如何将指向 base class 成员的指针转换为指向 derived class 相同成员的指针

Question

考虑以下示例：

struct foo {
    int bax;
};

struct fuu : foo {
};

template<int foo::*>
struct tox {};

template<int fuu::*>
struct tux {};

int foo::* xo = &foo::bax;
int fuu::* xu = &fuu::bax;   // works

typedef int foo::*boz;

typedef tox<&foo::bax> qox;
typedef tux<&fuu::bax> qux; // fails: 'int foo::*' cannot be converted to a value of type 'int fuu::*'
typedef tux<(boz)&fuu::bax> qux; // fails: non-type template argument of type 'boz' (aka 'int foo::*') cannot be converted to a value of type 'int fuu::*'

此示例在 http://coliru.stacked-crooked.com/a/15f3e7acd8de04a3 上也可用，clang++ 和 g++ 都会产生相同的错误。

如何转换 fuu::bax 使其被模板 tux 接受？

Answer 1

不幸的是，我认为目前还没有办法做到这一点。关于这个问题有一个open defect report。您将不得不以某种方式解决它，例如更改 tux 以便它接受 int foo::*，或者添加另一个模板参数：

template <typename T, int T::* arg>
struct tux {
    static_assert(std::is_convertible<int T::*, int fuu::*>::value,
                  "T must be an unambiguous accessible base of fuu");
    // note that this still works
    // however, you won't be able to use it as a template argument
    static constexpr int fuu::* pm = arg;
};

如何将指向 base class 成员的指针转换为指向 derived class 相同成员的指针

How to cast a pointer to member of base class to a pointer to the same member of derived class

c++

templates

pointer-to-member