function * = void *

[Bill Pursell]

The following code generates a compiler warning
when compiled with gcc -pedantic:

typedef (*FUNC)(int);
FUNC f;

void *
get_f(void)
{
return &f;
}

int main(void)
{
FUNC fp;
fp = (FUNC)get_f();
return 0;
}
~

The warning is:
a.c: In function `main':
a.c:14: warning: ISO C forbids conversion of object pointer to function
pointer type

If I remove the cast, the warning is:
a.c: In function `main':
a.c:14: warning: ISO C forbids assignment between function pointer and
`void *'
Surely I can use this construction? Can I safely ignore the warning?
I'd prefer to keep -pedantic in my CFLAGS---how can I
correctly suppress the warning?

Or am I doing something wrong?

[jaysome]

On Thu, 22 Jun 2006 07:42:20 GMT, Simon Biber <ne**@ralmin.cc > wrote:

jaysome wrote:

On Wed, 21 Jun 2006 18:22:35 -0000, go***********@b urditt.org (Gordon
Burditt) wrote:

There is no guarantee that a function pointer can FIT in a void *
without truncating it.

If that's the case, how should I printf() the value of a function
pointer?

Byte by byte.

#include <stdio.h>
typedef void (*func_ptr_type )(void);
int main(void)
{
func_ptr_type func_ptr = NULL;
printf("val = %p\n", (void *)func_ptr);
return 0;
}

Try this:

#include <stdio.h>
#include <limits.h>
typedef int (*func_ptr_type )(void);
int main(void)
{
func_ptr_type func_ptr = main;
size_t i;
printf("val = 0x");
for(i = 0; i < sizeof func_ptr; i++)
{
printf("%0*X",
(CHAR_BIT + 3) / 4,
(unsigned)((uns igned char *)&func_ptr)[i]);
}
printf("\n");
return 0;
}

This compiles cleanly and prints val = 0x50104000
on my system.

It prints val = 0x00104000 on my system, which has a 64-bit dual-core
CPU. I'll try running the program on our DEC Alpha tomorrow and report
back the results.

The "%p" conversion specifier requires a (void*), but if I convert a
function pointer to (void*) and that gets "truncated" , isn't there a
serious flaw in the standard?

Not really. The %p specifier is only for printing void* pointers. It so
happens that all object pointers can be converted to unique void* values.

So in other words, you can printf() the value of an object pointer,
but you can't necessarily printf() the value of a a function pointer.

Someday, I'll read the standard to see if what you're saying is true.
In the meantime, I'm gonna perform a test on every platform/compiler I
work with. I think I'll be busy for a couple of weeks, if not a month.

--
jay

[Keith Thompson]

jaysome <ja*****@spamco p.net> writes:

On Thu, 22 Jun 2006 07:42:20 GMT, Simon Biber <ne**@ralmin.cc > wrote: [...]

Not really. The %p specifier is only for printing void* pointers. It so
happens that all object pointers can be converted to unique void* values.

So in other words, you can printf() the value of an object pointer,
but you can't necessarily printf() the value of a a function pointer.

Right.
Someday, I'll read the standard to see if what you're saying is true.
Today would be a good day.
In the meantime, I'm gonna perform a test on every platform/compiler I
work with. I think I'll be busy for a couple of weeks, if not a month.

Sure, have fun. It's entirely possible that every platform you try
will allow function pointers to be converted to void*, but that won't
really tell you anything about the standard.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Al Balmer]

On Thu, 22 Jun 2006 01:08:43 -0700, jaysome <ja*****@spamco p.net>
wrote:

Someday, I'll read the standard to see if what you're saying is true.
In the meantime, I'm gonna perform a test on every platform/compiler I
work with. I think I'll be busy for a couple of weeks, if not a month.

It would be quicker to read the standard.

--
Al Balmer
Sun City, AZ