没有匹配函数用于调用...继承的C++类中缺少重载

Question

有人可以解释一下这里发生了什么.为什么编译器在A类中看不到没有参数的hello()？

struct A {
    virtual void hello() {}
    virtual void hello(int arg) {}
};
struct B : A {
    virtual void hello(int arg) {}
};

int main()
{
    B* b = new B();
    b->hello();
    return 0;
}

g ++ main.cpp

main.cpp: In function ‘int main()’:
main.cpp:13:11: error: no matching function for call to ‘B::hello()’
main.cpp:13:11: note: candidate is:
main.cpp:7:15: note: virtual void B::hello(int)
main.cpp:7:15: note:   candidate expects 1 argument, 0 provided

Answer 1

因为你的覆盖hello(int arg)隐藏了具有相同名称的其他功能.

你可以做的是明确地将这些基类函数引入子类:

struct B : A {
    using A::hello; // Make other overloaded version of hello() to show up in subclass.
    virtual void hello(int arg) {}
};