I've a little question. I'm trying to define an array of function pointers dynamically with calloc . But I don't know how to write the syntax. Thanks a lot.
21.8k 7 7 gold badges 64 64 silver badges 65 65 bronze badges asked Mar 30, 2011 at 15:38 955 2 2 gold badges 11 11 silver badges 17 17 bronze badges Can you show us what you wrote until now? Commented Mar 30, 2011 at 15:42The 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 ) 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.