Function returning a function-pointer - how to prototype ?
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Richard Harnde
#1 / 5
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 Function returning a function-pointer - how to prototype ? Hi,
I have a function:
void *foo(int id, int (*new_fp)(int id, void *param)
{
int (*old_fp)(int, void*);
...
return(old_fp);
Quote: };
My problem is that foo doesn't really return a void*, it returns the
original function-pointer.
So ... how do I write that ?
TIA
--
RH
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| Tue, 08 Nov 2005 18:53:54 GMT |
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Hallvard B Furuset
#2 / 5
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Function returning a function-pointer - how to prototype ? Quote:
> void *foo(int id, int (*new_fp)(int id, void *param) > { > int (*old_fp)(int, void*); > ...
> return(old_fp);
> };
> My problem is that foo doesn't really return a void*, it returns the
> original function-pointer.
> So ... how do I write that ?
int (* <foo(int id, int (*new_fp)(int id, void *param))> )(int, void *)
without the <>'s. Note how the type matches that of old_fp if you
replace the contents of <> with a variable name.
However, this is pretty unreadable. The best way is to typedef the
function pointer type to make it more readable:
typedef int (*My_function_t)(int, void*);
My_function_t foo (int id, My_function_t new_fp)
{
My_function_t old_fp;
...
return(old_fp);
Quote: };
--
Hallvard
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| Tue, 08 Nov 2005 19:08:49 GMT |
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Richard B
#3 / 5
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Function returning a function-pointer - how to prototype ? Quote:
> I have a function: > void *foo(int id, int (*new_fp)(int id, void *param)
> {
> int (*old_fp)(int, void*);
> ...
> return(old_fp);
> };
> My problem is that foo doesn't really return a void*, it returns the
> original function-pointer.
> So ... how do I write that ?
Have it return any old function pointer type. All function pointer types
can be safely cast between one another. As long as you cast it back to
the right type before calling it, you can use any function pointer type
for passing a function pointer around.
Note that is is true only for function pointers, not for data pointers,
and certainly not between data and function pointers. You cannot use a
void * (data pointer type) to hold a function pointer.
Richard
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| Tue, 08 Nov 2005 19:30:06 GMT |
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Emmanuel Delahay
#4 / 5
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Function returning a function-pointer - how to prototype ?
Quote: > Hi, > I have a function:
> void *foo(int id, int (*new_fp)(int id, void *param)
> {
> int (*old_fp)(int, void*);
> ...
> return(old_fp);
Quote: > }; > My problem is that foo doesn't really return a void*, it returns the
> original function-pointer.
> So ... how do I write that ?
typedef int fun_f(int id, void *param);
fun_f *foo(int id, fun_f *new_fp)
{
fin_f *old_fp;
return old_fp;
Quote: }
I massively use function pointers in my code for output functions. I always
use the typedef. It makes the code readable. It's a pity that the good guys
that have designed the C-language don't use it with 'signal()'.
void (*signal(int sig, void (*func) (int sig)))(int);
eek!
My proposal:
typedef void signal_f (int sig);
signal_f *signal (int sig, signal_f *func);
Let's breath...
--
-ed- emdel at noos.fr
The C-language FAQ: http://www.*-*-*.com/ ~scs/C-faq/top.html
C-library: http://www.*-*-*.com/
FAQ de f.c.l.c : http://www.*-*-*.com/ ~rumeau/fclc/
"Clearly your code does not meet the original spec."
"You are sentenced to 30 lashes with a wet noodle."
-- Jerry Coffin in a.l.c.c++
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| Wed, 09 Nov 2005 06:31:00 GMT |
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Jack Klei
#5 / 5
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Function returning a function-pointer - how to prototype ? On Fri, 23 May 2003 11:53:54 +0100, "Richard Harnden"
Quote: > Hi, > I have a function:
> void *foo(int id, int (*new_fp)(int id, void *param)
> {
> int (*old_fp)(int, void*);
> ...
> return(old_fp);
> };
> My problem is that foo doesn't really return a void*, it returns the
> original function-pointer.
> So ... how do I write that ?
> TIA
am sure you did before posting here:
http://www.eskimo.com/~scs/C-faq/q1.22.html
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
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| Wed, 09 Nov 2005 10:44:40 GMT |
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