What does “void SomeFunction(int, void*)” mean?

Question

I am trying out some OpenCV and are on a tutorial for a Canny edge detector Example .

In this tutorial, there is a function declared like this:

void CannyThreshold(int, void*)

and then called like this from inside main;

CannyThreshold(0, 0);

I don't understand the purpose og the (int, void*) part of the declaration since none of these arguments are used in the CannyThreshold funtion.

Why is it simply not just declared like this?

void CannyThreshold();

Answer 1

Note this line in the example:

createTrackbar( "Min Threshold:", window_name, &lowThreshold, max_lowThreshold, CannyThreshold );

Here, CannyThreshold is passed as a callback argument to createTrackbar . The signature of createTrackbar requires the callback to accept these two arguments, and so CannyThreshold does, even if it has no use for them.

Answer 2

If you declare the function as

void CannyThreshold();

that is the same as

void CannyThreshold(void);

In other words, it's a function that takes no argument.

If you want the function to take arguments, but never use those arguments, you must still declare the types of the arguments.

Answer 3

This is supposedly because this function is a callback , hence its signature is enforced by whatever API is used (here: OpenCV).

A callback is a name given to a function when this function is registered into an event system to be called later, when something happens. A function registering a callback is called a functor , it takes another function as argument, and do something with it. Exemple:

using callback_t = void(*)(int);
void register_callback(callback_t cb);

If you want to define a callback to be called (back) by register_callback , it needs to be a function taking an int and returning void , even if you don't need the integer. So the following definition is possible:

void my_callback(int) { std::cout << "done\n"; }
register_callback(my_callback);