zero length variadic expansion of ill-formed call

Question

Question for standard gurus.

Trying to respond to another question, I came to doubt about the well-formedness of a code.

As far I know, the following code is ill-formed

int main ()

 {

   std::tuple<>  a;



   std::get<0>(a);

 }

because a call to

std::get<I>(t)

, when

t

is a

std::tuple<Ts...>

, is ill-formed when

I

is outside the range

[0, sizeof...(Ts)[

.

In this case

sizeof...(Ts)

is zero, so the range

[0, 0[

is empty, so

std::get<I>(a)

is ill-formed for every index

I

.

But when

std::get<I>(a)

is expanded through an empty variadic pack?

I mean: the following code

#include <tuple>



template <typename ... Args>

void bar (Args const & ...)

 { }



template <std::size_t ... I>

void foo ()

 {

   std::tuple<> a;



   bar( std::get<I>(a) ... );

 }



int main ()

 {

   foo<>();

 }

that uses a ill-formed (?) call (

std::get<I>(a)

) but zero-time variadic expanded (

sizeof...(I)

is zero), is well-formed or ill-formed?

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Answer 1

[temp.res]/8:

The program is ill-formed, no diagnostic required, if:

[...]

every valid specialization of a variadic template requires an empty template parameter pack, or

[...]

zero length variadic expansion of ill-formed call

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