Antlr4 how to build a grammar allowed keywords as identifier

Question

This is a demo code

label:
var id
let id = 10
goto label

If allowed keyword as identifier will be

let:
var var
let var = 10
goto let

This is totally legal code. But it seems very hard to do this in antlr.

AFAIK, If antlr match a token let, will never fallback to id token. so for antlr it will see

LET_TOKEN :
VAR_TOKEN <missing ID_TOKEN>VAR_TOKEN
LET_TOKEN <missing ID_TOKEN>VAR_TOKEN = 10

although antlr allowed predicate, I have to control ever token match and problematic. grammar become this

grammar Demo;
options {
  language = Go;
}
@parser::members{
    var _need = map[string]bool{}
    func skip(name string,v bool){
        _need[name] = !v
        fmt.Println("SKIP",name,v)
    }
    func need(name string)bool{
        fmt.Println("NEED",name,_need[name])
        return _need[name]
    }
}

proj@init{skip("inst",false)}: (line? NL)* EOF;
line
    : VAR ID
    | LET ID EQ? Integer
    ;

NL: '\n';
VAR: {need("inst")}? 'var' {skip("inst",true)};
LET: {need("inst")}? 'let' {skip("inst",true)};
EQ: '=';

ID: ([a-zA-Z] [a-zA-Z0-9]*);
Integer: [0-9]+;

WS: [ \t] -> skip;

Looks so terrible.

But this is easy in peg, test this in pegjs

Expression = (Line? _ '\n')* ;

Line
  = 'var' _ ID
  / 'let' _ ID _ "=" _ Integer

Integer "integer"
  = [0-9]+ { return parseInt(text(), 10); }

ID = [a-zA-Z] [a-zA-Z0-9]*

_ "whitespace"
  = [ \t]*

I actually done this in peggo and javacc.

My question is how to handle these grammars in antlr4.6, I was so excited about the antlr4.6 go target, but seems I choose the wrong tool for my grammar ?

Answer 1

The simplest way is to define a parser rule for identifiers:

id: ID | VAR | LET;

VAR: 'var';
LET: 'let';
ID: [a-zA-Z] [a-zA-Z0-9]*;

And then use id instead of ID in your parser rules.

A different way is to use ID for identifiers and keywords, and use predicates for disambiguation. But it's less readable, so I'd use the first way instead.