CODEWITHSUNDEEP
c++templates
user34537
user34537

Reputation:

How do i get type from pointer in a template?

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?

Upvotes: 10

Views: 6179

Answers (2)

skyking
skyking

Reputation: 14400

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))>

Upvotes: 0

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361802

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.

Upvotes: 21

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