c++ - 为什么我必须显式定义继承类提供的方法？

Question

考虑以下几点:

#include <string>

struct animal
{
public:
    virtual std::string speak() = 0;
};

struct bird : public animal
{
public:
    std::string speak()
    {
        return "SQUAK!";
    }
};

struct landAnimal : public animal
{
    virtual int feet() = 0;
};


struct sparrow : public bird, public landAnimal
{
    int feet() { return 2; }
    // This solves it, but why is it necessary, doesn't bird provide this?
    // std::string speak(){ return this->speak(); } 
};

int main()
{
    sparrow tweety = sparrow();
}

编译它，你会得到:

1>ex.cpp(35): error C2259: 'sparrow': cannot instantiate abstract class
1>  ex.cpp(35): note: due to following members:
1>  ex.cpp(35): note: 'std::string animal::speak(void)': is abstract
1>  ex.cpp(10): note: see declaration of 'animal::speak'

为什么需要注释方法才能编译？

Answer 1

因为，与您标记的不同，您没有继承钻石。你的麻雀是两只动物，其中只有一只被bird具体化了。另一个通过 landAnimal 继承的不是。

要获得真正的钻石，您需要的是虚拟继承，但您会发现它附带了大量警告。

旁注，如Martin Bonner正确地指出:

It's probably worth pointing out that the "fix" isn't a fix at all. Any call to sparrow::speak() will cause infinite recursion. It would need to be std::string speak() { return Bird::speak(); }.