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c++ defining a member-function pointer without knowing the type of the object


By : , Category : c++

Yes, use decltype:

void (std::remove_reference<decltype(*this)>::type::*function)(int, int);
ReLated :

You should not be assigning a function pointer to a void pointer.

See http://stackoverflow.com/a/5579907/1351983.

There are lot of issues with the code:

class A {
    public:
      double * var = new double;
};  
  1. What is var?
  2. You cannot initialize data inside a class like this (Use a constructor)

This works:

Fct thef;
GenericFunc<Fct, double, double>::F phi = &Fct::F;
double x = (thef.*phi)(1);

Online Demo

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, ...).

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