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> i has automatic storage, and its address is never taken, so the compiler > might decide to allocate this object to some 'write-only-memory'. No, it may not. Modification of i is 'observable behaviour' as per [intro.execution]/6. So I want to observe this on the machine in front of me. If the compiler optimizes the loop away, I cannot observe it. If it would leave the loop, I could observe it, eg. by means of a debugger. Regards, Martin
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