CODEWITHSUNDEEP
c++c++11variadic-templates
Khurshid Normuradov
Khurshid Normuradov

Reputation: 1604

variadic template pack within decltype

I may write as

template< class T0> struct Last0
{  
  using type = decltype(T0{}); // OK compiles. `type = T0`
};


template< class T0, class T1> struct Last1
{
    using type = decltype(T0{}, T1{}); // OK, compiles. `type = T1`
};

template< class T0, class T1, class T2> struct Last3{
   using type = decltype(T0{}, T1{}, T2{}); // Ok, compiles. `type = T2`
};

But, when I use variadic templates, it's not compiled:

template< class ... T> struct Last{
   using type = decltype(T{} ... ); //<--- Error !!!
};

What's problem?

Upvotes: 6

Views: 5007

Answers (2)

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171127

There is a taxative list of language constructs where pack expansion can happen (C++11, 14.5.3§4). With the exception of sizeof..., it's always in constructs where the comma , is a grammatical separator of a list, and not an operator. An expression cannot be a pack expansion.

To get the last type in a pack, you can do this:

template <class Head, class... Tail>
struct Last {
  typedef typename Last<Tail...>::Type Type;
};

template <class Head>
struct Last<Head> {
  typedef Head Type;
};

Upvotes: 5

Kerrek SB
Kerrek SB

Reputation: 477080

You can only apply decltype to an expression, not to a pack. Packs are very special and basically always need to be expanded. You essentially have the same problem as not being able to store packs directly: using type = T... isn't allowed, either.

The standard solution is to store packs inside some "container template", typically tuple:

using T = std::tuple<decltype(T{})...>;

Upvotes: 3

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