure' ID NL procStatement*;
commonStatement: 'statement'? NL;
procStatement: 'proc' NL;
globalStatement: 'global' NL;
We will see ...
By your criteria, it is impossible for a statement to follow a procDef. You are well within your rights to design a language that way, but I hope you have an answer ready for the FAQ "How do I write a statement which comes after a procedure definition."
Writing the grammar is the easy part:
file: statement* procDef* EOF;

antlr tokenizer starts with the last token

I have the following grammar:
grammar Aligner;
line
: emptyLine
| codeLine
;
emptyLine
: ( KW_EMPTY KW_LINE )?
( EOL | EOF )
;
codeLine
: KW_LINE COLON
indent
CODE
( EOL | EOF )
;
indent
: absolute_indent
| relative_indent
;
absolute_indent
: NUMBER
;
relative_indent
: sign NUMBER
;
sign
: PLUS
| MINUS
;
COLON: ':';
MINUS: '-';
PLUS: '+';
KW_EMPTY: 'empty';
KW_LINE: 'line';
NUMBER
: DIGIT+
;
EOL
: ('\n' | '\r\n')
;
SPACING
: LINE_WS -> skip
;
CODE
: (~('\n' | '\r'))+
;
fragment
DIGIT
: '0'..'9'
;
fragment
LINE_WS
: ' '
| '\t'
| '\u000C'
;
when I try to parse - empty line I receive error: line 1:0 no viable alternative at input 'empty line'. When I debug what is going on, the very first token is from type CODE and includes the whole line.
What I am doing wrong?
ANTLR will try to match the longest possible token. When two lexer rules match the same string of a given length, the first rule that appears in the grammar wins.
You rule CODE is basically a catch-all: it will match whole lines of text. So here ANTLR has the choice of matching empty line as one single token of type CODE, and as no other rule can produce a token of length 10, the CODE rule will consume the whole line.
You should rewrite the CODE rule to make it match only what you mean by a code. Right now it's way too broad.

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