The strange behavior of the `operator string()` type conversion function

Question

I have a C class and I want to convert it to string by using the operator string() syntax, but as I wrote in the comment, something strange happens.

class C {
public:
    operator string()
    {
        return string("anything");
    }
};
int main()
{
    C c;
    cout << string("test:").append(c) << endl; //success
    //cout << string("test:")+c<< endl;//fail
    //cout << c.append("test:") << endl;//fail
    cout << string("test:") + static_cast<string>(c) << endl;//success
}

Answer 1

 //cout << string("test:")+c<< endl;//fail

std::operator+(std::basic_string) are templates. Implicit conversion won't be considered in template argument deduction , which causes the deduction failing on the 2nd operand c .

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

And

//cout << c.append("test:") << endl;//fail

C doesn't have member function named append , then the code fails. You can convert c to std::string explicitly then call append on it like static_cast<string>(c).append("test:") .

 cout << string("test:").append(c) << endl; //success

string("test:").append expects an std::string , then c is converted to std::string implicitly and passed to append . (Note that append are non-template member functions.)

 cout << string("test:") + static_cast<string>(c) << endl;//success

c is converted to std::string explicitly, then two std::string s are passed to std::operator+(std::basic_string) .