Passing argument to array of function pointers

Question

I am writing a program in C which uses an array of function pointers. If I do not include an argument I am able to call the functions with my code without an issue.

int (*functionsArray[2][3])() = {
  {functionOne,functionTwo,functionThree},
  {functionFour,functionFive,functionSix}
};

However when I try and pass the argument int x:

int (*functionsArray[2][3])(int x) = {
  {functionOne,functionTwo,functionThree},
  {functionFour,functionFive,functionSix}
};

I get an error:

invalid conversion from 'int (*)()' to 'int (*)(int)'

Also none of these functions return an int, shouldn't I be able to declare them as void?

void(*functionsArray[2][3])(int x) = {
  {functionOne,functionTwo,functionThree},
  {functionFour,functionFive,functionSix}
};

Trying this results in an error:

 error: invalid conversion from 'int (*)()' to 'void (*)(int)'

Thanks.

Show us at least functionOne and show us at which line you get the error. — Jabberwocky
– Jabberwocky, Commented Oct 6, 2014 at 6:45
functionOne, functionTwo, ... should have all exactly the same prototype. The prototype should be same as the prototype you use for the array. — j123b567
– j123b567, Commented Oct 6, 2014 at 6:53
Have you tried reorganising the parentheses, maybe something like (void(*functionsArray(int x)))[2][3]) = ... . Secondly, and far more importantly, I would STRONGLY recommend creating a typedef for the function pointer, so the usage can be straight firmware array declaration. — racraman
– racraman, Commented Oct 6, 2014 at 6:56

paxdiablo · Accepted Answer · 2014-10-06 07:00:25Z

1

That will work fine, provided you declare the functions correctly:

#include <stdio.h>

int functionOne(int x)   { return 1; }
int functionTwo(int x)   { return 2; }
int functionThree(int x) { return 3; }
int functionFour(int x)  { return 4; }
int functionFive(int x)  { return 5; }
int functionSix(int x)   { return 6; }

int (*functionsArray[2][3])(int x) = {
    {functionOne,  functionTwo,  functionThree},
    {functionFour, functionFive, functionSix}
};

int main (void) {
    printf ("%d\n", (functionsArray[0][1])(99));
    printf ("%d\n", (functionsArray[1][2])(99));
    return 0;
}

The output of that program is 2 and 6.

It will also work if you want no return value:

#include <stdio.h>

void functionOne(int x)   { puts ("1"); }
void functionTwo(int x)   { puts ("2"); }
void functionThree(int x) { puts ("3"); }
void functionFour(int x)  { puts ("4"); }
void functionFive(int x)  { puts ("5"); }
void functionSix(int x)   { puts ("6"); }

void (*functionsArray[2][3])(int x) = {
    {functionOne,  functionTwo,  functionThree},
    {functionFour, functionFive, functionSix}
};

int main (void) {
    (functionsArray[0][1])(99);
    (functionsArray[1][2])(99);
    return 0;
}

That program also outputs 2 and 6, as expected.

It all comes down to ensuring that the function declarations match the type given in the array.

edited Oct 6, 2014 at 7:00

answered Oct 6, 2014 at 6:51

paxdiablo

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1 Comment

twwyt Over a year ago

gee I feel dumb. That fixed it up, thanks very much!

R Sahu · Accepted Answer · 2014-10-06 06:58:06Z

0

When you declare an array:

int (*functionsArray[2][3])(int x) = 
{
   {functionOne,functionTwo,functionThree},
   {functionFour,functionFive,functionSix}
};

every element of the array has to be of type int (*)(int). Otherwise, the compiler correctly reports an error.

Take a simple case:

void foo()
{
}

int (*fp)(int x) = foo;

should result in the same compiler error because you are trying to initialize a variable of type int (*)(int) using foo, whose type is void (*)().

answered Oct 6, 2014 at 6:58

R Sahu

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