How can I call a function using a function pointer?

Question

You can do the following: Suppose you have your A,B & C function as the following:

bool A()
{
   .....
}

bool B()
{
   .....
}

bool C()
{

 .....
}

Now at some other function, say at main:

int main()
{
  bool (*choice) ();

  // now if there is if-else statement for making "choice" to 
  // point at a particular function then proceed as following

  if ( x == 1 )
   choice = A;

  else if ( x == 2 )
   choice = B;


  else
   choice = C;

if(choice())
 printf("Success\n");

else
 printf("Failure\n");

.........
  .........
  }

Remember this is one example for function pointer. there are several other method and for which you have to learn function pointer clearly.

I think your question has already been answered more than adequately, but it might be useful to point out explicitly that given a function pointer

void (*pf)(int foo, int bar);

the two calls

pf(1, 0);
(*pf)(1, 0);

are exactly equivalent in every way by definition. The choice of which to use is up to you, although it's a good idea to be consistent. For a long time, I preferred (*pf)(1, 0) because it seemed to me that it better reflected the type of pf, however in the last few years I've switched to pf(1, 0).

Declare your function pointer like this:

bool (*f)();
f = A;
f();

Initially define a function pointer array which takes a void and returns a void.

Assuming that your function is taking a void and returning a void.

typedef void (*func_ptr)(void);

Now you can use this to create function pointer variables of such functions.

Like below:

func_ptr array_of_fun_ptr[3];

Now store the address of your functions in the three variables.

array_of_fun_ptr[0]= &A;
array_of_fun_ptr[1]= &B;
array_of_fun_ptr[2]= &C;

Now you can call these functions using function pointers as below:

some_a=(*(array_of_fun_ptr[0]))();
some_b=(*(array_of_fun_ptr[1]))();
some_c=(*(array_of_fun_ptr[2]))();