define a grammar in Antlr - antlr

Question

I have defined the following grammar.
grammar Sample_1;
#header {
package a;
}
#lexer::header {
package a;
}
program
:
define*
implement*
;
define
: IDENT '=(' INTEGER',' INTEGER ')'
;
implement
:IDENT '=(' (IDENT ','?)* ')'
;
fragment LETTER : ('a'..'z' | 'A'..'Z') ;
fragment DIGIT : '0'..'9';
INTEGER : DIGIT+ ;
IDENT : LETTER (LETTER | DIGIT)*;
WS : (' ' | '\t' | '\n' | '\r' | '\f')+ {$channel = HIDDEN;};
COMMENT : '//' .* ('\n'|'\r') {$channel = HIDDEN;};
How to check in this grammar so that when I have the example
A=(1,1)
B=(1,2)
G=(A,B)
the result is successful but if I write
A=(1,1)
B=(1,2)
G=(A,E)
it gives an error that E is not defined
thanks
the result:
i got it working thanks a lot:
grammar Sample_1;
#members{
int level=0;
}
#header {
package a;
}
#lexer::header {
package a;
}
program
:
block
;
block
scope {
List symbols;
}
#init {
$block::symbols=new ArrayList();
level++;
}
#after {
System.err.println("Hello");
level--;
}
: (define* implement+)
;
define
: IDENT {$block::symbols.add($IDENT.text);} '=(' INTEGER',' INTEGER ')'
;
implement
:IDENT '=(' (a=IDENT
{if (!$block::symbols.contains($a.text)){
System.err.println("undefined");
}}','?)* ')'
;
fragment LETTER : ('a'..'z' | 'A'..'Z') ;
fragment DIGIT : '0'..'9';
INTEGER : DIGIT+ ;
IDENT : LETTER (LETTER | DIGIT)*;
WS : (' ' | '\t' | '\n' | '\r' | '\f')+ {$channel = HIDDEN;};
COMMENT : '//' .* ('\n'|'\r') {$channel = HIDDEN;};

Answer 1

Antlr supports actions, little snippets of code embedded in the grammar file.
An action for an assignment could store into a map. An action for a right-hand-side IDENT could try to pull a value from the map, and throw an exception if it fails.
Chapter 6 in Terrence Parr's "The Definitive ANTLR Reference" covers actions.