Can you explain the following C/C++ statement?

Question

void (*func)(int(*[ ])());

Answer 1

Yes:

$ cdecl
explain void (* x)(int (*[])());
declare x as pointer to function 
  (array of pointer to function returning int) returning void

Answer 2

It's not a statement, it's a declaration.

It declares func as a pointer to a function returning void and taking a single argument of type int (*[])(), which itself is a pointer to a pointer to a function returning int and taking a fixed but unspecified number of arguments.

---

cdecl output for ye of little faith:

cdecl> explain void (*f)(int(*[ ])());
declare f as pointer to function (array of pointer to function returning int) returning void

Answer 3

void (*func)(blah); is a pointer to a function taking blah as an argument, where blah itself is int(*[ ])() is an array of function pointers.

Answer 4

Here's a guide on reading C declarations:

http://www.ericgiguere.com/articles/reading-c-declarations.html

Answer 5

The general procedure for reading hairy declarators is to find the leftmost identifier and work your way out, remembering that [] and () bind before * (i.e., *a[] is an array of pointer, not a pointer to an array). This case is made a little more difficult by the lack of an identifier in the parameter list, but again, [] binds before *, so we know that *[] indicates an array of poitners.

So, given

void (*func)(int(*[ ])());

we break it down as follows:

       func                   -- func
      *func                   -- is a pointer
     (*func)(           )     -- to a function taking
     (*func)(     [ ]   )     --    an array
     (*func)(    *[ ]   )     --    of pointers
     (*func)(   (*[ ])())     --    to functions taking 
                              --      an unspecified number of parameters
     (*func)(int(*[ ])())     --    returning int
void (*func)(int(*[ ])());    -- and returning void

What this would look like in practice would be something like the following:

/**
 * Define the functions that will be part of the function array
 */
int foo() { int i; ...; return i; }
int bar() { int j; ...; return j; }
int baz() { int k; ...; return k; }
/**
 * Define a function that takes the array of pointers to functions
 */
void blurga(int (*fa[])())
{
  int i;
  int x;
  for (i = 0; fa[i] != NULL; i++)
  {
    x = fa[i](); /* or x = (*fa[i])(); */
    ...
  }
}    
...
/**
 * Declare and initialize an array of pointers to functions returning int
 */
int (*funcArray[])() = {foo, bar, baz, NULL};
/**
 * Declare our function pointer
 */
void (*func)(int(*[ ])());
/**
 * Assign the function pointer
 */
func = blurga;
/**
 * Call the function "blurga" through the function pointer "func"
 */
func(funcArray); /* or (*func)(funcArray); */

Answer 6

Geordi is a C++ bot, allowing to train this:

<litb> geordi: {} void (*func)(int(*[ ])());
<litb> geordi: -r << ETYPE_DESC(func)
<geordi> lvalue pointer to a function taking a pointer to a pointer to a nullary function 
         returning an integer and returning nothing

It can do many useful things, like showing all parameter-declarations (this, in fact, is just matching raw C++ grammar rule names):

<litb> geordi: show parameter-declarations
<geordi> `int(*[ ])()`.

Let's do it in the opposite direction:

<litb> geordi: {} int func;
<litb> geordi: make func a pointer to function returning void and taking array of pointer to 
       functions returning int
<litb> geordi: show
<geordi> {} void (* func)(int(*[])());

It executes anything you give to it, if you ask it. If you are trained but just forgot some of the scary parentheses rules, you may also mix C++ and geordi-style type descriptions:

<litb> geordi: make func a (function returning void and taking (int(*)()) []) *
<geordi> {} void (* func)(int(*[])());

Have fun!