scala - Type transformation with shapeless -


i have class similar this:

class myclass[t <: hlist] {   val x: ??? }

my problem type of x val. i'd have hlist each type u of t hlist replaced option[u]. i.e. if specify:

new myclass[int :: string :: hnil]

i x have type of option[int] :: option[string] :: hnil

is possible? how it?

you'll need mapped instance witnesses t , type of x have relationship:

import shapeless._, ops.hlist.mapped  abstract class myclass[t <: hlist, ot <: hlist](implicit   mapped: mapped.aux[t, option, ot] ) {   val x: ot }

unfortunately kind of inconvenient instantiate:

new myclass[int :: string :: hnil, option[int] :: option[string] :: hnil] {   val x = some(0) :: some("") :: hnil }

there ways around this, require additional changes. example, allow both type parameters inferred:

import shapeless._, ops.hlist.comapped  class myclass[t <: hlist, ot <: hlist](val x: ot)(implicit   mapped: comapped.aux[ot, option, t] )

and then:

new myclass(option(0) :: option("") :: hnil)

or can use closer original class custom constructor in companion object:

import shapeless._, ops.hlist.mapped  abstract class myclass[t <: hlist] {   type ot <: hlist   def mapped: mapped.aux[t, option, ot]   val x: ot }  object myclass {   class partiallyapplied[t <: hlist] {     def apply[ot0 <: hlist](x0: ot0)(implicit       mapped0: mapped.aux[t, option, ot0]     ): myclass[t] =       new myclass[t] {         type ot = ot0         val mapped: mapped.aux[t, option, ot] = mapped0         val x: ot = x0       }   }    def apply[t <: hlist]: partiallyapplied[t] = new partiallyapplied[t] }

and then:

myclass[int :: string :: hnil](option(0) :: option("") :: hnil)

which of these approaches more appropriate depend on how you're using class.


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