派生类通过基类定义函数 [英] Derived class defines function via base class

问题描述

考虑以下代码:

struct A
{
  virtual void f() = 0;
};

struct B
{
  void f();
};

struct C : public A, public B
{};

int main()
{
  A* a = new C();
  B* b = new C();
  C* c = new C();

  // All these calls should result in B::f
  a->f();
  b->f();
  c->f();
}

编译器指出C是抽象的. 如何解决这种情况? 这个问题似乎与钻石继承相似，但是我看不到解决方案.

The compiler states that C is abstract. How can this situation be resolved? The issue seems similar to diamond inheritance, but I fail to see the solution.

谢谢，这是工作示例:

#include "stdio.h"

struct A
{
  virtual void f() = 0;
};

struct B
{
  void f()
  {
      printf("B::f\n");
  }
};

struct C : public A, public B
{
  void f()
  {
      printf("C::f\n");
      B::f();
  }
};

int main()
{
  A* a = new C();
  B* b = new C();
  C* c = new C();

  printf("Calling from A\n");
  a->f();
  printf("Calling from B\n");
  b->f();
  printf("Calling from C\n");
  c->f();
}

输出:

Calling from A
C::f
B::f
Calling from B
B::f
Calling from C
C::f
B::f

推荐答案

问题在于，两个f()函数完全不相关，即使它们恰好具有相同的名称.

The issue is that the two f() functions are completely unrelated, even though they happen to have the same name.

如果所需的行为是C的virtual f()调用B::f()的行为，则必须明确地做到这一点:

If the desired behaviour is for the C's virtual f() to call B::f(), you have to do it explicitly:

struct C : public A, public B
{
  void f();
};

void C::f() {
  B::f();
}

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