Why isn't automatic downcasting applied to template functions?

Question

Someone asked this question about string appending. It's string s; s = s + 2; not compiling. People gave answers stating that operator+ is defined as a template function while operator+= is not, so auto downcasting (int(2) to char(2)) is not applied.

The prototypes are

template<typename _CharT, typename _Traits, typename _Alloc>
class basic_string{
    basic_string&
      operator+=(_CharT __c);
};

template<typename _CharT, typename _Traits, typename _Alloc>
  inline basic_string<_CharT, _Traits, _Alloc>
  operator+(const basic_string<_CharT, _Traits, _Alloc>& __lhs, _CharT __rhs);

Why can't the compiler just use this prototype and cast int(2) to char(2)?

basic_string<char, _T, _A> operator+(const basic_string<char, _T, _A>, char);

The compiler (G++ 6.3.0) complains that

[Note] deduced conflicting types for parameter '_CharT' ('char' and 'int')

Answer 1

The key difference is that for the operator += variant, the char type template argument for the std::basic_string, and thus the argument type for its RHS, is already fixed to char, while the operator+ template has to deduce that from its arguments.

Thus, for the += case, the compiler knows you "want" the int->char conversion, there is nothing to deduce there.

For the operator+ case on the other hand, the compiler is looking at the template

template<class CharT, class Traits, class Alloc>
    basic_string<CharT,Traits,Alloc>
        operator+( const basic_string<CharT,Traits,Alloc>& lhs,
                   CharT rhs );

and, when trying to determine what CharT is supposed to be, it gets CharT = char from the first operand (as std::string is std::basic_string<char>) and CharT = int from the second operand. Such a conflict is defined to be a compilation error by the standard.