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Re: movmemm pattern
On Oct 25, 2010, at 3:44 PM, Richard Guenther wrote:
> On Mon, Oct 25, 2010 at 11:26 AM, Paul Koning <Paul_Koning@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 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.
paul