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    <p>I find it hard to believe a <code>p</code> token is created based on the grammar you posted. Lexer rules that have <code>fragment</code> in front of them will not produce tokens: these rules are only used by other lexer rules.</p> <p>A simple demo shows this:</p> <pre><code>lexer grammar ExpressionLexer; @lexer::members { public static void main(String[] args) throws Exception { ExpressionLexer lexer = new ExpressionLexer(new ANTLRStringStream(args[0])); CommonTokenStream tokens = new CommonTokenStream(lexer); tokens.fill(); // remove this line when using ANTLR 3.2 or an older version System.out.println(tokens); } } COMMA : ','; LPAREN : '('; RPAREN : ')'; LBRACK : '['; RBRACK : ']'; PLUS : '+'; MINUS : '-'; MULT : '*'; DIV : '/'; PREV_VALUE : 'PreviousValue'; fragment DIGIT : ('0'..'9'); fragment LETTER : ('a'..'z'|'A'..'Z'|'_'); fragment TAB : ('\t') ; fragment NEWLINE : ('\r'|'\n') ; fragment SPACE : (' ') ; </code></pre> <p>Now generate the lexer and compile the <code>.java</code> source file:</p> <pre>java -cp antlr-3.3.jar org.antlr.Tool ExpressionLexer.g javac -cp antlr-3.3.jar *.java</pre> <p>and run a few tests:</p> <pre>java -cp .:antlr-3.3.jar ExpressionLexer p line 1:0 no viable alternative at character 'p'</pre> <p>which is correct since there is no (non-fragment) rule that starts with, or matches, a <code>"p"</code>.</p> <pre>java -cp .:antlr-3.3.jar ExpressionLexer P line 1:1 mismatched character '' expecting 'r'</pre> <p>which is correct since the only (non-fragment) rule that starts with a <code>"P"</code> expects an <code>"r"</code> to be the next character (which isn't there).</p>
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    1. USBart Kiers
    1. USBart Kiers
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    1. COThank you for good explanation! Now I see my mistake. All I need to do is to add new lexer rule for words (without fragment) to produce correct tokens.
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      1. USVilla F.
    2. COYou're welcome @VillaF.
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