Pointer to function returning function pointer

Question

I would like to declare a variable of type pointer to function returning pointer to function. Essentially what the following does, but without any typedefs

Answer 1

You can have a look at the declaration of signal() which is a function taking a void(*)() and returning one of those. The variable ptr can be declared like this:

void (*(*ptr)())()

The notation is a bit awkward and clearly inside out. It may be easier to use trailing return types:

auto (*ptr)() -> void (*)()

... or, of course, use trailing return types all the way through:

auto (*ptr)() -> auto (*)() -> void

Answer 2

The general rule of C (and C++) declarations is: if you type the declaration as an expression, it will have the declaration's type.

So, you want a pointer to function which returns pointer to function returning void.

Let's say we have such a pointer, ptr. How to get void out of it?

1. Dereference ptr, getting a function returning pointer to function returning void: *ptr

2. Call the function, getting a pointer to function returning void: (*ptr)()

3. Dereference that pointer, getting a function returning void: *(*ptr)()

4. Call that function, getting void: (*(*ptr)())()

Now just turn this into a declaration:

void (*(*ptr)())();

P.S. I know other have answered in the meantime (and I've upvoted). But I wanted to show the generic process to arrive at the declaration form.