> There is a quite convincing argument that such an access _is_ an
I find this argument completely convincing and retract the contrary argument
that I've made many times in this forum and others. You learn something new
every day.
Just in case it wasn't clear:
int i;
*(volatile int *)&i=2;
In this case, there *is* an access to a volatile object. This is the end
result of the the standard's definition of what it means to apply the
'volatile int *' cast to '&i' and then apply the '*' operator to the result
and use it as an lvalue.
C does not define the type of an object by how it is defined but by how it
is accessed!
DS
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