ANTLR4 如何创建允许除两个选定字符之外的所有字符的正则表达式？

Question

Hi for example I have this code for the g4 file:

a: [A-Z][A-Z];
b: [a-z]'3';

Now I want to add one line more, which recognizes all characters that do not belong to a or b

I tried:

a: [A-Z][A-Z];
b: [a-z]'3';
ALLOTHERCHARACTERS: ~[a]|~[b]

But i didn´t work.

For example the input 84209ddjio29 should now be in ALLOTHERCARACTERS, but i didn ´t work.

(The Lexer gives at the end a java file, but I think this is not important to know, for this "task")

Answer 1

There are many things going wrong here: inside parser rules, you cannot use character sets. So a: [AZ][AZ]; is not possible. Only a lexer rule can use character sets, so A: [AZ][AZ]; is valid.

So, to define a valid (lexer) grammar, you'd need to do this:

A : [A-Z] [A-Z];
B : [a-z] '3';

Now for your second problem: how to negate rules A and B ? Answer: you cannot. You can only negate single characters. So negating A: [AZ]; would be NA: ~[AZ]; (or NA: ~A; is also valid). But you cannot negate a rule that matches 2 characters like A: [AZ] [AZ]; .

If you want a rule that matches anything other than upper case letters, lower case letters and the digit 3, then you can so this:

ALLOTHERCHARACTERS : ~[A-Za-z3];

Answer 2

This is the proper syntax for "anything except":

[^ab]

so that will match any character that is not a or b.