CODEWITHSUNDEEP
c++templatespointer-to-member
hutorny
hutorny

Reputation: 873

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

Consider the following example :

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::*'

This example is also available on http://coliru.stacked-crooked.com/a/15f3e7acd8de04a3 , Both clang++ and g++ produce the same error.

How to cast fuu::bax so it is accepted by template tux?

Upvotes: 1

Views: 65

Answers (1)

Brian Bi
Brian Bi

Reputation: 119059

Unfortunately, I don't think there's any way to do this yet. There is an open defect report about this issue. You will have to work around it somehow, such as changing tux so it accepts int foo::* instead, or adding another template parameter:

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;
};

Upvotes: 3

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