On 22/07/2005, at 7:15 PM, Paul Schlie wrote:
Geoffrey Keating writes:
without 'volatile', then this object cannot be modified unknown to the implementation, even if someone also writes '(*(volatile int *)&i) = 1'.- merely means: treat the object being referenced as volatile qualified int object (as the standard specifies, although it may result in an undefined behavior, nothing more or less; as although the object may have not been initially declared as being volatile, the program within the context of this particular references has asserted that it must be treated as such, thereby implying it's value must be assigned, and/or presumed to have beenpossibly modified beyond the logical view of the program).
It doesn't imply that. All it implies is that *from this access* the compiler cannot assume that the object is not "modified in ways unknown to the implementation". From other accesses, including from the original declaration (and its initializer if any), the implementation may be able to make that deduction. If, and only if, the implementation can make that deduction, then it can perform optimizations. In this example:
int i = 0; while (*(volatile int *)&i == 0) ;then the implementation can make that assumption, and optimise the loop into an infinite loop that does not test 'i', because the '= 0;' performs a store to a non-volatile object 'i' which therefore cannot be modified in ways unknown to the implementation and therefore will always be zero.
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