Function call with template parameter

Question

I have written a simple class template:

Click to copy

template <class T>
class my_class
{
public:
    my_class(T value)
        : my_Value(value)
    {
    }
private:
    T my_Value;
};

Now I can use this template in a simple function signature like: my_function(my_class<std::string> my_string)

When I want to call the function I can easily use it:

Click to copy

auto my_instance = my_class<std::string>("my value");
my_function(my_instance);

But what I want to realize is a function call like this:

Click to copy

my_function("my value")

My class template should implicit do the conversion to the type of the template for me. I think I need some kind of operator overload.

std:optional can do this for example.

Answer 1

You can have only one implicit user-conversion, so your call with const char* is invalid.

There are several options,

• add another constructor for my_class

  Click to copy

  my_class(T value) : my_Value(value) {}
  
  template <typename U, std::enable_if_t<std::is_convertible<U, T>, int> = 0>
  my_class(U value) : my_Value(value) {}

• add overload for my_function,

  Click to copy

  void my_function(my_class<std::string> my_string)
  void my_function(const char* s) { return my_function(my_class<std::string>{s}); }

• change call site to call it with std::string:

  Click to copy

  my_function(std::string("my value"))
  
  using namespace std::string_literals;
  my_function("my value"s)