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

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

template
void foo_impl(array const&)
{
    cout << __PRETTY_FUNCTION__ << endl;
}

template
auto foo(Vals&&... vals) {
    foo_impl({ std::forward(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.