How to make an ambiguous call distinct in C++?
- by jcyang
void outputString(const string &ss) {
cout << "outputString(const string& ) " + ss << endl;
}
void outputString(const string ss) {
cout << "outputString(const string ) " + ss << endl;
}
int main(void) {
//! outputString("ambigiousmethod");
const string constStr = "ambigiousmethod2";
//! outputString(constStr);
} ///:~
How to make distinct call?
EDIT: This piece of code could be compiled with g++ and msvc.
thanks.
