C ++成员函数虚函数重载和重载同时

问题描述

如果我有这样的代码：

struct A {
  virtual void f(int) {}
  virtual void f(void*) {}
};

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

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


int main() {
  C c;
  c.f(1);

  return 0;
}

我得到一个错误，说我试图做一个无效的转换int to void *。为什么编译器不知道他必须调用B :: f，因为这两个函数都声明为虚拟？

I get an error that says that I am trying to do an invalid conversion from int to void*. Why can't compiler figure out that he has to call B::f, since both functions are declared as virtual?

---

阅读jalf的回答后，我进一步减少了。这一个不工作，以及。不太直观。

After reading jalf's answer I went and reduced it even further. This one does not work as well. Not very intuitive.

struct A {
  virtual void f(int) {}
};

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


int main() {
  B b;
  b.f(1);

  return 0;
}

推荐答案

这是C ++中的重载解析工作原理。

The short answer is "because that's how overload resolution works in C++".

编译器在C类中搜索函数F，如果找到任何函数，它停止搜索，其中一个候选人。

The compiler searches for functions F inside the C class, and if it finds any, it stops the search, and tries to pick a candidate among those. It only looks inside base classes if no matching functions were found in the derived class.

但是，您可以将基类函数显式地引入到派生类的命名空间中： / p>

However, you can explicitly introduce the base class functions into the derived class' namespace:

struct C : public B {
  void f(void*) {}
  using B::f; // Add B's f function to C's namespace, allowing it to participate in overload resolution
};

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