Function pointers?

Function pointers?

[The Devils Jester]

I can pass pointers as arguments to a function without issue unless it
is a function pointer.  When I try and pass a function pointer, it
gets jumbled during the pass.  What am I doing wrong?

Here is an example code that prints the value of the pointer before
its passed, and in the function its passed to.  Both values are
different, does anyone know why?

(I am running OS X 10.8.2)

#include <stdio.h>
#include <ffi.h>

int test_a(void*ptr)
{ return 0; }

int test_b(void*ptr)
{
    printf("Address of passed ptr: %p\n", ptr );
    return 0;
}

int main()
{
    ffi_cif cif;
    ffi_type *args[1];
    void *values[1];
    int rc;

    args[0] = &ffi_type_pointer;
    values[0] = &test_a;

    if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 1, &ffi_type_uint, args) == FFI_OK)
     {
       printf("Address of ptr to pass: %p\n", values[0] );
       ffi_call(&cif, test_b, &rc, values);
     }
     return 0;
}

RE: Function pointers?

[Nathan Rossi]

> -----Original Message-----
> From: libffi-discuss-owner@sourceware.org [mailto:libffi-discuss-
> owner@sourceware.org] On Behalf Of The Devils Jester
> Sent: Tuesday, March 05, 2013 4:06 PM
> To: libffi-discuss@sourceware.org
> Subject: Function pointers?
> 
> I can pass pointers as arguments to a function without issue unless it
> is a function pointer.  When I try and pass a function pointer, it
> gets jumbled during the pass.  What am I doing wrong?
> 
> Here is an example code that prints the value of the pointer before
> its passed, and in the function its passed to.  Both values are
> different, does anyone know why?
> 
> (I am running OS X 10.8.2)
> 
> #include <stdio.h>
> #include <ffi.h>
> 
> int test_a(void*ptr)
> { return 0; }
> 
> int test_b(void*ptr)
> {
>     printf("Address of passed ptr: %p\n", ptr );
>     return 0;
> }
> 
> int main()
> {
>     ffi_cif cif;
>     ffi_type *args[1];
>     void *values[1];
>     int rc;
> 
>     args[0] = &ffi_type_pointer;
>     values[0] = &test_a;
> 
>     if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 1, &ffi_type_uint, args) == FFI_OK)
>      {
>        printf("Address of ptr to pass: %p\n", values[0] );
>        ffi_call(&cif, test_b, &rc, values);
>      }
>      return 0;
> }

Hi,

The 'values' array you are passing to ffi_call needs to contain pointers to the associated data (in this case the data is the function pointer). Because of the C syntax "test_a == &test_a", and the & operator is ignored, and thus you are passing the address of the function as the pointer to the data.

You will need to store the function pointer on the stack/heap before passing it to the ffi_call. E.g.
..
void* arg_values[1];
arg_values[0] = (void*)test_a;
values[0] = &(arg_values[0]);
...

Regards,
Nathan

Re: Function pointers?

[The Devils Jester]

I am not sure how that makes a difference.  With the above change, the
value stored in values[0] and the actual ptr printed in test_b are
still different.  Am I missing something obvious here?

