How do i get type from pointer in a template?

Question

I know how to write something up but i am sure there is a standard way of passing in something like func() and using template magic to extract T

Answer 1

I think you want to remove the pointer-ness from the type argument to the function. If so, then here is how you can do this,

template<typename T>
void func()
{
    typename remove_pointer<T>::type type;
    //you can use `type` which is free from pointer-ness

    //if T = int*, then type = int
    //if T = int****, then type = int 
    //if T = vector<int>, then type = vector<int>
    //if T = vector<int>*, then type = vector<int>
    //if T = vector<int>**, then type = vector<int>
    //that is, type is always free from pointer-ness
}

where remove_pointer is defined as:

template<typename T>
struct remove_pointer
{
    typedef T type;
};

template<typename T>
struct remove_pointer<T*>
{
    typedef typename remove_pointer<T>::type type;
};

In C++0x, remove_pointer is defined in <type_traits> header file. But in C++03, you've to define it yourself.