C++ - 方法和重载中的 const 关键字

Question

为什么这个程序:

#include <iostream>
using namespace std;
class Base {
public:
  Base() { cout << "Costruttore Base" << endl; }
  virtual void foo(int) { cout << "Base::foo(int)" << endl; }
  virtual void bar(int) { cout << "Base::bar(int)" << endl; }
  virtual void bar(double) { cout << "Base::bar(double)" << endl; }
  virtual ~Base() { cout << "Distruttore Base" << endl; }
};
class Derived : public Base {
public:
  Derived() { cout << "Costruttore Derived" << endl; }
  void foo(int) { cout << "Derived::foo(int)" << endl; }
  void bar(int) const { cout << "Derived::bar(int)" << endl; }
  void bar(double) const { cout << "Derived::bar(double) const" << endl; }
  ~Derived() { cout << "Distruttore Derived" << endl; }
};
int main() {
  Derived derived;
  Base base;
  Base& base_ref = base;
  Base* base_ptr = &derived;
  Derived* derived_ptr = &derived;
  cout << "=== 1 ===" << endl;
  base_ptr->foo(12.0);
  base_ref.foo(7);
  base_ptr->bar(1.0);
  derived_ptr->bar(1.0);
  derived.bar(2);
  return 0;
}

在调用 base_ptr->bar(1.0); 中调用 Base::bar(double)，而不是在 derived_ptr->bar(1.0 ); 被称为 Derived::bar(double) const。

我知道这是关于 const 关键字的，但我不明白为什么编译器会选择不同的重载函数。

如果我删除 const 关键字，一切都会按预期工作，在这两种情况下都会调用 Derived::bar

Answer 1

那是因为const改变了函数的签名，所以它们是不同的。要么使基类和派生类都成为const，要么一个都不会覆盖另一个。