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<TheType*>() and using template magic to extract TheType for use in your code (maybe TheType::SomeStaticCall).

What is the standard way/function to get that type when a ptr is passed in?

Strepitous

You should add some code to make the question clear. What is it that you want to do, what are the inputs? – Aponte

@David: I had something but had the oversight of not realizing it was be filter out as a tag (and not show up). Does it make since ATM? – Strepitous

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.

Millimeter

so remove_pointer isnt in the std? This is exactly what i wanted but my gut says theres something in stl i can use (but maybe not?) – Strepitous

@acidzombie24: In C++0x, Yes. In C++03, No. – Millimeter

darn, what is it in C++0x? (header and i guess the struct name if not the same) I guess i was half right? – Strepitous

@acidzombie24: C++0x = incoming version of C++ :P – Millimeter

For C++03 it's in TR1 and Boost.TypeTraits – Pompey

Very nice indeed, although I think the C++0x remove_pointer only removes one pointer at a time. – Groundsheet

Answer 2

If you just want to remove one level of pointer and get the actual type of the pointee, that is it T is int** want to get int* instead of int you could perhaps work with decltype, but that requires you to be able to construct an expression of type T (which is to be dereferenced) is actually a pointer that shouldbe possible using reinterpret_cast if T is a pointer. So it boils down to something like:

std::remove_reference_t<decltype(*reinterpret_cast<T>(NULL))>

Allyce