5 
//#define NOT_WORKS
#define HOW(X) 0

struct A {

};

struct B {
    A a;
};

struct C {
    B b;
};

int main(int argc, char **argv) {
    A B::*ba = &B::a;       // ba is a pointer to B::a member
    B C::*cb = &C::b;       // cb is a pointer to C::b member

#ifdef NOT_WORKS

    { A C::*ca = &C::b::a; }    // error: Symbol a could not be resolved / error: ‘C::b’ is not a class or namespace
    { A C::*ca = cb + ba; }     // error: invalid operands of types ‘B C::*’ and ‘A B::*’ to binary ‘operator+’

    A C::*ca = HOW(???);        // is possible to reach C::b::a by a member pointer?

#endif

    C cptr;
    A aptr = cptr.*cb.*ba;  // is pointer inference chaining the only solution?

    return 0;
}

如果成员指针的推理链是到达内部成员的唯一解决方案，我可以使用模板将其封装在单一类型上吗？

现在代码可以用gcc编译了

谢谢大家

4

1 回答 1

2

可以通过成员指针到达 C::b::a 吗？

有点：

C c; 
A B::*ca = &B::a;  //    usage: c.b.*ca;

指针推理链接是唯一的解决方案吗？

是的

于 2013-05-08T10:46:01.997 回答