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Shallow copy of a record
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Date: -- (:)
From: Michal Moskal <michal.moskal@g...>
Subject: Re: [Caml-list] Shallow copy of a record
On Sat, 22 Jan 2005 11:55:19 -0500, Chris King <colanderman@gmail.com> wrote:
> On Sat, 22 Jan 2005 15:49:09 +0900 (JST), Jacques Garrigue
> <garrigue@math.nagoya-u.ac.jp> wrote:
> 
> > This does not exist, and with good reason: there is no way in the type
> > system to define a function which works on all records, but only
> > records. So to do the copy you need to know at least one of the labels
> > of the record, which in turn gives you its type.
> 
> Isn't it the foo in {foo with a=b} that determines the type, though?

No, consider:

  let copy foo = {foo with a = b}

and now, if there were no assign:

  let copy foo = {foo}

what should be the type of copy?

-- 
: Michal Moskal :: http://nemerle.org/~malekith/ :: GCS !tv h e>+++ b++
: No, I will *not* fix your computer............ :: UL++++$ C++ E--- a?