Re: Why isn't class a subset of class... ?
Am 05.12.2012 19:08, schrieb Tobias M??ller:
DeMarcus <use_my_alias_here_at_hotmail_com@tellus.orb.dotsrc.org> wrote:'
Yes I know, but it takes a template so it should be able to take
std::vector, right?
template<template<class> class T>
class TakeTemplate
{
};
template < class T, class Alloc = allocator<T> > class vector;
std::vector template takes two parameters.
This is correct.
Implementors are even permitted to add more, as long as they have default arguments.
That means that there is probably no portable way for matching std::
templates.
This is incorrect. Rumors abounded to that extend, but
http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-closed.html#94
clearly points out that this assumption is wrong. In fact, an
implementation is generally required to allow for user-provided
specializations of library template as described in [namespace.std] p1
(I'm leaving out the constraints in the following):
"[..] A program may add a template specialization for any standard
library template to namespace std only if the declaration depends on a
user-defined type and the specialization meets the standard library
requirements for the original template and is not explicitly
prohibited.(footnote 180: Any library code that instantiates other
library templates must be prepared to work adequately with any
user-supplied specialization that meets the minimum requirements of the
Standard.)"
This gives even more evidence for the general #94 resolution that an
implementation cannot provide further template parameters for standard
library templates, because that would make it impossible to specialize
such a template.
HTH & Greetings from Bremen,
Daniel Kr??gler
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