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Re: movmemm pattern
On Oct 25, 2010, at 9:28 PM, Dave Korn wrote:
> 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_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 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); }
That doesn't make any difference; I still get alignment 1.
paul