How to get a type in C++ when its template argument is the argument

First, we can't do :

//It's wrong.

template <typename U, typename T>
U TypeFunc (U<T> p);

template <typename U, typename T>U<T> TypeFunc2();//either is wrong__typeof (TypeFunc(b))<int> p1;__typeof (TypeFunc2(b)) p2

Under some certain compilers' C++ dialect:

//It's ok

template < template <typename T> class U, typename V>
U<V> typefunc( U<T> tmpl, V type);

vector<int>b;
char c;
__typeof (typefunc(b,c)) u;
//u's type is vector<char>

In C++0x

//It's simple
template <typename T>
using V<T> = std::vector<T>;

V<int> a;

How ever, in GCC without 0x support we can do this by:

template <typename T, template <typename T> class U, typename T2>

U<T2> typefunc(U<T> a, T2 b);

vector <int> a;
char b;
__typeof (typefunc(a, b)) c;
//c's type is vector<char> by adding extra function template argument T