How define an array of function pointers in C

Question

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.<

Answer 1

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.