Frustration with decltype of functions

Question

I am now writing wrapper for some Windows library functions so I need extract their prototype and write a new wrapper with a modified prototype. For example: the function

int recv(SOCKET, char*, int, int)

will be wrapped by

recv_wrapper(SOCKET, char*, int, int, THREADID)

To avoid erroneousness, I write a wrapper template:

template <typename F> struct wrapper;

template <typename R, typename T1, typename T2, typename T3, typename T4>
struct wrapper<R(T1,T2,T3,T4)> 
{
  typedef R result_type;
  typedef R (type)(T1,T2,T3,T4,THREADID);
};

and then the following wrapped type works:

typedef int recv_t(SOCKET,char*,int,int);
typedef wrapper<recv_t>::type recv_wrapper_t;

But I want to go a bit further using decltype to get the type of recv automatically:

typedef decltype(recv) recv_t;
typedef wrapper<recv_t>::type recv_wrapper_t;

But this does not work, I try several ways to modify that but I get always the error from VS2010 compiler:

error C2027: use of undefined type 'wrapper<F>'
with
        [
            F=int (SOCKET,char *,int,int)
        ]

and

error C2146: syntax error : missing ';' before identifier 'recv_wrapper_t'

I am quite frustrated with that because here F is nothing but the recv_t defined explicitly:

typedef int recv_t(SOCKET,char*,int,int);

but when get it automatically with decltype, it does not work anymore.

Many thanks for any consideration.

N.B. I have fix a typo error following the suggestion of n.m.; but that still does not work.

Answer 1

template <typename F> wrapper; is invalid. Did you mean template <typename F> struct wrapper;? If you fix this, the obvious syntax works:

typedef wrapper<decltype(recv)>::type recv_wrapper_t;