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 Function returning pointers to functions 
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Anthon
#1 / 7
 Function returning pointers to functions

Hi,

I need a function returning a pointer to an other function. For some reason
it does not work correctly.

BTW, what is the return type of a pointer to a function ??   Maybe void * ,
as that is the largest address any pointer fits in , or is the a special
type for pointers to functions ??

Program code looks a bit like :

extern int f1();
extern int f2();

static (void *) getFunc(int i)
{
   switch i:
  {
       case 1:
          return $f1;
          break:
      case 2:
          return $f2;
          break:
   }

Quote:
}

int main(void)
{
      int  (* fP)();       <-- correct declaration ???
      fP = getFunc(1);  <-- get pointer to f1
      *fP();                  <-- call to f1

Quote:
}

Anyone who can tell me what is wrong in this code ?

Thanks a lot !

Anthony



Tue, 15 Feb 2005 15:53:37 GMT  
Kevin Easto
#2 / 7
 Function returning pointers to functions

Quote:

> Hi,

> I need a function returning a pointer to an other function. For some reason
> it does not work correctly.

> BTW, what is the return type of a pointer to a function ??   Maybe void * ,
> as that is the largest address any pointer fits in , or is the a special
> type for pointers to functions ??

Pointers to functions need to be specific to the type of the function
that they are a pointer to.

Quote:

> Program code looks a bit like :

It'd be nice if you pasted exactly what it looked like, or at least a
simplified version that shows the problem, in future.

Quote:

> extern int f1();
> extern int f2();

> static (void *) getFunc(int i)

this should be:

static int (*getFunc(int i))()

Quote:
> {
>   switch i:
>  {
>       case 1:
>          return $f1;

huh? i think you mean

           return &f1;

Quote:
>          break:
>      case 2:
>          return $f2;

likewise here

Quote:
>          break:
>   }
> }

> int main(void)
> {
>      int  (* fP)();       <-- correct declaration ???

Yes.

Quote:
>      fP = getFunc(1);  <-- get pointer to f1
>      *fP();                  <-- call to f1

Just fP();

Quote:
> }

> Anyone who can tell me what is wrong in this code ?

> Thanks a lot !

> Anthony

        - Kevin.


Tue, 15 Feb 2005 16:27:22 GMT  
Eric G. Mille
#3 / 7
 Function returning pointers to functions

Quote:

> Hi,

> I need a function returning a pointer to an other function. For some reason
> it does not work correctly.

> BTW, what is the return type of a pointer to a function ??   Maybe void * ,
> as that is the largest address any pointer fits in , or is the a special
> type for pointers to functions ??

You can't use void * for a pointer to function, only for objects..

Quote:
> Program code looks a bit like :

> extern int f1();
> extern int f2();

> static (void *) getFunc(int i)

It's a bit weird to define functions that return pointers to function
without using typedefs, but you'd want:

static int ((*getFunc (int i))()) {

Easier to:

typedef int (*int_func)();

static int_func getFunc (int i) { ... }

Quote:
> {
>    switch i:

That ain't C --> switch (int_expression) { ... }

Quote:
>   {
>        case 1:
>           return $f1;

                   ^ Not allowed in ISO C
Quote:
>           break:

                 ^ s/:/;/
Quote:
>       case 2:
>           return $f2;

                   ^ Remove '$'
Quote:
>           break:

                 ^ s/:/;/
And add:
        default:
            return 0;

Quote:
>    }
> }

> int main(void)
> {
>       int  (* fP)();       <-- correct declaration ???

Yes.

Quote:
>       fP = getFunc(1);  <-- get pointer to f1

Yes.

Quote:
>       *fP();                  <-- call to f1

No.  Either:

        (*fP)();

or just:

         fP();



Tue, 15 Feb 2005 17:05:55 GMT  
Emmanuel Delahay
#4 / 7
 Function returning pointers to functions


Quote:
> I need a function returning a pointer to an other function. For some
> reason it does not work correctly.

Let's see the code...

Quote:
> BTW, what is the return type of a pointer to a function ??   Maybe

Depences on the function!

Quote:
> void * , as that is the largest address any pointer fits in , or is

void* is fine for objects. Certainely not (portably) for functions.

Quote:
> the a special type for pointers to functions ??

