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 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_kon...@dell.com
wrote:
Question on movmemm:
Given
extern int *i, *j; void foo (void) { memcpy
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 paul_kon...@dell.com
wrote:
Question on movmemm:
Given
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 int *i, *j; void foo
On Tue, Oct 26, 2010 at 1:12 PM, Paul Koning paul_kon...@dell.com 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
then
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
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
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
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
