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 Function returning pointers to functions (repost) 
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Anthon
#1 / 6
 Function returning pointers to functions (repost)

Hi,

Sorry guys for my latest code about functions returning pointers to
functions full with syntax errors.

I have repaired those syntax errors, and after some changes my code now
looks like:

<----------------- this is line #1
static void f1()
{
   printf(" this is function f1 \n");

Quote:
}

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

Quote:
}

static void (*getFunc(int i))
{
   switch (i)
  {
      case 1:
          return &f1;  <----------------- this is line #17
          break;
      case 2:
          return &f2; <----------------- this is line #20
          break;
   }

Quote:
}

int main(void)
{
      void (*fP)();
      fP = getFunc(2);  <----------------- this is line #28
      fP();

Quote:
}

and it does compile, but anyone who can explain what those warnings mean ?

"test.c", line 17: warning: return value type mismatch
"test.c", line 20: warning: return value type mismatch
"test.c", line 28: warning: assignment type mismatch:
        pointer to function() returning void "=" pointer to void

It even runs as expected,
[~/tst] sunray : cc  test.c
[~/tst] sunray : a.out
 this is function f2

It would be nice to understand what those warnings refer to in order to
implement a much more complicated function of that type.

BTW,  what does the line of code :
    static void (*getFunc(int i))
mean ?
Is that a function returning a pointer to a void and expecting an integer as
argument, or a pointer to a function expecting an integer as argument and
returning a normal void (no address).
And is that the same as :
   static (void *) getFunc(int i)
and if  not: what does this declaration mean ?

Thanks a lot !
Anthony



Tue, 15 Feb 2005 18:06:38 GMT  
Preben Tr?ru
#2 / 6
 Function returning pointers to functions (repost)
Quote:
> static void f2()
> {
>    printf(" this is function f2 \n");
> }

> static void (*getFunc(int i))
> {
>    switch (i)
>   {
>       case 1:
>           return &f1;  <----------------- this is line #17

The function about to return here has been declared void
(nothing returned)

Quote:
>           break;
>       case 2:
>           return &f2; <----------------- this is line #20

Do.

Quote:
>           break;
>    }
> }

> int main(void)
> {
>       void (*fP)();
>       fP = getFunc(2);  <----------------- this is line #28

Since the function you're calling has been declared void,
you are better off not assigning anything to the "return value"

HINT K&R2: 5.11 Pointers to functions



Tue, 15 Feb 2005 19:04:40 GMT  
Emmanuel Delahay
#3 / 6
 Function returning pointers to functions (repost)


Quote:
> Sorry guys for my latest code about functions returning pointers to
> functions full with syntax errors.

> I have repaired those syntax errors, and after some changes my code
> now looks like:

> <----------------- this is line #1
> static void f1()
> {
>    printf(" this is function f1 \n");
> }

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

> static void (*getFunc(int i))

No. Please take some time to read the answers you have already received.

Quote:
> BTW,  what does the line of code :
>     static void (*getFunc(int i))
> mean ?

It's weird. I'm not sure it means anything...

Quote:
> Is that a function returning a pointer to a void and expecting an
> integer as argument, or a pointer to a function expecting an integer
> as argument and returning a normal void (no address).
> And is that the same as :
>    static (void *) getFunc(int i)
> and if  not: what does this declaration mean ?

As told before, using pointers to functions is much easier with typedefs.

--
-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 21:47:44 GMT  
malcolm ka
#4 / 6
 Function returning pointers to functions (repost)

Quote:

> Hi,

> Sorry guys for my latest code about functions returning pointers to
> functions full with syntax errors.

> I have repaired those syntax errors, and after some changes my code now
> looks like:

> <----------------- this is line #1
> static void f1()

Ok but, I prefer full prototypes:
   static void f1(void)

Quote:
> {
>    printf(" this is function f1 \n");
> }

> static void f2()

Ok but, I prefer full prototypes:
  static void f2(void)

Quote:
> {
>    printf(" this is function f2 \n");
> }

> static void (*getFunc(int i))

This defines a function returning a void pointer which
it seems by C99 standard is compatible only with object
pointers -- not function pointers. I think you are realy
looking for:

   static void (*getfunc(int i))(void)

Which (I hope) turns it into a function returning a pointer to
a function with an empty argument list and returning nothing (void).

