alt Hacker NewsIf code has undefined behavior, the entire execution path that leads to that UB has no assigned semantics in the C model. So there are no volatile accesses in this code according to the C abstract machine - the entire execution path is UB, so it can be assumed it doesn't happen at all.
> An object that has volatile-qualified type may be modified in ways unknown to the implementation or have other unknown side effects. Therefore any expression referring to such an object shall be evaluated strictly according to the rules of the abstract machine
The execution path has unknown side effects, and so the execution path must be strictly followed. That's uh... The entire point of that section in the C standard. Its why volatile is called out, in the semantic model for the abstract machine.
Otherwise... Why call it out, at all? It must be strictly followed, not lazily, as in other areas of the standard.