是否可以编写这样的C++模板owner_of<...>:

struct X { int y; }
owner_of<&X::y>::typeX吗?

最佳答案

您几乎可以做到这一点(或者至少到目前为止我找不到更好的解决方案):

#include <string>
#include <type_traits>

using namespace std;

template<typename T>
struct owner_of { };

template<typename T, typename C>
struct owner_of<T (C::*)>
{
    typedef C type;
};

struct X
{
    int x;
};

int main(void)
{
    typedef owner_of<decltype(&X::x)>::type should_be_X;
    static_assert(is_same<should_be_X, X>::value, "Error" );
}

如果您介意使用decltype,也许宏可以做到:
#define OWNER_OF(p) owner_of<decltype( p )>::type

int main(void)
{
    typedef OWNER_OF(&X::x) should_be_X;
    static_assert(is_same<should_be_X, X>::value, "Error" );
}

基于decltype:的替代解决方案
template<typename T, typename C>
auto owner(T (C::*p)) -> typename owner_of<decltype(p)>::type { }

int main(void)
{
    typedef decltype(owner(&X::x)) should_be_X;
    static_assert(is_same<should_be_X, X>::value, "Error" );
}

关于c++ - 有没有办法从C++中的成员指针类型派生对象类型,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/14963502/

10-10 14:23
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