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Shallow copy of a record
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Date: | 2005-01-22 (17:55) |
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?