Re: Haskell 1.3 (newtype)

[Sebastian Hunt <seb@cs.city.ac.uk>]

Resent-Message-Id: <199509131449.QAA17987@animal.cs.chalmers.se>
From: Sebastian Hunt <seb@cs.city.ac.uk>
Subject: Re: Haskell 1.3 (newtype)
Date: Wed, 13 Sep 1995 15:47:12 +0100 (BST)
To: wadler@dcs.gla.ac.uk
Old-Resent-From: haskell-request@dcs.gla.ac.uk
Errors-To: haskell-request@dcs.gla.ac.uk
Approved: haskell@dcs.gla.ac.uk
Resent-Date:  Wed, 13 Sep 1995 15:47:47 +0100
Resent-From: kh@dcs.gla.ac.uk
Resent-To: haskell-list-dist@dcs.gla.ac.uk


On Wed, 13 Sep 1995 wadler@dcs.gla.ac.uk wrote:

> Well, I'm glad to see I provoked some discussion!
...
> Why should foo evaluate its argument?  It sounds to me like
> Lennart is right, and I should not have let Simon lead me astray!
... 
> This is assuming I have understood Lennart correctly, and that
> 
> 	foo (Age n) = (n, Age (n+1))
>  	foo' a = (n, Age (n+1)) where (Age n) = a
> 
> are equivalent when Age is declared as a strict datatype. Unlike
> Sebastian or Simon, I believe it would be a disaster if for a newtype
> one had to distinguish these two definitions.

I don't see how these two can be equivalent, unless a special case is 
made in the semantics for data types with a single constructor when the
constructor happens to be strict. Consider

	data G = F !Int | D Int

	f :: G -> Bool
	f (D _) = True
	f (F _) = False

If Lennart is right about foo, doesn't it follow that

	f (D undefined) = True?

In which case, since D is strict, we have

	f undefined = f (D undefined) = True

and so, by monotonicity of f, f v = True for all v and, in particular,

	f (F e) = True

This can't be right, surely?

Sebastian