It looks like you're trying to create an alias for a type. I'd use a
constant for this, not a subset, for reasons Brad has already explained.
Your code runs fine for me when DEF is written like my constant DEF = ABC.
> On 10 Jul 2020, at 23:37, Brad Gilbert wrote:
> I honestly think that there is an argument to be made that it shouldn't even
> be possible to write a `subset` without a `where` clause.
Making the "where" clause non-optional, is remarkably simple: removing a '?'
However, there appear to be qui
On 2020-07-10 23:37, Brad Gilbert wrote:
Subset types are not object types.
A subset is basically a bit of checking code and base type associated
with a new type name.
In something like:
my ABC $a .= new;
That is exactly the same as:
my ABC $a = ABC.new;
Well there is no functiona
Subset types are not object types.
A subset is basically a bit of checking code and base type associated with
a new type name.
In something like:
my ABC $a .= new;
That is exactly the same as:
my ABC $a = ABC.new;
Well there is no functional `.new` method on any subset types, so
`DEF.
