Re: Why is static member initialization inconsistent?

From:
Dave Abrahams <dave@boostpro.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 5 Jan 2012 09:03:02 -0800 (PST)
Message-ID:
<m2ehve3of0.fsf@boostpro.com>
on Wed Jan 04 2012, Gene Bushuyev <publicfilter-AT-gbresearch.com> wrote:

It's a very fundamental rule that members of a template are not
implicitly instantiated until they are used. That reduces the
compilation time, size of libraries, cuts on dependencies, and makes
sfinae possible.


I don't think this has anything to do with SFINAE, which applies to
ordinary function templates that aren't class members. You may be
thinking of the a similar advantage: it's easy to make class templates
with optional features that are supported depending on details of the
template parameters. For example, if T is LessThanComparable, you can
call std::list<T>::sort without arguments.

   l.sort()

This overload of std::list<T>::sort() contains code
that evaluates

   x < y

where x and y are of type T.

If the overload of list<T>::sort that takes no arguments was eagerly
instantiated, list<T> would not be usable at all with a T that isn't
LessThanComparable:

   struct X {};

   std::list<X> l; // <== compilation would fail in sort()

Since list<T>::sort is instantiated only when used, we can make a
list<X>, and we're fine as long as we don't try to sort it.

--
Dave Abrahams
BoostPro Computing
http://www.boostpro.com

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