Why does outputting a class with a conversion operator not work for std::string?

Question

This works, printing 1:

#include 

struct Int {
    int i;
    operator int() const noexcept {return i;}
};

int main() {
    Int i;
    i.i          


        

Answer 1

That operator is a free template function. User defined conversions do not get checked when matching against a template function arguments, it instead uses type pattern matching (substitution).

In theory a SFINAE overload using std::is_convertable<> would be able to do what you want, but that technique was not used when operator<< that outputs a std::string to a basic_ostream was defined.

A manual overload to output your class to basic_ostream<...> will fix your problem.

I would do this:

struct String {
  std::string s;
  operator std::string() const {return s;}
  friend std::ostream& operator<<( std::ostream& os, String const& self) {
    return os<

which has the added benefit of not creating a wasted copy.