Re: memcpy to an unaligned address

[Andrew Pinski <pinskia@physics.uc.edu>]

> 
> On Aug 2, 2005, at 1:15 PM, Shaun Jackman wrote:
> > There is no padding. The structure is defined as
> > __attribute__((packed)) to explicitly remove the padding. The result
> > is that gcc knows the unaligned four byte member is at an offset of
> > two bytes from the base of the struct, but uses a four byte load at
> > the unaligned address of base+2. I don't expect...
> >     p->unaligned = n;
> > ... to work,
> 
> Actually, that works just fine, with:
> 
> typedef struct {
>    unsigned short int a;
>    unsigned int b;
> } __attribute__((packed)) st;
> 
> void foo(st *s, int n)
> {
>    s->b = n;
> }
> 
> Ah, I was having trouble getting it to fail for me...  Now I can:
> 
> #include <memory.h>
> 
> typedef struct {
>    unsigned short int a;
>    unsigned int b;
> } __attribute__((packed)) st;
> 
> void foo(st *s, int n)
> {
>    memcpy(&s->b, &n, sizeof n);
> }
> 
> Yes, this is a compiler bug in the expansion of memcpy, please file a  
> bug report.  The solution is for the compiler to notice the memory  
> alignment of the destination and `do-the-right-thing' when it isn't  
> aligned.

No it is not, once you take the address (which should be rejected), it
is of type "unsigned int *" and not unaligned variable, passing it to
memcpy assumes the type alignment is the natural alignment.

-- Pinski