Why can an unnamed struct not be used as a trailing return type?

Question

struct { int a, b; } f(int x, int y) // OK
{
    return { x, y };
}

auto g(int x, int y) -> struct { int a, b; } // error C2332
{
    return { x, y };
}

int main()
{
    auto n = f(1, 2).a; // OK
}

My compiler is VC++ 2013 RC.

Why is g wrong while f is OK?

Is this a bug of VC++?

Answer 1

Actually, in C++, it's illegal to define a type in a parameter or return type, named or not. See C++11[diff.decl]:

Change: In C++, types may not be defined in return or parameter types. In C, these type definitions are allowed

So the actual problem is the first case being accepted, not the second one being rejected.