CODEWITHSUNDEEP
c++pointersc++11
Ryan
Ryan

Reputation: 14659

Is defining function pointer as a new type with a template type as an argument not possible in C++?

In C, I can do this.

typdef void(TRAVERSAL_CALLBACK*)(int a);

That would then allow me to pass function pointers to other functions as arguments, with that function having 1 argument of type int.

I've been working with templates, and currently, passing the function pointer to the method works like this:

void my_function(void(TRAVERSAL_CALLBACK*)(T &data));

Now, is it possible if I define a new type with argument of type T. I have tried but it has failed to compile.

typedef void(TRAVERSAL_CALLBACK*)(T &data);

I'm using C++11. Is this something that I have to accept as not possible, or is there an idomatic way to do this in C++11 that I am not aware of?

Upvotes: 1

Views: 78

Answers (2)

IdeaHat
IdeaHat

Reputation: 7881

In C++11, there is template type aliasing using the using statement, example shown below:

#include <iostream>

using namespace std;

template <typename T>
T foo() {return 10;}

template <typename T>
using fptr = T (foo*)(void);
int main()
{
    fptr<int> x1 = foo<int>;
    fptr<double> x2 = foo<double>;
   cout << "Hello World" << endl; 

   return 0;
}

However, in your case you most likely don't need to go that far. If you define the type within the scope of the template, it should work fine:

template <typename T>
struct bar
{
  typedef T (*qiz)(void);

  void baz(qiz q)
  {
    std::cout << q << std::endl;
  }

};

Upvotes: 3

Matteo Italia
Matteo Italia

Reputation: 126927

Yes, in C++11 it's possible, thanks to the "new" using:

template<typename T>
using TRAVERSAL_CALLBACK = void(*)(T &data);

Notice however that in C++ you'd probably just make the whole type of the callback a template, to allow the usage of other types of callable objects (functors, lambdas, ...).

Upvotes: 6

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