How does the ANTLR lexer unambiguously eliminate its rules (or why does my parser create "input mismatch" errors)?

Question

Note. This is a question with an answering machine , the purpose of which is to provide information on one of the most common errors made by ANTLR users.

---

When I test this very simple grammar:

grammar KeyValues;

keyValueList: keyValue*;
keyValue: key=IDENTIFIER '=' value=INTEGER ';';

IDENTIFIER: [A-Za-z0-9]+;
INTEGER: [0-9]+;

WS: [ \t\r\n]+ -> skip;

With the following input:

foo = 42;

I end up with the following runtime error:

line 1: 6 mismatch input '42', waiting for INTEGER
line 1: 8 inconsistent input ';' expecting '='

Why does ANTLR not recognize 42as INTEGERin this case?
It must exactly match the pattern [0-9]+.

, INTEGER IDENTIFIER, , , , ?

Answer 1

ANTLR , , , ( " INTEGER ", ). . , -, , .

, lexer . :

• ,
• (, '='),
• lexer , ,

, , ANTLR.

---

, keyValue: IDENTIFIER '=' INTEGER ';', '=' ';' .

42 INTEGER, IDENTIFIER, IDENTIFIER, : IDENTIFIER '=' IDENTIFIER ';', ' t keyValue. , lexer, , " INTEGER next".

, . :

• IDENTIFIER [A-Za-z] [A-Za-z0-9]* ( ). , , , .
• INTEGER IDENTIFIER. , , , .
• , :
  INTEGER IDENTIFIER, INTEGER. id: IDENTIFIER | INTEGER;, IDENTIFIER , keyValue key=id '=' value=INTEGER ';'.

---

:

:

grammar LexerPriorityRulesExample;

// Parser rules

randomParserRule: 'foo'; // Implicitly declared token type

// Lexer rules

BAR: 'bar';
IDENTIFIER: [A-Za-z]+;
BAZ: 'baz';

WS: [ \t\r\n]+ -> skip;

:

aaa foo bar baz barz

:

IDENTIFIER 'foo' BAR IDENTIFIER IDENTIFIER EOF

• aaa IDENTIFIER

  IDENTIFIER , .

• foo 'foo'

  randomParserRule 'foo', IDENTIFIER.

• BAR BAR

  BAR, IDENTIFIER .

• baz IDENTIFIER

  baz, IDENTIFIER. , BAR.

  , baz , IDENTIFIER , baz.

• barz IDENTIFIER

  BAR 3 (BAR), IDENTIFIER 4 . IDENTIFIER , BAR.

• EOF ( ) - , .

, . , , .

, 'foo', , lexer. , . , , .