Re: Array assignment via struct

On Sat, 06 Aug 2005 18:22:50 -0400, Joe Wright wrote:
> Netocrat wrote:
>>>>>Jack Klein wrote:
>>>>>>Joe Wright wrote:
<snip>
>>>>>>>#define LEN 20
>>>>>>>
>>>>>>>typedef struct {
>>>>>>> char a[LEN];
>>>>>>>} S;
>>>>>>>
>>>>>>>int main(void) {
>>>>>>> S sa;
>>>>>>> char A[LEN];
>>>>>>>
>>>>>>> S *ps = malloc(LEN);
<snip>
>>>>>>> *(S*)A = sa;
>>>
>>>[snip]
>>>
>>>>>>Here is where you invoke undefined behavior, since A isn't dynamically
>>>>>>allocated. There is no guarantee that A meets the alignment
>>>>>>requirements for an S. The compiler might generate code that assumes
>>>>>>that A is, causing some sort of trap on some platforms, or possible
>>>>>>misaligned data or overwriting the destination array.
<snip>
>> So to expand on
>> Jack's explanation of specific code being generated, perhaps something
>> like this:
>>
>> 4 padding bytes are added after the array of 20 char in the struct so
>> that it can be placed on an 8-byte boundary. The compiler generates
>> code to retrieve the elements of the array 8-bytes at a time and unaligned
>> access to 8-byte-wide data on this particular implementation is not
>> allowed.
>>
>> The automatic char[20] variable A is not aligned on an 8-byte boundary, so
>> when it's accessed through the struct, unaligned access occurs and our
>> implementation spits the dummy.
>>
>> So Joe - no go. Thou code be fraught.

Correction: thy code be fraught. Thou codest flawed source.

> I think not. Consider..
>
> struct {
> char a[17];
> } sa;
>
> ..and explain any case for sizeof sa not being 17. Annecdotes of long
> forgotten DEC Stations don't count.

Why not? Were they not valid C implementation hosts?

That's a contrived choice because being (2 pow 4) + 1 it's impossible to
minimise accesses. Using my example above, consider a 20-byte array
accessed in 8-byte chunks. When aligned on an 8-byte (or 4-byte)
boundary, it will take 3 accesses to read/write the entire array in
8-byte chunks. It will take 4 accesses to do the same when its alignment
is 1, 2 or 3 bytes off an 8-byte alignment. That's a supportable reason
for properly aligning the struct on an 8-byte boundary and hence requiring
4 padding bytes.

As for your 17 byte example, well, this implementation may pad out 7
bytes but more likely it would pad 3 and it would access 4 bytes at a time
on a 4-byte boundary.

Totally hypothetical but for all I know (I don't have a lot of varied
hardware experience) a machine like this does exist. Not the machine that
I work from though (Intel P4) because unaligned access whilst slower is
not an error.

.

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