c++ overload "<<" doesn't work for output "obja + objb" - c++

Question

I overload "<<" and "+" operators for my class. It works fine if I output single object using cout<<a;. But it fails when I output cout<<a+b;
g++ version: g++ (GCC) 9.2.0
class UTF8string{
private:
string data;
public:
UTF8string();
UTF8string(string a);
int bytes();
int find(string a);
int length();
void replace(string a, string b);
UTF8string operator+(UTF8string &a);
UTF8string operator*(int a);
UTF8string operator+=(const UTF8string& rhs){
data.append(rhs.data);
return *this;
}
friend UTF8string operator!(UTF8string &a);
friend std::ostream& operator<<(std::ostream& os, UTF8string &a);
friend UTF8string operator+(int a, UTF8string &other );
};
std::ostream& operator<<(std::ostream& os, UTF8string &a){
os << a.data;
return os;
}
UTF8string UTF8string::operator+(UTF8string &a){
string b = data ;
return UTF8string(b.append(a.data));
}
my test code:
cout << "test + test2: " << test + test2 << endl; // doesn't work
cout << "test contains: " << test << endl; // works well
Error
testUTF8string.cpp:38:38: 错误：cannot bind non-const lvalue reference of type ‘UTF8string&’ to an rvalue of type ‘UTF8string’
38 | cout << "test + test2: " << test + test2 << endl;
| ~~~~~^~~~~~~

Answer 1

You're using references-to-non-const throughout.
That is a bad idea.
If you don't need to modify the object through a reference, the reference should be to const.
If your reference is not to const, temporaries cannot bind to it.
The return-by-value result of your operator+ is a temporary.

Answer 2

Result of your overloaded operator+ in test + test2 is temporary value. Temporary value cannot be passed to function by non-const reference.
It can be fixed by using const reference:
std::ostream& operator<<(std::ostream& os, const UTF8string &a){
os << a.data;
return os;
}
UTF8string UTF8string::operator+(const UTF8string &a){
string b = data ;
return UTF8string(b.append(a.data));
}
It is good practice to always use const references when passing arguments which are meant to be read-only.

Answer 3

Apart from the solution what #Lightness Races with Monica have suggested you can also overload the operator << for rvalue references like:-
friend std::ostream& operator<<(std::ostream& os, const UTF8string &&a);
It is sometimes useful to overload like this when moving is cheaper then copying.
Or for some reason you want a different implementation in two cases.
NOTE: In case of operator << it doesn't make sense to do the oveloading as the data/object being passed to stream should be const