How can I call a function using a function pointer?

Question

Suppose I have these three functions:

bool A();
bool B();
bool C();

How do I call one of these functions conditionally using a function poi

Answer 1

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.

Answer 2

You can declare the function pointer as follows:

bool (funptr*)();

Which says we are declaring a function pointer to a function which does not take anything and return a bool.

Next assignment:

funptr = A;

To call the function using the function pointer:

funptr();

Answer 3

//Declare the pointer and asign it to the function
bool (*pFunc)() = A;
//Call the function A
pFunc();
//Call function B
pFunc = B;
pFunc();
//Call function C
pFunc = C;
pFunc();

Answer 4

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]))();