Can the parser return the source text without additional spaces?
Yes, you need to define a lexer rule that captures these spaces, and then skip()them:
Space
: (' ' | '\t') {skip();}
;
which will ignore spaces and tabs.
PS. , Java . skip() (skip() #, ). \r \n.
, . , ANTLR, :
grammar T;
parse
: stat* EOF
;
stat
: Type Identifier '=' Int ';'
;
Type
: 'int'
| 'double'
| 'boolean'
;
Identifier
: ('a'..'z' | 'A'..'Z' | '_') ('a'..'z' | 'A'..'Z' | '_' | '0'..'9')*
;
Int
: '0'..'9'+
;
Space
: (' ' | '\t' | '\n' | 'r')+ {skip();}
;
:
int x = 5 ; double y =5;boolean z = 0 ;
:
int x=5;
double y=5;
boolean z=0;
( ():
grammar T;
parse returns [String str]
@init{StringBuilder buffer = new StringBuilder();}
@after{$str = buffer.toString();}
: (stat {buffer.append($stat.str).append('\n');})* EOF
;
stat returns [String str]
: Type Identifier '=' Int ';'
{$str = $Type.text + " " + $Identifier.text + "=" + $Int.text + ";";}
;
Type
: 'int'
| 'double'
| 'boolean'
;
Identifier
: ('a'..'z' | 'A'..'Z' | '_') ('a'..'z' | 'A'..'Z' | '_' | '0'..'9')*
;
Int
: '0'..'9'+
;
Space
: (' ' | '\t' | '\n' | 'r')+ {skip();}
;
:
import org.antlr.runtime.*;
public class Main {
public static void main(String[] args) throws Exception {
String source = "int x = 5 ; double y =5;boolean z = 0 ;";
ANTLRStringStream in = new ANTLRStringStream(source);
TLexer lexer = new TLexer(in);
CommonTokenStream tokens = new CommonTokenStream(lexer);
TParser parser = new TParser(tokens);
System.out.println("Result:\n"+parser.parse());
}
}
:
Result:
int x=5;
double y=5;
boolean z=0;