SFINAE overload operator<< to call 'print' method if it exists - c++

Question

I've been trying to understand SFINAE and was trying to write a simple overloaded operator << that would call the 'print' method on any class that contains such a method. I read through the answers on the question Is it possible to write a template to check for a function's existence? and tried writing:
template<class T, class = decltype(void(std::declval<T>().print), std::true_type{})>
inline std::ostream &operator<<(std::ostream &out, const T &obj) {
obj.print(out); return out; }
and
template<class T, class = decltype(void(std::declval<T>().print(std::declval<std::ostream &>())), std::true_type{})>
inline std::ostream &operator<<(std::ostream &out, const T &obj) {
obj.print(out); return out; }
but this simply doesn't work -- the compiler seems to have no problem instantiating the template for any type, so gives slews of 'ambiguous overload' errors when I try to print things like string literals...

Answer 1

Your "function exists" expression is incorrect. Try this:
template <typename T,
typename = decltype(
void(std::declval<T>().print(std::declval<std::ostream&>())),
std::true_type{})> // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
std::ostream & operator<<(std::ostream & out, const T & obj)
{
obj.print(out);
return out;
}
You might also consider this alternative:
template <typename T>
auto operator<<(std::ostream & out, const T & obj)
-> decltype(obj.print(out), (void)0, out)
{
obj.print(out);
return out;
}

Answer 2

I don't understand what you've meant your second type parameter
class = decltype(void(std::declval<T>().print), std::true_type{})
to mean. What is this supposed to evaluate to?
I think you can make it work by using the following.
#include <iostream>
#include <type_traits>
template<typename T,
typename = decltype(std::declval<const T>().print(std::cout))>
std::ostream&
operator<<(std::ostream& out, const T& obj)
{
obj.print(out);
return out;
}
struct A
{
void
print(std::ostream& out) const
{
out << "A";
}
};
int
main()
{
A a {};
std::cout << "And the winner is: " << a << std::endl;
}
It will correctly output And the winner is: A but there are probably some corner cases I've overlooked.
The expression
decltype(std::declval<const T>().print(std::cout))
will evaluate to the return type of the print(std::ostream&) const member function if such function is declared and a type error otherwise.