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sharwell avatar sharwell commented on May 5, 2024

As far as the actual input symbols which could be matched by such a rule, it's easy to see the following rule would match exactly the same thing:

expr2
    : 'x'+
    ;

The problem lies in how to build the parse tree. If the input is x x, the only possible parse tree from the expr rule is:

(expr x x)

However, if the input is x x x, either of the following two trees is possible, and without an operator in place to specify an associativity there's no clear choice:

(expr (expr x x) x)
(expr x (expr x x))

Depending on the intended associativity and desired tree structure, the rule expr would need to be written in one of the following forms.

expr
    : 'x'
    | expr 'x'
    ;

expr
    : 'x'
    | 'x' expr
    ;

expr
    : 'x'+
    ;

from antlr4.

sharwell avatar sharwell commented on May 5, 2024

You may also be able to produce a similar rule by moving the closure outside the expr rule:

exprs
    : expr+
    ;

expr
    : 'x'
    ;

from antlr4.

sharwell avatar sharwell commented on May 5, 2024

This issue has been resolved; pull request parrt#105 contains a regression test.

from antlr4.

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