C++: inheriting overloaded non-virtual method and virtual method both with the same name causes problem

I am trying to inherit two equally named methods with different parameter lists to a derived class. One of them is virtual and overridden in the derived class, the other one is non-virtual. Doing so, i get a compile error while trying to access the non-virtual method of the base class from an derived class object.

Here is the code snippet

class Base {
public: 
    void f() {
        cout << "[Base::f()]" << endl;
    }

    virtual void f(int arg) {
        cout << "[Base::f(" << arg << ")]" << endl;
    }
};


class Deriv : public Base {
public:
    virtual void f(int arg) {
        cout << "[Deriv::f(" << arg << ")]" << endl;
    }
};


int main() {
    Deriv d;
    d.f(-1);    
    d.f();  // <<-- compile error   
    return 0;
}

which produces the following compile error:

error: no matching function for call to ‘Deriv::f()’
note: candidates are: virtual void Deriv::f(int)

I am not an expert in C++, but until now I thought to be right in making the assumption that member methods can be completely distinguished by their signatures. Thus, the non-virtual method Base::f() should not be overridden and should remain accessible. Am I wrong with this?

Here are some interesting/additional comments on that:

- the overriding method Deriv::f(int arg) could be non-virtual as well; the error occurs in either way
- the error disappears/can be circumvented...
... by casting the Deriv object to the Base class
... when not overriding Base::f(int arg) in Deriv
... by adding the command "Base::f;" to the public part of Deriv

So, since I already know how to avoid this compile error, I am mainly interested in why this error happens!

In Deriv, add this:

using Base::f;

In addition to the link given by @DumbCoder, you can find more details in my answer to a similar question: Overriding a Base's Overloaded Function in C++