Is i = 0, ++i defined?

Question

I recently learned about the , operator and the fact that it introduces a sequence point.

I also learned that the following code led to undefined behavior:

i = ++i; 

Because i was modified twice between two sequence points.

But what about the following codes ?

i = 0, ++i; i = (0, ++i); 

While I know the rules, I can't get to a conclusion. So is it defined behavior or not ?

edit: Just as @paxdiablo mentions, defined or not, this is really a bad practice which should be avoided. This question is asked solely for educational purposes and better understanding of the "rules".

Answer 1

Yes. = has higher precedence than ,, so this expression is equivalent to (i = 0), ++i. , is a sequence point, so it's guaranteed that the ++i occurs after the assignment.

I'm not sure whether i = (0, ++i) is defined though. My guess would be no; there's no sequence point between the increment and the assignment.

Answer 2

i = 0, ++i; 

As the other answer pointed out it is not Undefined Behaviour.

i = (0, ++i); 

The behaviour is undefined in this case because there is no sequence point between ++i and assignment to i.

i = (0, ++i, 0) 

The behaviour is well defined¹ in C++03, IMHO.

_{¹ See extended discussion for a similar expression.}