Unusable constructorof for a type with a type-parameter that is unused #58

Consider a type like:

struct Foo{D}
       x::Int
end

Which could be used as Foo{:KeyFacts}(25).
constructorof returns just Foo

Then the following issue occurs:

julia> constructorof(Foo{:ABC})
Foo

julia> constructorof(Foo{:ABC})(22)
ERROR: MethodError: no method matching Foo(::Int64)

Which is a problem.
That informatiuon is lost and i can't construct it anymore.
If the type param had been used it could be deduced from the type by the default constructor, but where it isn't it can not

I assume the reason for this is that uses like Setfield.jl might change the type of one of the fields and so for type-parameters that are used it wants them to removed to hit the default constructor.

I am not sure on the solution to this.

My own use case is same type in, same type out.
So one option is to expose a duplicate of setproperties and of constructorof that always has the same type.
It is a bit ugly though.

For the special case where all the type parameters are unused, we could detect that by them never occuring in fieldtypes.
And then we know that it is safe to generate the constructor that has the type parameter set.
But that doens't work for cases where some are used and some are not.
Though for many of these case that i have seen in the wild they have just one unused type parameter and the rest used, and expose a constructor FooBar{C}(a::A, b::B) where {A,B,C}