Re: strange pointer behavior


>
> Why do you not use %p which works on all platforms instead of %x which
> potentially invokes undefined behaviour?
i didn't know that one. thanks.

i also think i get the rest of your explanation. thanks for your reply.

kind regards,
Bruno.





>
>> array, ptrArray[0], element, *element);
>
> Note that you have a function with variable argument list: If you
> want a specific pointer type (here: void*), cast.
>
>>
>> printf("array[9] = %d, (*element)[9] = %d, element[9] = %d\n",
>> array[9], (*element)[9], element[9]);
>
> Note: element[9] is a pointer and points to storage you do not
> own. Bang.
>>
>> return 0;
>> }
>>
>> now according to the print statements, the different pointers all point
>> to
>> the same thing, even though 'element' is dereferenced at one place, and
>> not
>> dereferenced at another place. if i try to print a value from the array,
>> i
>> get 2 different results.
>> ??
>
> For reference:
>
> #include <stdio.h>
>
> typedef int t_Array[10];
>
> int main (void)
> {
> t_Array array;
> array[0] = 123;
> array[9] = 321;
>
> t_Array *ptrArray[] = {&array};
> t_Array *element = ptrArray[0];
>
> printf("array = %10p, ptrArray[0] = %10p, element = %10p, "
> "*element = %10p\n", (void *)array, (void *)ptrArray[0],
> (void *)element, (void *)*element);
>
> printf("array[9] = %d, (*element)[9] = %d, element[9] = %p\n",
> array[9], (*element)[9], (void *)element[9]);
>
> return 0;
> }
>
>
> ptrArray[0] aka element is a t_Array* containing &array.
> The array starts at the same storage location as array[0],
> so ptrArray[0][0] aka element[0] aka *element aka array
> is the same.
> array[9] and (*element)[9] are the same.
> element[9] is the same as *(element+9) is
> *(t_Array *)((unsigned char*)element + 9*sizeof *element) is
> a t_Array starting at the location
> ((unsigned char*)&array + 9*sizeof array) which certainly is
> not an int value and certainly not within the memory you own.
>
>
>> ----------------------------------------------
>> example 2:
>> typedef int t_Array[10];
>> void function(t_Array Array)
>> {
>> t_Array *ptrArray[] = {&Array}; //error
>> }
>> int main(int argc, char* argv[])
>> {
>> t_Array array;
>> t_Array *ptrArray[] = {&array}; //no error
>> return 0;
>> }
>> the 2 declarations of the arrays look the same to me, but 1 gives an
>> error, and the other one doesn't. i have looked up the documentation of
>> that
>> error, but that didn't shine much light.
>> ??
>>
>> ----------------------------------------------
>> example 3:
>> typedef int t_Array[10];
>> void function(t_Array Array)
>> {
>> t_Array *element = NULL;
>> void * ptrArray[] = {NULL};
>>
>> ptrArray[0] = &Array; //first element should be pointer to a t_Array
>> element = (t_Array *)ptrArray[0];
>> printf("Weird Failure: (*element)[9] = %d\n", (*element)[9]);
>>
>> ptrArray[0] = Array; //first element should be Array itself
>> element = (t_Array *)ptrArray[0];
>> printf("Weird Success: (*element)[9] = %d\n", (*element)[9]);
>> }
>> int main(int argc, char* argv[])
>> {
>> t_Array array;
>> array[9] = 321;
>> function(array);
>> return 0;
>> }
>> this is what really caused my headaches. i used void pointers to get rid
>> of
>> the error, but that caused weird behavior. in the first situation in
>> 'function', i expect to dereference a pointer to end up with an array,
>> but
>> that clearly is not what happens.
>> in the second situation i expect to do something illegal: ie to use a
>> pointer to an array as the array itself, and somehow that seems to work.
>> what do i miss here?
>>
>> i would be very grateful for some light on this murky behavior (or at
>> least
>> my murky understanding).
>
> Your problem with examples 1, 2 and 3 is that you
> a) do not understand fully the difference between pointer and array.
> See the comp.lang.c FAQ to help further your understanding.
> Your problem with examples 2 and 3 is that you seem unaware of
> the fact that
> b) you cannot pass arrays to functions -- instead you always pass
> pointers to the first array element.
> c) typedefs are just a way to write types conveniently or create
> some abstraction by logical markup (naming a type's role instead of
> its physical type) but not a way to generate new types.
>
> Cheers
> Michael
> --
> E-Mail: Mine is an /at/ gmx /dot/ de address.


.

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