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[QUOTE="Old Wolf, post: 2466834"]
Correct. But the evaluation of the left-hand operand of the
assignment operator does not cause the value of x to be
read or written, so it does not fall foul of the rule about UB
due to multiple reads/writes between sequence points.

I didn't mention this as it is a red herring compared to
the main issue.

However, this code would certainly be undefined, for the reason
you mention:

*p = (p = q, 11);

where p and q are valid pointers to int.

The evaluation of an operator's operands, does not mean that
the evaluation of the operator has started. In fact, my lemma
is that the assignment operator cannot be evaluated until
its right-hand operand has been evaluated (although the left
hand one may be evaluated at any time).

Note that 'evaluating' the left hand operand of the assignment
operator means determining the adddress at which the value
will be stored (it doesn't mean reading or writing that value).

Yes, x is modified twice between sequence points.
[/QUOTE]

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