|
Boost : |
Subject: Re: [boost] Help! Detecting that two template aliases are equal.
From: Nathan Ridge (zeratul976_at_[hidden])
Date: 2012-11-03 02:58:26
> > I haven't checked this, but it's the behavior that I would expect.
>
> That seems surprising to me. It's a simple alias here, so I would expect it
> to be exactly that -- an alias. If I make a typedef of a type, I expect it
> to alias that type such that is_same tells me that are the same (which it,
> of course, does). An alias should, whenever possible, be exactly that, an
> alias.
Suppose you have an alias of the form
template <typename T>
using alias = typename some_metafunction<T>::type;
It could very well be that some_metafunction<T>::type is foo<T> for
all types T (where foo is some class template), but it's undecidable for
the compiler to determine that in the general case.
So, in the general case, it is not reasonable to expect a compiler to
deem 'alias' and 'foo', as template *template* parameters, to be equal
(of course for any T, the template *type* parameters alias<T> and
foo<T> *will* be equal).
Given that this is the case, it would be strange if, as a special case,
the compiler did deem 'alias' and 'foo', as template template parameters,
to be equal if 'alias' happened to be defined as follows:
template <typename T>
using alias = foo<T>;
I think the behaviour you're experiencing, while undesirable for your
particular use case, is consistent.
Regards,
Nate
Boost list run by bdawes at acm.org, gregod at cs.rpi.edu, cpdaniel at pacbell.net, john at johnmaddock.co.uk