movmemm pattern

[Paul Koning]

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 larger alignment as expected.

	paul