为什么需要重新声明重载的虚函数？

我有一个带有两个重载函数的基类f(void) and f(int)。班上Derived实施f(int)通过致电f(void). Derived2实施f(void) only.

编译器拒绝执行Derived::f(int)因为它想打电话f(int)但我没有提供任何论据，因为我想打电话f(void)。为什么编译器会拒绝它？为什么添加行virtual int f(void) = 0;解决我的问题吗？

class Base
{
public:
  explicit Base(void) {}
  virtual ~Base(void) {}

  virtual int f(void) = 0;
  virtual int f(int i) = 0;
};

class Derived : public Base
{
public:
  // provide implementation for f(int) which uses f(void). Does not compile.
  virtual int f(int i) {puts("Derived::f(int)"); return f();}
  // code only compiles by adding the following line.
  virtual int f(void) = 0;
};

class Derived2 : public Derived
{
public:
  // overwrite only f(void). f(int) is implemented by Derived.
  virtual int f(void) {puts("Derived2::f(void)"); return 4;}
};

int main(void)
{
  Base * p = new Derived2();
  int i0 = p->f();  // outputs Derived2::f(void) and returns 4
  int i1 = p->f(1); // outputs "Derived::f(int) Derived2::f(void)" and return 4
  delete p;
  return 0;
}

Derived::f hides Base::fs。给定return f();在体内Derived::f(int)， 名字f发现在范围内Derived, then 姓名查询停止。中的名字Base不会被发现并参与重载决策。

名称查找按如下所述检查范围，直到找到至少一个任何类型的声明，此时查找停止并且不再检查其他范围。

你可以加using Base::f;介绍名字来自Base纳入范围Derived.

class Derived : public Base
{
public:
  using Base::f;

  // provide implementation for f(int) which uses f(void).
  virtual int f(int i) {puts("Derived::f(int)"); return f();}
};