struct C
{
int Foo(int i) { return i; }
typedef decltype(C::Foo) type;
};
Since there is no such type as a member function type (there isn
The code is ill-formed: there are only a few ways that a member function name (e.g. C::Foo) can be used, and this is not one of them (the complete list of valid uses can be found in the C++ language standard, see C++11 §5.1.1/12).
In the context of your example, the only thing you can really do is take the address of the member function, &C::Foo, to form a pointer to the member function, of type int (C::*)(int).
Since the code is ill-formed, the compiler should reject it. Further, it yields inconsistent results depending on how C::Foo is used; we'll look at the inconsistency below.
Please report a bug on Microsoft Connect. Alternatively, let me know and I am happy to report the issue.
If you have a type but you don't know what the type is, you can find out the name of the type by using it in a way that causes the compiler to emit an error. For example, declare a class template and never define it:
template
struct tell_me_the_type;
Then later, you can instantiate this template with the type in which you are interested:
tell_me_the_type x;
Since tell_me_the_type hasn't been defined, the definition of x is invalid. The compiler should include the type T in the error it emits. Visual C++ 2012 RC reports:
error C2079: 'x' uses undefined struct 'tell_me_the_type_name'
with
[
T=int (int)
]
The compiler thinks that C::Foo is of type int (int). If that is the case, then the compiler should accept the following code:
template
struct is_the_type_right;
template <>
struct is_the_type_right { };
is_the_type_right x;
The compiler does not accept this code. It reports the following error:
error C2079: 'x' uses undefined struct 'is_the_type_right'
with
[
T=int (int)
]
So, C::Foo both is of type int (int) and is not of type int (int), which violates the principle of noncontradiction. :-)