C2440 cannot convert from 'void (_cdecl*)(int)' to 'void(_cdecl&)(int) - c++

Question

I am using C++ in native mode with Visual Studio 2017 and I am trying to compile and run the example code found at Debugging a Parallel Application in Visual Studio.
For the record, I program in C not C++. I am clueless when it comes to method declarations (among many other things). I suspect correcting the error is simple but, I simply don't know how.
In other words, I am currently RTFineM. I simply copied and pasted the example given in the url above and ran into 2 problems. First it complained about something being deprecated but a simple define took care of that problem. Second it complained about not being able to convert a type into another as stated in the title.
The RunFunc class causing the problem is declared as follows:
class RunFunc
{
Func& m_Func;
int m_o;
public:
RunFunc(Func func,int o):m_Func(func),m_o(o)
{
};
void operator()()const
{
m_Func(m_o);
};
};
My question/request is: how does the declaration of RunFunc need to be in order for the example to compile and run properly ?
Thank you, much appreciate the help.

Answer 1

In this constructor
RunFunc(Func func,int o):m_Func(func),m_o(o)
{
};
the prameter Func func is adjusted by the compiler to the type Func *func. On the other hand the data member m_Func is declared as a referenced type.
Func& m_Func;
And the error message says about incompatibility of the types.
C2440 cannot convert from 'void (_cdecl*)(int)' to 'void(_cdecl&)(int)
Try to declare the constructor like
RunFunc(Func &func,int o):m_Func(func),m_o(o)
{
};
Or declare the data member like
Func *m_Func;
without changing the constructor.
Here are two demonstrative programs
#include <iostream>
typedef void Func( int );
class RunFunc
{
Func& m_Func;
int m_o;
public:
RunFunc(Func &func,int o):m_Func(func),m_o(o)
{
};
void operator()()const
{
m_Func(m_o);
};
};
int main() {
return 0;
}
and
#include <iostream>
typedef void Func( int );
class RunFunc
{
Func *m_Func;
int m_o;
public:
RunFunc(Func func,int o):m_Func(func),m_o(o)
{
};
void operator()()const
{
m_Func(m_o);
};
};
int main() {
return 0;
}

Answer 2

In your code you are tyring to bound a reference to a temporary, namely to copy of argument passed to the constructor. You can try to run the following code snippet to see the difference:
struct Func {
int _i;
void operator()(int i) { cout << i*_i << endl; }
};
class RunFunc
{
Func& m_Func;
int m_o;
public:
RunFunc(Func &func, int o) :m_Func(func), m_o(o)
// RunFunc(Func func, int o) :m_Func(func), m_o(o)
{
};
void operator()()const
{
m_Func(m_o);
};
};
int main() {
Func f{ 5 };
RunFunc rf(f, 2);
rf();
return 0;
}

Answer 3

This is a legacy approach. You can use standard library functor and binder instead. For example:
#include <functional>
#include <iostream>
static void my_callback(int i) {
std::cout<< i << std::endl;
}
int _tmain(int argc, _TCHAR* argv[])
{
std::function<void()> functor;
functor = std::bind(my_callback, 1);
functor();
return 0;
}