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