为什么需要重新声明重载的虚函数? [英] Why do I need to redeclare overloaded virtual functions?

问题描述

我有一个带有两个重载函数 f(void)和 f(int)的基类. Derived 类通过调用 f(void)来实现 f(int). Derived2 仅实现 f(void).

I have a base class with two overloaded functions f(void) and f(int). The class Derived implements f(int) by calling f(void). Derived2 implements f(void) only.

编译器拒绝实现 Derived :: f(int)，因为它想调用 f(int)，但是我没有提供参数，因为我想调用> f(void).为什么编译器拒绝它?为什么添加行 virtual int f(void)= 0; 可以解决我的问题?

The compiler rejects the implementation Derived::f(int) because it wants to call f(int) but I provided no arguments because I want to call f(void). Why does the compiler reject it? Why does adding the line virtual int f(void) = 0; fix my problem?

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;
}

推荐答案

派生的:: f 隐藏 Base :: f s.给定 return f(); 在 Derived :: f(int)的正文中，名称 f 在的范围内>派生，然后名称查找停止.找不到 Base 中的名称并参与重载解析.

Derived::f hides Base::fs. Given return f(); in the body of Derived::f(int), the name f is found in the scope of Derived, then name lookup stops. The names in Base won't be found and participate in overload resolution.

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为止，这时查找停止，并且不再检查任何范围.

name lookup examines the scopes as described below, until it finds at least one declaration of any kind, at which time the lookup stops and no further scopes are examined.

您可以使用Base :: f; 添加，以将 Base 中的名称引入 Derived 的范围内.

You can add using Base::f; to introduce the name from Base into the scope of 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();}
};

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