c++ - initializer_list and template type deduction

Question

Consider the function:

template<typename T>
void printme(T&& t) {
  for (auto i : t)
    std::cout << i;
}

or any other function that expects one parameter with a begin()/end() - enabled type.

Why is the following illegal?

printme({'a', 'b', 'c'});

When all these are legitimate:

printme(std::vector<char>({'a', 'b', 'c'}));
printme(std::string("abc"));
printme(std::array<char, 3> {'a', 'b', 'c'});

We can even write this:

const auto il = {'a', 'b', 'c'};
printme(il);

or

printme<std::initializer_list<char>>({'a', 'b', 'c'});

question from:https://stackoverflow.com/questions/12431495/initializer-list-and-template-type-deduction

Answer 1

Your first line printme({'a', 'b', 'c'}) is illegal because the template argument T could not be inferred. If you explicitly specify the template argument it will work, e.g. printme<vector<char>>({'a', 'b', 'c'}) or printme<initializer_list<char>>({'a', 'b', 'c'}).

The other ones you listed are legal because the argument has a well-defined type, so the template argument T can be deduced just fine.

Your snippet with auto also works because il is considered to be of type std::initializer_list<char>, and therefore the template argument to printme() can be deduced.

---

The only "funny" part here is that auto will pick the type std::initializer_list<char> but the template argument will not. This is because § 14.8.2.5/5 of the C++11 standard explicitly states that this is a non-deduced context for a template argument:

A function parameter for which the associated argument is an initializer list (8.5.4) but the parameter does not have std::initializer_list or reference to possibly cv-qualified std::initializer_list type. [Example:

template<class T> void g(T);
g({1,2,3}); // error: no argument deduced for T

— end example ]

However with auto, § 7.1.6.4/6 has explicit support for std::initializer_list<>

if the initializer is a braced-init-list (8.5.4), with std::initializer_list<U>.