Hi, I have an function pointer, example:

typedef bool (*MkErrorResetOnExitF) (MK_EXTENSION_ARGS, MK_OBJ const fmtobj)

and I want to use the *return* type of the function pointer in a macro,
example:

#define MkWalkErrExtReturn(_err, _fct, ...) ({ \
typeof(_fct) ret = 0; \
if (_err->mkerrext) { \

MK_EXT ext, head=MkEXT(_err->mkerrext); \

DL_FOREACH(head,ext) { \

MK_ERREXT errext = MkErrExt(ext); \

if (errext && errext->_fct) { \

ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__); \

if (ret) break; \

} \

} \

} \

ret; \

})

used in:

> if (MkWalkErrExtReturn(err, fErrorResetOnExit, fmtobj)) return;

the problem is that:

> typeof(_fct) = typeof(fErrorResetOnExit)

return the type of the entire function pointer and *not* the type of the
function pointer return.

question: it is possible to get in C (gcc) the return typ of a function
pointer with some "typeof" extension?

mfg

Am 04.06.2022 um 10:41 schrieb aotto1968:

> #define MkWalkErrExtReturn(_err, _fct, ...) ({                 \
>   typeof(_fct) ret = 0;                        \
>   if (_err->mkerrext) {                                        \
>     MK_EXT ext, head=MkEXT(_err->mkerrext);                    \
>     DL_FOREACH(head,ext) {                                     \
>       MK_ERREXT errext = MkErrExt(ext);                        \
>       if (errext && errext->_fct) {                            \
>         ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__);   \
>         if (ret) break;                                        \
>       }                                                        \
>     }                                                          \
>   }                                                            \
>   ret;                                                         \
> })

Disgusting.

aotto1968 <aotto1968@t-online.de> writes:
>question: it is possible to get in C (gcc) the return typ of a function
>pointer with some "typeof" extension?

#include <stdio.h>

int main( void )
{ char f(); printf( "%d\n", ( int )( sizeof( typeof( f() )))); }

On Saturday, June 4, 2022 at 5:42:06 AM UTC-3, aotto1968 wrote:
> Hi, I have an function pointer, example:
>
> typedef bool (*MkErrorResetOnExitF) (MK_EXTENSION_ARGS, MK_OBJ const fmtobj)
>
> and I want to use the *return* type of the function pointer in a macro,
> example:
>
> #define MkWalkErrExtReturn(_err, _fct, ...) ({ \
> typeof(_fct) ret = 0; \
> if (_err->mkerrext) { \
>
> MK_EXT ext, head=MkEXT(_err->mkerrext); \
>
> DL_FOREACH(head,ext) { \
>
> MK_ERREXT errext = MkErrExt(ext); \
>
> if (errext && errext->_fct) { \
>
> ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__); \
>
> if (ret) break; \
>
> } \
>
> } \
>
> } \
>
> ret; \
>
> })
>
> used in:
>
> > if (MkWalkErrExtReturn(err, fErrorResetOnExit, fmtobj)) return;
>
> the problem is that:
>
> > typeof(_fct) = typeof(fErrorResetOnExit)
>
> return the type of the entire function pointer and *not* the type of the
> function pointer return.
>
> question: it is possible to get in C (gcc) the return typ of a function
> pointer with some "typeof" extension?
>
>
> mfg

You need to create a expression calling the function. (Maybe dummy arguments)

#include <stdio.h>
double f(int a) { return 1; }

int main()
{ printf("%d\n", (int)sizeof(typeof(f)));
printf("%d\n", (int)sizeof(typeof(f(0))));
} this program prints using gcc:
1 8

I was not expecting 1 for in the first line.. but sizeof function pointer.
I was expecting 8 for the second line.

On Saturday, June 4, 2022 at 10:06:57 AM UTC-3, Thiago Adams wrote:
> On Saturday, June 4, 2022 at 5:42:06 AM UTC-3, aotto1968 wrote:
> > Hi, I have an function pointer, example:
> >
> > typedef bool (*MkErrorResetOnExitF) (MK_EXTENSION_ARGS, MK_OBJ const fmtobj)
> >
> > and I want to use the *return* type of the function pointer in a macro,
> > example:
> >
> > #define MkWalkErrExtReturn(_err, _fct, ...) ({ \
> > typeof(_fct) ret = 0; \
> > if (_err->mkerrext) { \
> >
> > MK_EXT ext, head=MkEXT(_err->mkerrext); \
> >
> > DL_FOREACH(head,ext) { \
> >
> > MK_ERREXT errext = MkErrExt(ext); \
> >
> > if (errext && errext->_fct) { \
> >
> > ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__); \
> >
> > if (ret) break; \
> >
> > } \
> >
> > } \
> >
> > } \
> >
> > ret; \
> >
> > })
> >
> > used in:
> >
> > > if (MkWalkErrExtReturn(err, fErrorResetOnExit, fmtobj)) return;
> >
> > the problem is that:
> >
> > > typeof(_fct) = typeof(fErrorResetOnExit)
> >
> > return the type of the entire function pointer and *not* the type of the
> > function pointer return.
> >
> > question: it is possible to get in C (gcc) the return typ of a function
> > pointer with some "typeof" extension?
> >
> >
> > mfg
> You need to create a expression calling the function. (Maybe dummy arguments)
>
> #include <stdio.h>
> double f(int a) { return 1; }
>
> int main()
> {
> printf("%d\n", (int)sizeof(typeof(f)));
> printf("%d\n", (int)sizeof(typeof(f(0))));
> }
> this program prints using gcc:
> 1
> 8
>
> I was not expecting 1 for in the first line.. but sizeof function pointer.
> I was expecting 8 for the second line.

