Comment by SpaceNugget
7 hours ago
The print example has no defined order of accesses, function parameters can be evaluated in any order. But further, the entire problem with UB is that it supercedes the regular guarantees that you get (like with volatile) when it's encountered. Yes gcc and clang do the obvious thing that makes the most sense in this example, but what people are trying to tell you is that they could just not do that and they would still be complying with the standard. For example, you can imagine a more serious example of UB that causes the program to fail to compile completely, and then do you emit the correct number of in order reads of volatile variables? Obviously not.
Function parameters cannot be evaluated in any order, when one of them is a volatile.
> The initialization shall occur in initializer list order, each initializer provided for a particular subobject overriding any previously listed initializer for the same subobject
And what I am trying to tell people, is the standard has expectations around the volatile keyword, that the compilers took into account when designing how they would work - it isn't just kindness, its compliance. But no one is actually talking about the quotes from the standard, and just quoting themselves and their own understandings.
That quote doesn't have anything to do with parameter evaluation order. There is no order for function parameter evaluation.
And no, there is no exception for undefined behavior. There can't be, otherwise the behavior would be... defined. It's in the name. Again, what do you think the compiler emits when the undefined behavior causes the program to not compile altogether?