Deduce template argument for size of initializer list

Question

I have the following (not compilable) code:

template< size_t N >
void foo( std::array )
{
  // Code, where \"N\" is used.
}

int main()
{
  f         


        

Answer 1

Thanks to core issue 1591, you can use

template <std::size_t N>
void foo( int const (&arr)[N] )
{
  // Code, where "N" is used.
}

foo({1, 2, 3});

Answer 2

If using an initializer list is not a must, you can use variadic template parameter packs:

template<size_t S>
void foo_impl(array<int, S> const&)
{
    cout << __PRETTY_FUNCTION__ << endl;
}

template<typename... Vals>
auto foo(Vals&&... vals) {
    foo_impl<sizeof...(vals)>({ std::forward<Vals>(vals)... });
}

you'd call it as follows:

foo(1,2,3,4,5);

This defers common type checks until the initialization point of std::array (unless you add some admittedly ugly asserts), so you probably should prefer Columbo's answer.