Quote:
> {
>    switch (i)
>   {
>       case 1:
>           return &f1;  <----------------- this is line #17
>           break;
>       case 2:
>           return &f2; <----------------- this is line #20
>           break;
>    }

Usually considered undesirable to leave a random result
if the switch expresion is not matched. In this case
the function runs off the end without a return value.
Some compilers will issue a warning.

Quote:
> }

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

The full prototype would be:
         void (*fP)(void);

Quote:
>       fP = getFunc(2);  <----------------- this is line #28
>       fP();
> }

> and it does compile, but anyone who can explain what those warnings mean ?

> "test.c", line 17: warning: return value type mismatch
> "test.c", line 20: warning: return value type mismatch
> "test.c", line 28: warning: assignment type mismatch:
>         pointer to function() returning void "=" pointer to void

> It even runs as expected,
> [~/tst] sunray : cc  test.c
> [~/tst] sunray : a.out
>  this is function f2

> It would be nice to understand what those warnings refer to in order to
> implement a much more complicated function of that type.

> BTW,  what does the line of code :
>     static void (*getFunc(int i))
> mean ?
> Is that a function returning a pointer to a void and expecting an integer as
> argument, or a pointer to a function expecting an integer as argument and
> returning a normal void (no address).
> And is that the same as :
>    static (void *) getFunc(int i)

I believe it is equivalent to:
      static void getFunc(int i)
If you have copied the line from somewhere I suspect you've lost a pair
of parentheses:
      static void (*getFunc(int i))()

Malcolm Kay



Tue, 15 Feb 2005 22:28:13 GMT  
Simon Bibe
#5 / 6
 Function returning pointers to functions (repost)

Quote:

> > BTW,  what does the line of code :
> >     static void (*getFunc(int i))
> > mean ?

Emmanuel Delahaye replied:

Quote:
> It's weird. I'm not sure it means anything...

It means "declare getFunc as function (argument i as int), returning
pointer to void". The outer parentheses are unnecessary.

Quote:
> > Is that a function returning a pointer to a void and expecting an
> > integer as argument,

Yes.

Quote:
> > or a pointer to a function expecting an integer
> > as argument and returning a normal void (no address).

No. To declare getFunc as pointer to function (expecting one int argument)
and returning void, you'd do this:
  void (*getFunc)(int);

Quote:
> > And is that the same as :
> >    static (void *) getFunc(int i)
> > and if  not: what does this declaration mean ?

No, this one is a syntax error as far as I can tell.

Perhaps what you really want is a
  function expecting [int] returning [pointer to function expecting [void]
returning [void]]

void f1(void) {puts("Function 1");}
void f2(void) {puts("Function 2");}

void (*getFunc(int i))(void)
{
  void (*p)(void);
  if(i == 1)
  {
    p = f1;
  }
  else
  {
    p = f2;
  }
  return p;

Quote:
}

--
Simon.


Wed, 16 Feb 2005 01:25:45 GMT  
pete
#6 / 6
 Function returning pointers to functions (repost)

Quote:

> Hi,

> Sorry guys for my latest code about functions returning pointers to
> functions full with syntax errors.

> I have repaired those syntax errors, and after some changes my code now
> looks like:

> <----------------- this is line #1
> static void f1()
> {
>    printf(" this is function f1 \n");
> }

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

/*
** I couldn't help but wonder if an array of function pointers
** might be useful to you.
*/

#include <stdio.h>

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

Quote:
}

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

Quote:
}

static void f0(void)
{
   printf(" this is neither function f1 nor f2 \n");

Quote:
}

int main(void)
{
    int index;
    static void (*array[])(void) = {f0, f1, f2};

    index = sizeof array / sizeof *array;
    while (index--) {
        array[index]();
    }
    return 0;

Quote:
}

--
 pete


Wed, 16 Feb 2005 08:08:37 GMT  
 
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