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class - Why is the Manifest not available in the constructor? - Stack Overflow

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Original source (stackoverflow.com)

Tags: scala manifests stackoverflow.com

Clipped on: 2012-09-25

up vote 5 down vote favorite

Consider this code:

class Foo[T : Manifest](val id: String = manifest[T].erasure.getName)

I basically want to store an identifier in Foo, which is often just the class name.

Subclass which do not need a special identifier could then easily use the default value.

But this doesn't even compile, the error message is:

error: No Manifest available for T.

Is there another approach which will work?

EDIT:

Why does this work if the manifest isn't available until the primary constructor?

class Foo[T: Manifest](val name: String) { 
  def this() = this(manifest[T].erasure.getName)
}

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edited Sep 6 '11 at 19:35

asked Sep 3 '11 at 17:39

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Not an answer, but you'll have much cleaner code if you use manifest[T] instead of implicitly[Manifest[T]]. – Kevin Wright Sep 4 '11 at 7:55

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1 Answer

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up vote 7 down vote accepted

When the syntactic sugar is removed from that context bound, it gets rewritten as:

class Foo[T]
  (val id: String = implicitly[Manifest[T]].erasure.getName)
  (implicit ev$1: Manifest[T]) = ...

So the Manifest evidence simply isn't available when determining the default value of id. I'd instead write something like this:

class Foo[T : Manifest](id0: String = "") {
  val id = if (id0 != "") id0 else manifest[T].erasure.getName
}

In your second approach (which is a great solution, by the way!), expect a rewrite similar to:

class Foo[T](val name: String)(implicit x$1: Manifest[T]) { 
  def this()(implicit ev$2: Manifest[T]) = this(manifest[T].erasure.getName)
}

So yes, the manifest is available before the call to manifest[T].erasure

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edited Sep 4 '11 at 8:07

answered Sep 3 '11 at 18:02

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But doesn't create that a superfluous field? – soc Sep 3 '11 at 18:29

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To avoid the superfluous field, remove the val modifier from the definition of id0? – Kipton Barros Sep 3 '11 at 18:34

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Mhhh ... I think I just solve it by adding a – soc Sep 3 '11 at 18:35

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Kipton - True that, consider it gone – Kevin Wright Sep 4 '11 at 7:51

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@soc - It's usually best to avoid underscores, a heavily overloaded symbol, unless the spec demands it. – Kevin Wright Sep 4 '11 at 9:41

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