On Tue, Mar 5, 2013 at 1:46 AM, Nathan Rossi <nathan.rossi@xilinx.com> wrote:
>> -----Original Message-----
>> From: libffi-discuss-owner@sourceware.org [mailto:libffi-discuss-
>> owner@sourceware.org] On Behalf Of The Devils Jester
>> Sent: Tuesday, March 05, 2013 4:06 PM
>> To: libffi-discuss@sourceware.org
>> Subject: Function pointers?
>>
>> I can pass pointers as arguments to a function without issue unless it
>> is a function pointer.  When I try and pass a function pointer, it
>> gets jumbled during the pass.  What am I doing wrong?
>>
>> Here is an example code that prints the value of the pointer before
>> its passed, and in the function its passed to.  Both values are
>> different, does anyone know why?
>>
>> (I am running OS X 10.8.2)
>>
>> #include <stdio.h>
>> #include <ffi.h>
>>
>> int test_a(void*ptr)
>> { return 0; }
>>
>> int test_b(void*ptr)
>> {
>>     printf("Address of passed ptr: %p\n", ptr );
>>     return 0;
>> }
>>
>> int main()
>> {
>>     ffi_cif cif;
>>     ffi_type *args[1];
>>     void *values[1];
>>     int rc;
>>
>>     args[0] = &ffi_type_pointer;
>>     values[0] = &test_a;
>>
>>     if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 1, &ffi_type_uint, args) == FFI_OK)
>>      {
>>        printf("Address of ptr to pass: %p\n", values[0] );
>>        ffi_call(&cif, test_b, &rc, values);
>>      }
>>      return 0;
>> }
>
> Hi,
>
> The 'values' array you are passing to ffi_call needs to contain pointers to the associated data (in this case the data is the function pointer). Because of the C syntax "test_a == &test_a", and the & operator is ignored, and thus you are passing the address of the function as the pointer to the data.
>
> You will need to store the function pointer on the stack/heap before passing it to the ffi_call. E.g.
> ..
> void* arg_values[1];
> arg_values[0] = (void*)test_a;
> values[0] = &(arg_values[0]);
> ...
>
> Regards,
> Nathan
>
>

Re: Function pointers?

[The Devils Jester]

I think it works now with the recommended changes. I was just under
the impression that values[0] and the value of ptr in test_b should
agree.  Since I am actually able to call the function with the ptr
passed, I take it that that was a false assumption?

On Tue, Mar 5, 2013 at 6:44 AM, The Devils Jester
<thedevilsjester@gmail.com> wrote:
> I am not sure how that makes a difference.  With the above change, the
> value stored in values[0] and the actual ptr printed in test_b are
> still different.  Am I missing something obvious here?
>
> On Tue, Mar 5, 2013 at 1:46 AM, Nathan Rossi <nathan.rossi@xilinx.com> wrote:
>>> -----Original Message-----
>>> From: libffi-discuss-owner@sourceware.org [mailto:libffi-discuss-
>>> owner@sourceware.org] On Behalf Of The Devils Jester
>>> Sent: Tuesday, March 05, 2013 4:06 PM
>>> To: libffi-discuss@sourceware.org
>>> Subject: Function pointers?
>>>
>>> I can pass pointers as arguments to a function without issue unless it
>>> is a function pointer.  When I try and pass a function pointer, it
>>> gets jumbled during the pass.  What am I doing wrong?
>>>
>>> Here is an example code that prints the value of the pointer before
>>> its passed, and in the function its passed to.  Both values are
>>> different, does anyone know why?
>>>
>>> (I am running OS X 10.8.2)
>>>
>>> #include <stdio.h>
>>> #include <ffi.h>
>>>
>>> int test_a(void*ptr)
>>> { return 0; }
>>>
>>> int test_b(void*ptr)
>>> {
>>>     printf("Address of passed ptr: %p\n", ptr );
>>>     return 0;
>>> }
>>>
>>> int main()
>>> {
>>>     ffi_cif cif;
>>>     ffi_type *args[1];
>>>     void *values[1];
>>>     int rc;
>>>
>>>     args[0] = &ffi_type_pointer;
>>>     values[0] = &test_a;
>>>
>>>     if (ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 1, &ffi_type_uint, args) == FFI_OK)
>>>      {
>>>        printf("Address of ptr to pass: %p\n", values[0] );
>>>        ffi_call(&cif, test_b, &rc, values);
>>>      }
>>>      return 0;
>>> }
>>
>> Hi,
>>
>> The 'values' array you are passing to ffi_call needs to contain pointers to the associated data (in this case the data is the function pointer). Because of the C syntax "test_a == &test_a", and the & operator is ignored, and thus you are passing the address of the function as the pointer to the data.
>>
>> You will need to store the function pointer on the stack/heap before passing it to the ffi_call. E.g.
>> ..
>> void* arg_values[1];
>> arg_values[0] = (void*)test_a;
>> values[0] = &(arg_values[0]);
>> ...
>>
>> Regards,
>> Nathan
>>
>>