c – 这与函数重载有什么关系？

#include <iostream>
#include <complex>
using namespace std;
class Base
{
public:
    virtual void f(int);
    virtual void f(double);
    virtual ~Base() {};
};

void Base::f(int) { cout << "Base::f(int)" << endl; }
void Base::f( double ) { cout << "Base::f(double)" << endl; }

class Derived: public Base {
public:
    void f(complex<double>);
};

void Derived::f(complex<double>) { cout << "Derived::f(complex)" << endl; }

int main()
{
    Base* pb = new Derived;
    pb->f(1.0);
    delete pb;
}

代码打印Base :: f(double),我没有问题.但我无法理解作者在第122页的顶部给出的解释(重点是我的)：

Interestingly,even though the Base* pb is pointing to a Derived
object,this calls Base::f( double ),because overload resolution is
done on the static type (here Base),not the dynamic type (here
Derived).

我的理解是,调用pb-> f(1.0)是一个虚拟调用,Base :: f(double)是Derived中f(double)的最终覆盖.这与函数重载有什么关系？