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Force prodecural cosmetic filtering when explicitly stated
Related issue: - https://github.com/uBlockOrigin/uBlock-issues/issues/2228 Using `#?#` (instead of `##` for a procedural cosmetic filter will prevent uBO from trying to convert the filter into a declarative one.
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@ -1453,7 +1453,7 @@ Parser.prototype.SelectorCompiler = class {
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}
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// Procedural selector?
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const compiled = this.compileProceduralSelector(raw);
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const compiled = this.compileProceduralSelector(raw, asProcedural);
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if ( compiled === undefined ) { return false; }
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out.compiled =
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@ -1531,8 +1531,8 @@ Parser.prototype.SelectorCompiler = class {
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return true;
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}
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compileProceduralSelector(raw) {
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const compiled = this.compileProcedural(raw, true);
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compileProceduralSelector(raw, asProcedural = false) {
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const compiled = this.compileProcedural(raw, true, asProcedural);
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if ( compiled !== undefined ) {
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compiled.raw = this.decompileProcedural(compiled);
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}
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@ -1757,7 +1757,7 @@ Parser.prototype.SelectorCompiler = class {
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return raw.join('');
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}
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compileProcedural(raw, root = false) {
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compileProcedural(raw, root = false, asProcedural = false) {
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if ( raw === '' ) { return; }
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const tasks = [];
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@ -1813,6 +1813,7 @@ Parser.prototype.SelectorCompiler = class {
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// "forgiving", we also need to validate that the argument itself
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// is also a valid CSS selector.
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if (
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asProcedural === false &&
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this.querySelectable(raw.slice(opNameBeg, i)) &&
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this.querySelectable(oparg)
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) {
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