C ++ 0x decltype和范围解析运算符

Question

With a class such as Foo:

struct Foo { static const int i = 9; };

I find that GCC 4.5 will reject the following

Foo f;
int x = decltype(f)::i;

It will work if I use an intermediate typedef, such as:

typedef decltype(f) ftype;
int x = ftype::i;

but I prefer to keep the namespace clean. I thought precedence may be an issue, so I've also tried parentheses, but no luck. Is it impossible as presented, or is there a piece of syntax that can help me?

Answer 1

It is valid C++0x to say decltype(f)::i . GCC just doesn't support it yet. You can work it around with an identity template

template<typename T> struct identity { typedef T type; };
int x = identity<decltype(f)>::type::i;

identity is part of the boost::mpl namespace.