Question

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.

Answer 1

C++ does not allow you to overload functions where the only difference in the function signature is that one takes an object and another takes reference to an object. So something like:

void foo(int);

and

void foo(int&);

is not allowed.

You need to change the number and/or the type of the parameter.

In your case the function that accepts a reference, you can make it accept a pointer, if you want to allow the function to change its argument.