On Thu, 5 Oct 2006, Jinesh K J wrote:
In this case we only need to write this as:
*( (volatile char *) 0);
But unlike the value, the address is not volatile. It's not a variable.
I'm not sure I agree. A 'volatile char *' means 'a pointer to something
that is volatile'. However ...
&(*( (volatile char *) 0));
will still have no side effect. Only the char is volatile, not the
address, right? So,
... in this case you are not referencing location zero, mainly
(re-)computing its address. What 'volatile' says, at least to my
understanding, is that the compiler may not optimize away an access to the
variable (memory location) in question. If however other rules of C do not
raise the need for accessing the location in the first place, there will
be no access, regardless of the 'volatile' qualifier.
I would say, in this case, that the memory location is considered
volatile, but the rules of C (taking the address) mean that there is no
access to it, even though initially it may look that way.
s->foo; // fetches the value of foo(ideally)
&(s->foo); // will not fetch the value of foo - only compute its address
- since, computing the address of foo (here, by reading the variable s)
has no operation over the value of foo. So, this expression must have
side effect only upon 's', if there ever is any.
Agreed.
...
So I believe that in a recommended implementation,
a;
must fetch the value of 'a', though it should be discarded. But then
the compiler in our case optimises it (by not reading 'a'), unless 'a'
is declared as 'volatile'.
That's my interpretation tool.
/Ricard
--
Ricard Wolf Wanderlöf ricardw(at)axis.com
Axis Communications AB, Lund, Sweden www.axis.com
Phone +46 46 272 2016 Fax +46 46 13 61 30
