calling a function from a pointer

spx2 · 05-25-2006, 02:03 PM

i have done this

Code:

#include <iostream.h>
void *a[20];
int adunare(){cout<<"AsdA";

return 5;};
void main()
{
 a[0]=adunare;
 cout <<a[0];
 a[0]();

 cin>>int();
 }

i am trying to make an array of pointers to functions
which is in this case a.
i am trying to call a[0] which is actually adunare
by calling a[0] like this : a[0]();
why is it not working and how should it have been done ?

paulsm4 · 05-25-2006, 02:08 PM

This should help:
http://www.newty.de/fpt/fpt.html

Here's a simple example (this syntax applicable to either C or C++):

Code:

#include <stdio.h>

typedef int (*fnptr)(int i);

int
a(int i)
{
  printf ("a(%d)= %d...\n", i, i * 2);
  return (i*2);
}

int
b(int i)
{
  printf ("b(%d)= %d...\n", i, i * 10);
  return (i*10);
}

int
main ()
{
   fnptr p = a;
   (*p)(3);

   p = b;
   (*p)(3);
   return 0;
}

The interesting thing about this example is (if you happen to be coding in C++), it really would make more sense if "a" and "b" were each classes (instead of function pointers).

spx2 · 05-25-2006, 02:47 PM

dear paulsm4 the question is if one can call p(3).
a(3),b(3) we can obviously call.
thank you!
awaiting your answer

paulsm4 · 05-25-2006, 04:52 PM

Sorry -

Code:

int
main ()
{
   fnptr p = a;
   (*p)(3);

   p = b;
   (*p)(3);
   return 0;
}

Is that what you were looking for?

Quote:

OUTPUT:
a(3)= 6...
b(3)= 30...

Again, this is a great tutorial:
http://www.newty.de/fpt/fpt.html