On Tue, Oct 26, 2010 at 1:12 PM, Paul Koning wrote:
>
> On Oct 26, 2010, at 1:27 PM, Dave Korn wrote:
>
>> On 26/10/2010 17:16, Paul Koning wrote:
>>> On Oct 25, 2010, at 9:28 PM, Dave Korn wrote:
...
What happens if you dereference i and j before the memcpy in foo? Do you
the
On Oct 26, 2010, at 1:27 PM, Dave Korn wrote:
> On 26/10/2010 17:16, Paul Koning wrote:
>> On Oct 25, 2010, at 9:28 PM, Dave Korn wrote:
>>> ...
>>>
>>> What happens if you dereference i and j before the memcpy in foo? Do you
>>> then get int-sized shared alignment in movmemM?
>>>
>>> extern i
On 26/10/2010 17:16, Paul Koning wrote:
> 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
wrote:
> Question on movmemm:
>
> G
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
>>> wrote:
Question on movmemm:
Given
extern int *i, *j; void foo (void
On Mon, 25 Oct 2010, Richard Guenther wrote:
> Because the int * could point to unaligned data and there is no access
> that would prove otherwise (memcpy accepts any alignment).
As previously discussed, in ISO C storing a pointer in a particular
pointer type or converting to / through that type
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
>> wrote:
>>> Question on movmemm:
>>>
>>> Given
>>>
>>> extern int *i, *j; void foo (void) { memcpy (i, j, 10); }
>>>
>>> I would expect to see arg
On Oct 25, 2010, at 3:44 PM, Richard Guenther wrote:
> On Mon, Oct 25, 2010 at 11:26 AM, Paul Koning 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)
>
On Mon, Oct 25, 2010 at 11:26 AM, Paul Koning 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.
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?
If I have
extern int i[10], j[10];
then I do get la
