Re: movmemm pattern

[Dave Korn <dave dot korn dot cygwin at gmail dot com>]

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); }

    cheers,
      DaveK