Interestingly, at least in C++, this was changed in the recent past. It used to be that evaluation of arguments was not sequenced at all and if any evaluation touched the same variable, and at least one was a write, it was UB.

It was changed as part of the C++11 memory model and now, as you said, there is a sequenced-before order, it is just unspecified which one it is.

I don't know much about C, but I believe it was similarly changed in C11.

▲

tialaramex 5 days ago | parent | next [-]

Sure, prior to the C++ 11 memory model there just isn't a memory ordering model in C++ and all programs in either C or C++ which would need ordering for correctness did not have any defined behaviour in the language standard.

This is very amusing because that means in terms of the language standard Windows and Linux, which both significantly pre-date C++ 11 and thus its memory model, were technically relying on Undefined Behaviour. Of course, as operating systems they're already off piste because they're full of raw assembly and so on.

Linux has its own ordering model as a result, pre-dating the C++ 11 model. Linus is writing software for multi-processor computers more than a decade before the C++ 11 model so obviously he can't wait around for that.

[Edit: Corrected Linux -> Linux when talking about the man]

	▲	gpderetta 5 days ago \| parent [-]
		It is not so much that windows and linux were relying on UB, but that these platforms, with their compilers, provided guarantees beyond the standard. e.g. GCC not only aims for C/C++ standard compliance, but also POSIX. Of course these guarantees were often not fully written down nor necessarily self consistent (but then again, neither is the current standard).

▲

gliptic 5 days ago | parent | prev [-]

Yes, but that's just a subset of expressions where unspecified sequencing applied. For instance, the example with two `print()` as parameters would have a sequence point (in pre-C++11 terminology) separating any reads/writes inside the `print` due to the function calls. It would never be UB even though the order in which the prints are called is still unspecified.

▲

gpderetta 5 days ago | parent [-]

IIRC the point was that there was no sequence point between argument evaluation, so for example f(++i, ++i) was UB. Or maybe it was only for builtin operators?

Cppreference is not authoritative[1], but seems to support my recollection. In fact it states that the f(++i, ++i) was UB till C++17.

[1] https://en.cppreference.com/w/cpp/language/eval_order.html, Pre C++11 Ordering Rules, point (2).

	▲	gliptic 5 days ago \| parent [-]
		`f(++i, ++i)` is/was indeed UB, but the example in munificent's comment was `foo(print(1), print(2))` which as far as I know is not even if both `print` calls read/write the same memory.