C中的函数指针数组

Question

I'm having really hard time comprehending the syntax for function pointers. What I am trying to do is, have an array of function pointers, that takes no arguments, and returns a void pointer. Can anyone help with that?

Answer 1

1. First off, you should learn about cdecl :

    cdecl> declare a as array 10 of pointer to function(void) returning pointer to void void *(*a[10])(void ) 

2. You can do it by hand - just build it up from the inside:

   a

   is an array:

   a[10]

   of pointers:

   *a[10]

   to functions:

   (*a[10])

   taking no arguments:

   (*a[10])(void)

   returning void * :

   void *(*a[10])(void)

3. It's much better if you use typedef to make your life easier:

    typedef void *(*func)(void); 

   And then make your array:

    func a[10]; 

Answer 2

Whenever compound syntax gets too complicated, a typedef usually clears things up.

Eg

typedef void *(* funcPtr)(void);

funcPtr array[100];

Which without the typedef I guess would look like:

void *(* array[100])(void);

Answer 3

Start with the array name and work your way out, remembering that [] and () bind before * ( *a[] is an array of pointer, (*a)[] is a pointer to an array, *f() is a function returning a pointer, (*f)() is a pointer to a function):

        farr               -- farr
        farr[N]            -- is an N-element array
       *farr[N]            -- of pointers
      (*farr[N])(    )     -- to functions
      (*farr[N])(void)     --   taking no arguments
     *(*farr[N])(void)     --   and returning pointers
void *(*farr[N])(void);    --   to void

Answer 4

Use typedef s

typedef void* func(void);
func *arr[37];

Answer 5

查看http://www.newty.de/fpt/fpt.html#arrays ，了解C和C ++函数指针数组的示例和解释。