but function pointer it prints 8 ..so it makes sense.
printf("%d", (int)sizeof(void(*)()));
(MS compiler gives an error sizeof function)

On 2022-06-04, Stefan Ram <ram@zedat.fu-berlin.de> wrote:
> aotto1968 <aotto1968@t-online.de> writes:
>>question: it is possible to get in C (gcc) the return typ of a function
>>pointer with some "typeof" extension?
>
> #include <stdio.h>
>
> int main( void )
> { char f(); printf( "%d\n", ( int )( sizeof( typeof( f() )))); }

I'm not sure what the aim is here, but, obviously, the above
example is greatly improved by

printf("%d\n", (int) sizeof f());

--
TXR Programming Language: http://nongnu.org/txr
Cygnal: Cygwin Native Application Library: http://kylheku.com/cygnal

Kaz Kylheku <480-992-1380@kylheku.com> writes:

> On 2022-06-04, Stefan Ram <ram@zedat.fu-berlin.de> wrote:
>> aotto1968 <aotto1968@t-online.de> writes:
>>>question: it is possible to get in C (gcc) the return typ of a function
>>>pointer with some "typeof" extension?
>>
>> #include <stdio.h>
>>
>> int main( void )
>> { char f(); printf( "%d\n", ( int )( sizeof( typeof( f() )))); }
>
> I'm not sure what the aim is here,

I think the aim was to show that typeof can be applied to a function
call expression, rather than to a bare function designator, and SR
wanted to confirm the type using the size -- 1 being very obviously the
size of a char.

Unfortunately that confirmation does not work because function types
have no "natural" size and gcc seems to just pick 1 (and why not?) so
had the OP changed the source to see what size typeof(f) is they would
probably be confused by the result also being 1.

For the OP: typeof (f) is a function type -- exactly the type of f.
It's not the type of pointer to f (for that you could use typeof (&f))
nor is it the return type of f (for that use typeof (f()), though in
your case you will have to provide dummy arguments).

Function types, like array types, can be useful, but they often confuse
beginners who have been told (erroneously) that C has no arrays and no
functions -- that both are "really" pointers. They are not!

--
Ben.

aotto1968 <aotto1968@t-online.de> writes:

> Hi, I have an function pointer, example:
>
> typedef bool (*MkErrorResetOnExitF) (MK_EXTENSION_ARGS, MK_OBJ const fmtobj)
>
> and I want to use the *return* type of the function pointer in a
> macro, example:
>
> #define MkWalkErrExtReturn(_err, _fct, ...) ({ \
> typeof(_fct) ret = 0; \
> if (_err->mkerrext) { \
> MK_EXT ext, head=MkEXT(_err->mkerrext); \
> DL_FOREACH(head,ext) { \
> MK_ERREXT errext = MkErrExt(ext); \
> if (errext && errext->_fct) { \
> ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__); \
> if (ret) break; \
> } \
> } \
> } \
> ret; \
> })
>
> used in:
>
>> if (MkWalkErrExtReturn(err, fErrorResetOnExit, fmtobj)) return;
>
> the problem is that:
>
>> typeof(_fct) = typeof(fErrorResetOnExit)
>
> return the type of the entire function pointer and *not* the type
> of the function pointer return.
>
> question: it is possible to get in C (gcc) the return typ of a
> function pointer with some "typeof" extension?

The pattern to follow is for the 'typeof' expression to be the
same as an expression that is to be assigned to the variable in
question. Hence,

#define MkWalkErrExtReturn(_err, _fct, ...) ({ \
typeof( _fct (MK_EXTENSION_CALL, __VA_ARGS__) ) ret = 0; \
if (_err->mkerrext) { \
MK_EXT ext, head=MkEXT(_err->mkerrext); \
DL_FOREACH(head,ext) { \
MK_ERREXT errext = MkErrExt(ext); \
if (errext && errext->_fct) { \
ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__); \
if (ret) break; \
} \
} \
} \
ret; \
})

If you try this modified definition I expect you will find that
it does what you want.

On 05.06.22 16:04, Tim Rentsch wrote:
…
>
> The pattern to follow is for the 'typeof' expression to be the
> same as an expression that is to be assigned to the variable in
> question. Hence,
>
> #define MkWalkErrExtReturn(_err, _fct, ...) ({ \
> typeof( _fct (MK_EXTENSION_CALL, __VA_ARGS__) ) ret = 0; \
> if (_err->mkerrext) { \
> MK_EXT ext, head=MkEXT(_err->mkerrext); \
> DL_FOREACH(head,ext) { \
> MK_ERREXT errext = MkErrExt(ext); \
> if (errext && errext->_fct) { \
> ret = _fct (MK_EXTENSION_CALL, __VA_ARGS__); \
> if (ret) break; \
> } \
> } \
> } \
> ret; \
> })
>
> If you try this modified definition I expect you will find that
> it does what you want.

This looks good, thanks.