Lpc2000

Jaya,

The "vanilla" cast is only to be treated as a byte packed cast if the
object that is being cast is itself byte packed. 

In thinking about my original case, the compiler should at least warn
since, with;

char buf[]={...};
int *ip;

ip = (int *)&buf[i];

the left and right side of this assignment are not of the same type,
even though the declaration would have you think so. ip is a pointer
to a default word aligned object, and (int *)&buf[i] is a pointer to a   
byte aligned object. These are really not the same data types. 

But I'm afraid I may be getting ahead of myself here. If I'm not
careful someone is going to smack me with the C99 spec :-).

I'll stick with my original assertion that the compiler could deal
with this issue in a more useful way than it does.

Dave S.




--- In lpc2000@yahoogroups.com, "jayasooriah" <jayasooriah@...> wrote:
>
> Dave,
> 
> If I understand you (now more correctly): because byte packing is
> default, you want a vanilla cast (without __packed prefix) to be
> treated as byte packed cast.
> 
> How would a programmer then indicate a cast to a vanilla (not packed)
> pointer?
> 
> If the compiler always assumes byte packed no matter what, arguably
> the compiler cannot take advantage of the faster method of dealing
> with composite objects.
> 
> Jaya
> 
> --- In lpc2000@yahoogroups.com, "dsidlauskas1" <dsidlauskas@> wrote:
> >
> > Jaya,
> > 
> > I don't think I'm asking it to connect the two statements.
> > 
> > What I'm expecting it to do is to recognize that buf is a byte packed
> > object (the defalt for the compler)and when casting byte packed
> > objects the compiler should cast them to other byte packed objects.
> > 
> > in otherwords;
> > 
> > char buf[10];
> > int ip*;
> > 
> > ip = (int *)&buf[5];
> > 
> > should be treated by the compiler as:
> > 
> > ip = (__packed int *)&buf[5];
> > 
> > and this is because char buf[10] is a byte packed object for this
> > compiler.
> > 
> > Dave
>