Add a method to create a new type symbol

[nicolasstucki]

How can I get a type symbol for a type alias to use TypeDef.apply?

You can get it from an existing type alias. We currently do not have a method to create new type symbol. We might want to add this as Symbol.newType or Symbol.newTypeAlias.

Originally posted by @nicolasstucki in #19344 (comment)

[jchyb]

I was asked by @Gedochao to elaborate a bit on the requirements here.

The idea is to add Symbol.newType in scala.quoted.Quotes.reflectModule.SymbolModule implementation, which would generate a type alias/type member. This method would likely have a signature like:

def newType(parent: Symbol, name: String, tpe: TypeRepr)

It would have to work with the preexisting TypeDef.apply(symbol: Symbol), which means:

1. the aliased type should be included in the Symbol definition
2. the resulting type alias/member either cannot be made abstract, or we should include that information in Symbol.newType(parent: Symbol, name: String, tpe: Option[TypeRepr]).

I would personally prefer if it could be made abstract for completeness' sake, but the original issue raised in the discussion linked above does not require that.

[pweisenburger]

the resulting type alias/member either cannot be made abstract, or we should include that information in Symbol.newType(parent: Symbol, name: String, tpe: Option[TypeRepr]).

Generally, type members define a lower and an upper bound. Maybe newType should take a TypeBounds instead of a TypeRepr (abstract types without explicit bounds could then be created using TypeBounds.empty):

def newType(parent: Symbol, name: String, bounds: TypeBounds)

I think, this would also be more consistent with the rest of the API.

Please also allow for passing a flags and a privateWithin argument, like for newMethod and newVal (unfortunately, newClass is missing this feature but it should also have it).