Create an instance of a type parameter in macro
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Paul Butcher
Apr 22, 2012, 6:20:17 PM4/22/12
to scala-user
I want to create an instance of something passed as a type parameter to a macro. For instance, if I wanted to create a macro called "myNew" which does exactly what new does, I want:
def myNew[T] = macro myNewImpl[T]def myNewImpl[T: c.TypeTag](c: Context) = reify(<[ new T ]>)
But all my attempts to work out what needs to go in the reify have failed :-(
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Miles Sabin
Apr 22, 2012, 6:40:44 PM4/22/12
to Paul Butcher, scala-user
On Sun, Apr 22, 2012 at 11:20 PM, Paul Butcher <pa...@paulbutcher.com> wrote:
> I want to create an instance of something passed as a type parameter to a
> macro. For instance, if I wanted to create a macro called "myNew" which does
> exactly what new does, I want:
>
> def myNew[T] = macro myNewImpl[T]
>
> def myNewImpl[T: c.TypeTag](c: Context) = reify(<[ new T ]>)
>
> But all my attempts to work out what needs to go in the reify have failed
> :-(
> I want to create an instance of something passed as a type parameter to a
> macro. For instance, if I wanted to create a macro called "myNew" which does
> exactly what new does, I want:
>
> def myNew[T] = macro myNewImpl[T]
>
> def myNewImpl[T: c.TypeTag](c: Context) = reify(<[ new T ]>)
>
> But all my attempts to work out what needs to go in the reify have failed
> :-(
I think in this case you'll need to construct the tree by hand ... the
following works for me,
def myNew[T] = macro myNewImpl[T]
def myNewImpl[T: c.TypeTag](c: Context) : c.Expr[Any] = {
import c.mirror._
Apply(
Select(
New(TypeTree().setType(c.tag[T].tpe)),
newTermName("<init>")),
List())
}
Cheers,
Miles
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Paul Butcher
Apr 23, 2012, 4:24:29 AM4/23/12
to Miles Sabin, scala-user
Cool - thanks Miles.
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