is
open fun isElaboratedTypeSpecifier(specifier: String, language: Language<out LanguageFrontend>): Boolean
Returns whether the specifier is part of an elaborated type specifier. This only applies to C++ and can be used to declare that a type is a class / struct or union even though the type is not visible in the scope.
Return
true, if it is part of an elaborated type. false, otherwise
Parameters
specifier
the specifier