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Glen Low · Guest

Hi folks, it's your friendly neighborhood SFINAE pest again.

Qn of the week: does SFINAE work with incomplete types?

Consider the following:

struct X;

template <typename U> void f (typename U::inner* i)
{
std::cout << "ho";
}

template {
std::cout << "hie";
}

int main ()
{
f }

According to gcc (and possibly online Comeau), this is valid and
outputs "hie". However the Standard does say in its SFINAE passage
(14.8.2/2):

"Attempting to use a type in the qualifier portion of a qualified name
that names a type when that type does not contain the specified
member, or if the specified member is not a type where a type is
required."

X is an incomplete type, so you *could* argue that it is a "type that
does not contain the specified member", but is this a correct
interpretation?

Cheers,
Glen Low, Pixelglow Software
www.pixelglow.com

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