Typedef with c++ pointer

Question

the normal typedef syntax i know is:
typedef /*real type name*\ /*alias name*\;

but when i trying to create array of function that returning void pointer, stackoverflow say i must use "typedef void (**function_ptr)();"

is that a special syntax only for function pointer?
why its not using normal typedef syntax?

its not homework or something, so you freely to answer it

Answer 1

The typedef syntax is a little different for function pointers.

In pointers, the name given alongside the pointer in the typedef declaration becomes the type's name.

Like for a normal array, the typedef statement will be: typedef char _string[50];

Now _string is a type to represent a 50 character array, and you can do :
_string a = "Hello";

Similarly, to declare a typename for a function pointer, you do :

typedef int(*fx)(int,int);

The type, fx, can be used to hold functions that accept two integers and return an integer.

'fx' can be used like this : https://code.sololearn.com/cUNvlu5w8M6B/?ref=app

Answer 2

oh well. very good answer. lot of thanks to you.