Does this paragraph allow an implementation to optimize out volatile accesses in C? -

this question has answer here:

• why volatile needed in c? 15 answers

the c standard says

an actual implementation need not evaluate part of expression if can deduce value not used , no needed side effects produced (including caused calling function or accessing volatile object)

when volatile variable not needed? according paragraph, volatile appears become subject as-if rule other non-volatile object.

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the answers given in non-duplicate linked question not helpful me not address above quoted paragraph

• when value considered "used"? appears different "reading value object" because corresponding access can omitted, according above quote.
• what "needed side effect"?

refer comments below.

i think they're referring situations like

void test(int volatile *y) {     int x = *y; }  int main(void) {     test(some_volatile_y);     return 0; } 

where load memory optimized away (unevaluated), because result not needed on every implementation. (of course, on implementations memory load can trigger event, of course result "needed" , cannot optimized away.)

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thanks alex's comment below pointing out footnote in c standard (c99: 6.7.3 type qualifiers: footnote 114):

a volatile declaration may used describe object corresponding memory-mapped input/output port or object accessed asynchronously interrupting function. actions on objects declared shall not ‘‘optimized out’’ implementation or reordered except permitted rules evaluating expressions.

note says may -- it's giving example. clause j.3 mentions constitutes access object has volatile-qualified type implementation-defined, makes sense: previous clause example of "access" could mean, mean else depending on implementation.

so, on implementations system knows can't cause needed side effect, volatile access may removed.