4 ms·
No, it does not. It's still UB. UB is defined by the standard, not by your compiler's implementation. Certain behaviors may be implementation defined by the sta
by pertymcpert 8y ago
No, it does not. It's still UB. UB is defined by the standard, not by your compiler's implementation. Certain behaviors may be implementation defined by the standard, those can be defined by your compiler.
But if the standard says it's UB, it's UB. End of story.
- mehrdadn 8y agoWhere/how do you obtain such confidence in something so wrong? The standard not only doesn't prohibit the implementation from defining something it leaves undefined (surely you don't think even possible behavior becomes invalid as soon as it is documented??), it in fact explicitly permits this possibility to occur -- I suppose to emphasize just how nuts the notion of some kind of 'enforced unpredictability' is: > Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner...