It's not the concept as a whole, but rather one of the methods it uses to determine if a class has an n
data-member. Here is the full code; an ordinary use of SFINAE for member detection.
template <typename T>
struct has_X {
struct Fallback { int X; };
struct Derived : T, Fallback {};
template <typename U, U> struct S;
template <typename C> static char (&f(S<int Fallback::*, &C::X> *))[1];
template <typename C> static char (&f(...))[2];
public:
const static bool value = sizeof(f<Derived>(0)) == 2;
};
The part where Derived
inherits from both Fallback
and T
confuses me because when we do the overload of f
, &C::X
is &Derived::X
. But shouldn't this overload always be chosen because isn't Derived
guaranteed to have X
since it inherits from Fallback
which has that data-member?
Maybe I'm overlooking something. However, this single piece of code has shown and taught me things I never knew, so maybe there is something to this. What I would expect is for that overload to always be chosen (not the one with the ...
) because Derived
should always have X
since it inherits from Fallback
. But this is not the case. Can someone please explain why?
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Fallback
has one data member named X
, but Derived will have two if T
also has a member named X
, in which case Derived::X
cannot be taken unambiguously. So if T
does not have X
, the first overload is used, and if T
has X
, the second more general version is used. This is why you can tell these cases apart depending on the size of their return types.
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