#include <iostream>
using namespace std;
class Base {
public:
virtual void doSomething(double a) {
cout << "This is double from Base: " << a << endl;
}
virtual void doSomething(int a) {
cout << "This is int from Base: " << a << endl;
}
virtual ~Base() {}
};
class Derived : public Base {
public:
virtual void doSomething(double a) {
cout << "This is double from Derived: " << a << endl;
}
virtual ~Derived() {}
};
int main() {
Derived d;
int i = 2;
double j = 0.1;
d.doSomething(i);
d.doSomething(j);
return 0;
}
We would expect the output to be:This is int from Base: 2 This is double from Derived: 0.1
Strangely, the output is:
This is double from Derived: 2 This is double from Derived: 0.1
The way overload resolution works in C++ is that the compiler will first search for function doSomething() inside the Derived class. If it finds it, it will stop there; otherwise it will look the one in the Base class for a matching function. To solve this problem, we need to add using Base::doSomething in order let it participate in the overload resolution.
#include <iostream>
using namespace std;
class Base {
public:
virtual void doSomething(double a) {
cout << "This is double from Base: " << a << endl;
}
virtual void doSomething(int a) {
cout << "This is int from Base: " << a << endl;
}
virtual ~Base() {}
};
class Derived : public Base {
public:
virtual void doSomething(double a) {
cout << "This is double from Derived: " << a << endl;
}
using Base::doSomething;
virtual ~Derived() {}
};
int main() {
Derived d;
int i = 2;
double j = 0.1;
d.doSomething(i);
d.doSomething(j);
return 0;
}
The output is:
This is int from Base: 2 This is double from Derived: 0.1
