基类的成员指针 [英] Member Pointer to Base Class
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问题描述
所有。我无法理解为什么波纹管代码需要一个演员才能工作。有人可以解释一下吗？
all. I can't undestand why the bellow code need a cast to work. Someone can explain it?
class Base {
};
class Derived : public Base {
};
class Class {
public:
Derived member;
};
...
Derived obj;
Base *ptrObj = &obj; // ok, no cast needed
Derived Class::* ptr = &Class::member; // ok
Base Class::* ptr = &Class::member; // wrong, need cast, why?
推荐答案
因为如果 Base （协变），你可以这样做，这是一个禁忌：
Because if Base were allowed (covariant), you could then do this, which is a no-no:
Base Class::* ptr = &Class::member;
Class obj;
obj.*ptr = Base(); // <-- assigned a Base into a Derived field?!
同时，指向成员的指针也不能逆变，因为否则你可以做到这一点，这也是禁忌：
At the same time, pointers-to-members cannot be contravariant either, because otherwise you could do this, which is also a no-no:
struct Class2 {
Base member;
};
Derived Class2::* ptr2 = &Class2::member;
Class2 obj2;
obj2.member = Base();
Derived& d = obj2.*ptr2; // <-- assigned a Base into a Derived
因此，指向成员的指针既不是协变或逆变，但不变：类型必须完全匹配。
So, pointers-to-members are neither covariant nor contravariant, but are invariant: the type must match exactly.
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