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Re: c++/7181: foo<n>::bar = foo<n-1>::bar + foo<n-2>::bar evaluatesto zero at compile time

From: Paolo Carlini <pcarlini at unitus dot it>
To: Mark Mitchell <mark at codesourcery dot com>
Cc: "gcc-gnats at gcc dot gnu dot org" <gcc-gnats at gcc dot gnu dot org>, "gcc-prs at gcc dot gnu dot org"<gcc-prs at gcc dot gnu dot org>, "dobrynin at bigfoot dot com" <dobrynin at bigfoot dot com>, "gcc-bugs at gcc dot gnu dot org"<gcc-bugs at gcc dot gnu dot org>, Nathan Sidwell <nathan at codesourcery dot com>
Date: Tue, 02 Jul 2002 22:44:41 +0200
Subject: Re: c++/7181: foo<n>::bar = foo<n-1>::bar + foo<n-2>::bar evaluatesto zero at compile time
References: <64180000.1025641986@warlock.codesourcery.com>

Mark Mitchell wrote:

--On Tuesday, July 02, 2002 10:28:13 PM +0200 Paolo Carlini <pcarlini@unitus.it> wrote:

Hi,

from a very practical point of view, would be difficult to restore the
behaviour of 2.95.x? Note that Intel and Comeau adopts that "particular"
initialization order and the current "equivalent" one ;-) breaks a whole
body of literature on template metaprogramming...


Perhaps.  It may also be that picking one order makes this example work,
but some similar example fail.

Thanks for your feedback. This one of yours is an important point, indeed. Anyone attempting to restore the old behaviour should first prove that it's really the "best" one in some non trivial sense. I had always believed it is, but...

Thanks again,
Paolo.

References:
- Re: c++/7181: foo<n>::bar = foo<n-1>::bar + foo<n-2>::bar evaluatesto zero at compile time
  - From: Mark Mitchell

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