> Program code looks a bit like :

> extern int f1();
> extern int f2();

It appears that the type of the function is

typedef int func_f();

Quote:

> static (void *) getFunc(int i)

weird!

static func_f *getFunc(int i)

Quote:
> {
>    switch i:

Huh! Why not compiling the code before posting it...

    switch (i)

Quote:
>   {
>        case 1:
>           return $f1;

Assuming you meant & instead of $, it's useless.

           return f1;

Quote:
>           break:

           break;

Note that due to the return, this break is never reached.

Quote:
>       case 2:
>           return $f2;

ditto.

Quote:
>           break:

ditto.

Quote:
>    }

What if i is not 1 or 2 ?

Quote:
> }

> int main(void)
> {
>       int  (* fP)();       <-- correct declaration ???

Yes, or

   func_f *fP;

Quote:
>       fP = getFunc(1);  <-- get pointer to f1
>       *fP();                  <-- call to f1

or
        fP();

Note that comments are supposed to be wrapped by /*  */

If you meant that f1 and f2 have no parameters, say it:

extern int f1(void);
extern int f2(void);
typedef int func_f(void);

Note also that once the typedef is defined, the prototypes can be defined
as follow:

extern func_f f1;
extern func_f f2;

Quote:
> }

> Anyone who can tell me what is wrong in this code ?

The returned type was weird, not to mention the typos...

Try that:

#ifndef ED_FUN_H
#define ED_FUN_H

/* fun.h */

typedef int func_f (void);
func_f f1;
func_f f2;

#endif /* ED_FUN_H */
/* GUARD (c) AETA 2000 Dec 16 2000 Ver. 1.2 */

/* fun.c */

#include "fun.h"
#include <stdio.h>

int f1 (void)
{
   puts ("f1");
   return 0;

Quote:
}

int f2 (void)
{
   puts ("f2");
   return 0;

Quote:
}

/* main.c */

#include "fun.h"

#include <stddef.h>

static func_f *getFunc (int i)
{
   func_f *pf;

   switch (i)
   {
   case 1:
      pf = f1;
      break;

   case 2:
      pf = f2;
      break;

   default:
      pf = NULL;
   }
   return pf;

Quote:
}

int main (void)
{
   func_f *fP;
   int i;

   for (i = 0; i < 3; i++)
   {
      fP = getFunc (i);
      if (fP)
      {
         fP ();
      }
   }
   return 0;

Quote:
}

--
-ed- emdel at noos.fr
FAQ de f.c.l.c : http://www.isty-info.uvsq.fr/~rumeau/fclc/
C-library: http://www.dinkumware.com/htm_cl/index.html
"Mal nommer les choses c'est ajouter du malheur au monde."
Albert Camus.


Tue, 15 Feb 2005 17:11:10 GMT  
Carl McGuir
#5 / 7
 Function returning pointers to functions


Quote:
> Hi,

> I need a function returning a pointer to an other function. For some
reason
> it does not work correctly.

> BTW, what is the return type of a pointer to a function ??   Maybe void *
,
> as that is the largest address any pointer fits in , or is the a special
> type for pointers to functions ??

> Program code looks a bit like :

> extern int f1();
> extern int f2();

> static (void *) getFunc(int i)
> {
>    switch i:
>   {
>        case 1:
>           return $f1;
>           break:
>       case 2:
>           return $f2;
>           break:
>    }
> }

> int main(void)
> {
>       int  (* fP)();       <-- correct declaration ???
>       fP = getFunc(1);  <-- get pointer to f1
>       *fP();                  <-- call to f1
> }

> Anyone who can tell me what is wrong in this code ?

> Thanks a lot !

> Anthony

This is my best shot without rewriting the code:

#include <stdio.h>

typedef void (*pChecksub)(); /* Adapted from Pete Becker. */

static void f1()
{
   printf(" this is function f1 \n");

Quote:
}

static void f2()
{
   printf(" this is function f2 \n");

Quote:
}

static pChecksub getFunc(int i)
{
  switch (i)
  {
      case 1:
          return &f1;
          break;
      case 2:
          return &f2;
          break;
   }

Quote:
}

int main(void)
{
      void (*fP)();

      fP = getFunc(2);
      fP();
      return 0;

- Show quoted text -

Quote:
}



Tue, 15 Feb 2005 19:02:55 GMT  
Dave Near
#6 / 7
 Function returning pointers to functions
On Fri, 30 Aug 2002 09:53:37 +0200, Anthony said:

Quote:
> I need a function returning a pointer to an other function. For
> some reason it does not work correctly.

