14.5/3 of N2800 notes that template aliases cannot be specialized, and
N1489 shows how to achieve the same effect using traits classes.  If I
want to have a template alias for a vector with an allocator of my
choice, where I have different allocators for pointer and non-pointer
types, I can do this:

template<typename T>            // primary
struct VecAllocator {           // template
 typedef MyAllocator type;
};

template<typename T>              // specialized
struct VecAllocator<T*> {         // template
 typedef MyPtrAllocator> type;
};

template<typename T>
using MyAllocVec = std::vector<T, VecAllocator<T>::type>;

My knowledge of concepts is sketchy, but I'm wondering if I could
achieve the same effect by overloading my alias on concepts and thus
get rid of the traits class.  Could I?

Thanks,

Scott



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On 22 Mai, 10:07, Scott Meyers <use...@aristeia.com> wrote:
> 14.5/3 of N2800 notes that template aliases cannot be specialized, and
> N1489 shows how to achieve the same effect using traits classes.  If I
> want to have a template alias for a vector with an allocator of my
> choice, where I have different allocators for pointer and non-pointer
> types, I can do this:
>
> template<typename T>            // primary
> struct VecAllocator {           // template
>  typedef MyAllocator type;
> };
>
> template<typename T>              // specialized
> struct VecAllocator<T*> {         // template
>  typedef MyPtrAllocator> type;
> };
>
> template<typename T>
> using MyAllocVec = std::vector<T, VecAllocator<T>::type>;

You need the typename here

template<typename T>
using MyAllocVec = std::vector<T, typename VecAllocator<T>::type>;

but otherwise this alias definition is correct.

> My knowledge of concepts is sketchy, but I'm wondering if I could
> achieve the same effect by overloading my alias on concepts and thus
> get rid of the traits class.  Could I?

There is currently an active issue about this, see

http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#851

Even though the grammar would currently allow such a
constrained form of an alias, the tendency is to consider
this as an artifact of the current spec.

Or did you mean something differently here?

Greetings from Bremen,

Daniel


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daniel.kruegler@googlemail.com wrote:
>>
>> template<typename T>
>> using MyAllocVec = std::vector<T, VecAllocator<T>::type>;
>
> You need the typename here

Duh, right, thanks.

>> My knowledge of concepts is sketchy, but I'm wondering if I could
>> achieve the same effect by overloading my alias on concepts and thus
>> get rid of the traits class.  Could I?
>
> There is currently an active issue about this, see
>
> http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#851
>
> Even though the grammar would currently allow such a
> constrained form of an alias, the tendency is to consider
> this as an artifact of the current spec.
>
> Or did you mean something differently here?

I think that's what I meant.  In general, I view traits as a
roundabout way of expressing things, so if there is a more
straightforward mechanism (e.g., concepts), I'd prefer that.
Something like:

 template<PointerType T>
 using MyAllocVec = std::vector<T, MyPtrAllocator<T>>;

 template<typename T>
 using MyAllocVec = std::vector<T, MyAllocator<T>>;

Scott

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