Antlr4 how to build a grammar allowed keywords as identifier - parsing

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.