Make garbage parameters in SyntaxFactory unlabeled

If the `garbage` parameters in the SyntaxFactory are labelled, it’s not possible to specify e.g. the second garbage node but leave the others to their default (swiftlang/swift#60274).

We could either give these `garbage` parameters distinct names or make them unlabeled to work around that issue. I chose to go for the unlabeled approach because it should be clear that these parameters contain garbage because of their type.

Author: ahoppen

Sources/SwiftSyntax/SyntaxFactory.swift.gyb

@@ -57,7 +57,9 @@ public enum SyntaxFactory {
 %          param_type = param_type + "?"
 %       end
 %       if child.is_garbage_nodes():
-%         child_params.append("garbage %s: %s = nil" % (child.swift_name, param_type))
+%         # It would be nice if we could label all of these arguments 'garbage'.
+%         # But if we do this, we hit https://github.com/apple/swift/issues/60274.
+%         child_params.append("_ %s: %s = nil" % (child.swift_name, param_type))
 %       else:
 %         child_params.append("%s: %s" % (child.swift_name, param_type))
 %       end

Sources/SwiftSyntax/gyb_generated/SyntaxFactory.swift

@@ -117,7 +117,15 @@ public struct ${type.buildable()}${conformance_clause(conformances)} {
   /// - Returns: The built `${type.syntax()}`.
   func build${type.base_name()}(format: Format, leadingTrivia additionalLeadingTrivia: Trivia? = nil) -> ${type.syntax()} {
     let result = SyntaxFactory.make${type.base_name()}(
-      ${',\n      '.join(['%s: %s' % (child.factory_parameter_name(), child.generate_expr_build_syntax_node(child.name(), 'format')) for child in children])}
+    % parameters = []
+    % for (index, child) in enumerate(children):
+    %   comma = ',' if index != len(children) - 1 else ''
+    %   if child.child.is_garbage_nodes():
+      ${child.generate_expr_build_syntax_node(child.name(), 'format')}${comma}
+    %   else:
+      ${child.name()}: ${child.generate_expr_build_syntax_node(child.name(), 'format')}${comma}
+    %   end
+    % end
     )
     let combinedLeadingTrivia = leadingTrivia + (additionalLeadingTrivia ?? []) + (result.leadingTrivia ?? [])
     return result.withLeadingTrivia(combinedLeadingTrivia)