Sure it does, you're just not doing it right.

return_type (*function_name(arg_list))(arg list);

So, for example, a function called foo taking a char as an
argument which returns a pointer to
a function which takes two integer arguments and returns int
would be declared as...
int (*foo(char))(int, int);

Quote:
> BTW, what is the return type of a pointer to a function ??

A pointer to a function points to a function. And functions can
return anything. So pointer to functions point at functions which
return anything.

Quote:
> Maybe void * ,
> as that is the largest address any pointer fits in , or is the a
> special type for pointers to functions ??

No - pointers to functions can usually look just like functions,
thankfully.

Here's a code snippet which might help...

#include <stdio.h>
#include <stdlib.h>

/* For ease of reading, typedef a function accepting 2 int
 * arguments and returning int
 */
typedef int foo(int a, int b);

/* Declare two functions which will do a particularly complicated
 * calculation
 */
foo add;
foo subtract;
/* We could also leave this as
 * int add(int a, int b);
 * int subtract(int a, int b);
 */

/* Define a function which accepts an int, and returns a pointer
 * to a function accepting 2 int arguments and returning int
 */
foo *grommit(int i)
{
  switch(i)
  {
    case 0:
      return add;
      break;
    case 1:
      return subtract;
      break;
    default:
      return NULL;
  }

Quote:
}

int main(void)
{
  foo *func; /* Declare pointer to our function type */

  func = grommit(0);
  printf("4 + 17 = %d\n", func(4, 17));
  func = grommit(1);
  printf("17 - 4 = %d\n", func(17, 4));

  return 0;

Quote:
}

int add(int a, int b)
{
  return a + b;

Quote:
}

int subtract(int a, int b)
{
  return a - b;

Quote:
}

Hope this helps,
Dave.

--
           David Neary,
     E-Mail: bolsh at gimp dot org
CV: http://www.redbrick.dcu.ie/~bolsh/CV/CV.html



Tue, 15 Feb 2005 20:35:55 GMT  
Daniel Romi
#7 / 7
 Function returning pointers to functions

Quote:



> > Hi,

> > I need a function returning a pointer to an other function. For some
>  reason
> > it does not work correctly.

> > BTW, what is the return type of a pointer to a function ??   Maybe void *
>  ,
> > as that is the largest address any pointer fits in , or is the a special
> > type for pointers to functions ??

> > Program code looks a bit like :

> > extern int f1();
> > extern int f2();

> > static (void *) getFunc(int i)
> > {
> >    switch i:
> >   {
> >        case 1:
> >           return $f1;
> >           break:
> >       case 2:
> >           return $f2;
> >           break:
> >    }
> > }

> > int main(void)
> > {
> >       int  (* fP)();       <-- correct declaration ???
> >       fP = getFunc(1);  <-- get pointer to f1
> >       *fP();                  <-- call to f1
> > }

> > Anyone who can tell me what is wrong in this code ?

> > Thanks a lot !

> > Anthony

> This is my best shot without rewriting the code:

> #include <stdio.h>

> typedef void (*pChecksub)(); /* Adapted from Pete Becker. */

> static void f1()
> {
>    printf(" this is function f1 \n");
> }

> static void f2()
> {
>    printf(" this is function f2 \n");
> }

> static pChecksub getFunc(int i)
> {
>   switch (i)
>   {
>       case 1:
>           return &f1;
>           break;
>       case 2:
>           return &f2;
>           break;
>    }
> }

> int main(void)
> {
>       void (*fP)();

>       fP = getFunc(2);
>       fP();
>       return 0;
> }

Give this a try (this is done under GCC 3.1 compiler):

void
f1(int a)
{
   printf("F1: %d\n", a);

Quote:
}

void (*f2())(int a)  // Here's the trick, defining just like a function pointer
{
   printf("F2\n");
   return(f1);       // returns the f1 function

Quote:
}

int
main(int argc, char **argv)
{
   void (*f0)(int a);

   f0=f2();     // gets a pointer to the f1 function
   f0(111111);  // calls the f1 function

Quote:
}

The syntax is cool for f2()--

Dan



Wed, 16 Feb 2005 02:04:30 GMT  
 
 Page 1 of 1
  [ 7 post ] 

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