RegexExpressionHelper - Support all bracket style delimiters

[staabm]

refs #3271 (comment)

I was not aware these variants exist

[staabm]

//cc @Seldaek

I was not able to add a test for [(\d+)(?P<num>\d+)]n

function (string $s): void {
	if (preg_match('[(\d+)(?P<num>\d+)]n', $s, $matches)) {
		assertType('array{0: string, num: numeric-string, 1: numeric-string}', $matches);
	}
};

because the hoa ast ist different for it in comparison to the other bracket style delims

<?php

require __DIR__.'/vendor/autoload.php';

// 1. Read the grammar.
$grammar  = new Hoa\File\Read(__DIR__.'/resources/RegexGrammar.pp');

// 2. Load the compiler.
$compiler = Hoa\Compiler\Llk\Llk::load($grammar);

// 3. Lex, parse and produce the AST.
$ast      = $compiler->parse('[(\d+)(?P<num>\d+)]n');

// 4. Dump the result.
$dump     = new Hoa\Compiler\Visitor\Dump();
echo $dump->visit($ast);

[phpstan-bot]

This pull request has been marked as ready for review.

[Seldaek]

Yeah that's because the parser has no concept of delimiters.. And supporting those delimiters would probably be a pain tbh. So I guess let's leave it as is for now.

[Seldaek]

On second thought I guess actually what you should do is probably parse delimiters first, remove delimiters/modifiers, then pass the actual pattern to the parser. That'd probably be the more accurate way to use it. Because otherwise when using () as delimiters, aren't you ending up with one match group too many?

[staabm]

On second thought I guess actually what you should do is probably parse delimiters first, remove delimiters/modifiers, then pass the actual pattern to the parser.

nice catch. looking at hoa-regex README the example actually also doesn't use modifiers/delimiters

[ondrejmirtes]

Thank you!

BTW: Could you please look into \Q...\E? https://stackoverflow.com/questions/20518758/how-the-preg-match-handles-the-delimiter-when-q-e-used

Can't find the issue right now but someone tried to use it in ignoreErrors too, and the validator rejected it.