Calling functions using an array of pointers

Question

I have the following files in my program, a header with certain function definitions, and a program with the body of the functions.

something.h

typedef struct _foo {
 int id;
 int lucky_number;
} foo;

typedef void (*pointer_fc)(foo *);

void first(foo *);

void second(foo *);

void third(foo *);

extern pointer_fc fc_bases[3];

something.c

pointer_fc fc_bases[] = {first, second, third};

/* body of functions */

Note that in the header I had defined an array of pointers to functions, and in the something.c program, the functions are associated with every element of the array.

Let's suppose in certain moment I need in the main.c program to call all 3 functions. With this, how I can use the extern array of pointers to call this functions in my main.c.

fc_bases[n](params); (and the 3 is not needed in the extern decl.) — WhozCraig
– WhozCraig, Commented Apr 16, 2013 at 1:46
The question about the function without parameters was irrelevant to the problem itself, so I edited it. — SealCuadrado
– SealCuadrado, Commented Apr 16, 2013 at 1:50

luser droog · Accepted Answer · 2013-04-16 02:00:55Z

1

Function pointers are automatically dereferenced when you call them, so it's as simple as

foo f;
fc_bases[0](&f);
fc_bases[1](&f);
fc_bases[2](&f);

answered Apr 16, 2013 at 2:00

luser droog

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2 Comments

SealCuadrado Over a year ago

Both answers solve my question, but the main.c program had declared the structure this way, so it was easier to implement.

luser droog Over a year ago

If you want some reading practice with function-pointers, I've got some example code here that puts them in an array indexed by an enum, tied together with macros. I've also got them in arrays of structs for a finite-state machine here.

Deepu · Accepted Answer · 2013-04-16 01:52:31Z

1

If f1 is declared as follows as a structure pointer variables of the structure foo,

foo *f1;

Then you can call the functions first() and second() as follows,

pointer_fc fc_bases[] = {first, second};

(*fc_bases[0])(f1);

(*fc_bases[1])(f1);

answered Apr 16, 2013 at 1:52

Deepu

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