What does T::* signify in the declaration of a function parameter list?

what is the meaning of T::* inside the registration function declaration

This is the syntax of a pointer to member. Let's take a look at the whole type and name of the parameter:

void(T::*callback)(const pcl::visualization::KeyboardEvent&, void*)

This is the declaration of a variable named callback. It's a pointer to member function. More precisely, it's a pointer to member function of the class T.

If we take the name out of the type, we see things more clearly:

// class name ---v     v------- parameters
            void(T::*)(const pcl::visualization::KeyboardEvent&, void*)
//          ^---- return type

It's in fact, a pointer to function member of the class T that returns void. It's a function that takes strictly two parameters: a const pcl::visualization::KeyboardEvent& and a void*.

why am I not allowed to pass this

It's simple. Look at the type of your function:

using func_type = decltype(keyboardEventCallback);
// hint: the type is: void(*)(const pcl::visualization::KeyboardEvent&, void*, void*)

Let's compare the two types side by side:

void(*)(const pcl::visualization::KeyboardEvent&, void*, void*)
void(T::*)(const pcl::visualization::KeyboardEvent&, void*)

First, your function is not a member function, it's a plain function pointer. It's not the same type. Then, you got three arguments, as the type of the parameter only ask for two. This is different.

---

Now, how can you fix this??

You could use a lambda:

auto myCallback = [](const pcl::visualization::KeyboardEvent& e, void* c) { /* ... */ }

using lambdaType = decltype(myCallback);

// Be careful here, we don't want our lambda to go out of scope when it is called.
m_vis->registerKeyboardCallback(&lambdaType::operator(), myCallback, this);

---

Or even simpler: just define keyboardEventCallback inside your class, and send it:

// don't forget: keyboardEventCallback must receive the same parameter as asked.
m_vis->registerKeyboardCallback(&MyClass::keyboardEventCallback, *this, this);

This is the syntax for member functions.

Example:

class A{
  int giveMe5();
};

&A::giveMe5;  // will be of type int(A::*)()

Why does the type differ from free functions and static member functions ? Because member functions have an implicit parameter that points to the object on which the function gets called.

https://isocpp.org/wiki/faq/pointers-to-members#fnptr-vs-memfnptr-types says:

The type of this function is different depending on whether it is an ordinary function or a non-static member function of some class:

- Its type is int (*)(char,float) if an ordinary function

- Its type is int (Fred::*)(char,float) if a non-static member function of class Fred