How define an array of function pointers in C
The type of a function pointer is just like the function declaration, but with "(*)" in place of the function name. So a pointer to:
int foo( int )
would be:
int (*)( int )
In order to name an instance of this type, put the name inside (*), after the star, so:
int (*foo_ptr)( int )
declares a variable called foo_ptr that points to a function of this type.
Arrays follow the normal C syntax of putting the brackets near the variable's identifier, so:
int (*foo_ptr_array[2])( int )
declares a variable called foo_ptr_array which is an array of 2 function pointers.
The syntax can get pretty messy, so it's often easier to make a typedef to the function pointer and then declare an array of those instead:
typedef int (*foo_ptr_t)( int );
foo_ptr_t foo_ptr_array[2];
In either sample you can do things like:
int f1( int );
int f2( int );
foo_ptr_array[0] = f1;
foo_ptr_array[1] = f2;
foo_ptr_array[0]( 1 );
Finally, you can dynamically allocate an array with either of:
int (**a1)( int ) = calloc( 2, sizeof( int (*)( int ) ) );
foo_ptr_t * a2 = calloc( 2, sizeof( foo_ptr_t ) );
Notice the extra * in the first line to declare a1 as a pointer to the function pointer.
I put a small example here that may help you
typedef void (*fp)(int); //Declares a type of a void function that accepts an int
void test(int i)
{
printf("%d", i);
}
int _tmain(int argc, _TCHAR* argv[])
{
fp function_array[10]; //declares the array
function_array[0] = test; //assings a function that implements that signature in the first position
function_array[0](10); //call the cuntion passing 10
}