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[英]Cannot cast “member pointer to derived class” to “member pointer to base class”
[英]How to cast a pointer to member of base class to a pointer to the same member of derived class
请考虑以下示例:
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::*'
此示例也可在http://coliru.stacked-crooked.com/a/15f3e7acd8de04a3上获得 ,clang ++和g ++都会产生相同的错误。
如何投射fuu :: bax所以它被模板tux接受?
不幸的是,我认为还没有办法做到这一点。 有关此问题的公开缺陷报告 。 您将不得不以某种方式解决它,例如更改tux
以使其接受int foo::*
,或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;
};
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