On 26/10/2010 01:53, Paul Koning wrote:
> On Oct 25, 2010, at 3:44 PM, Richard Guenther wrote:
>
>> On Mon, Oct 25, 2010 at 11:26 AM, Paul Koning <paul_kon...@dell.com>
>> wrote:
>>> Question on movmemm:
>>>
>>> Given
>>>
>>> extern int *i, *j; void foo (void) { memcpy (i, j, 10); }
>>>
>>> I would expect to see argument 4 (the shared alignment) to be
>>> sizeof(int) since both argument are pointers to int. What I get
>>> instead is 1. Why is that?
>> Because the int * could point to unaligned data and there is no access
>> that would prove otherwise (memcpy accepts any alignment).
>
> Ok, but if I do a load on an int*,
I think that is what Richard meant by an "access that would prove otherwise".
> I get an aligned load, not an unaligned
> load, so in all those other cases there *is* an assumption that an int*
> contains a properly aligned address.
This is a bit like GCC optimising away a null-pointer check if it knows
you've already dereferenced the pointer. Either you've already crashed by
then, or it doesn't matter.
What happens if you dereference i and j before the memcpy in foo? Do you
then get int-sized shared alignment in movmemM?
extern int *i, *j; void foo (void) { *i; *j; memcpy (i, j, 10); }
cheers,
DaveK
