Is i=++i; undefined behavior?

Question

This is undefined behavior, right? Double assignment in the same sequence point?

int i = 0;
i = ++i;

I think we need to reopen this. The duplicate was for post-increment. And the rules do differ on the UB. (I won't answer as per dupe hammer reopen etiquette). — Bathsheba, May 02 '19 at 13:49
@Bathsheba if you hypothetically had an answer, what would that answer be? — Carbon, May 02 '19 at 13:51
@Carbon: My answer waffled on about operator overloading which deviated from relevance. — Bathsheba, May 02 '19 at 13:52
@Bathsheba: If that's the case, I'm sure you can find a duplicate that talks about pre-increment issues. There are plenty of such questions on this site. — Nicol Bolas, May 02 '19 at 13:52
In that hypothetical answer, was the conclusion that it was or was not defined? — Carbon, May 02 '19 at 13:52
I *think* `i = ++i;` is defined in C++11, but the "dupe" is to an expression that isn't defined until C++17. — Bathsheba, May 02 '19 at 13:53
There, dupe target changed to a Q that answers it all. `i = ++i` is UB in C++98/03, and okay in C++11 and above. It also covers all of the other cases. — NathanOliver, May 02 '19 at 13:53
Sorry folks but this needs an explicit answer since there is a lot of nonsense in the comments. — Bathsheba, May 02 '19 at 13:56

Bathsheba · Accepted Answer · 2019-05-02T13:55:04.657

2

i = ++i; is undefined in all C++ standards prior to C++11.

Note that cohorts of similar expressions using the postfix increments are not defined until C++17.

Both are defined if ++ is overloaded, which is a moot point here as the incrementee is a primitive.

edited May 02 '19 at 13:55

answered May 02 '19 at 13:48

Bathsheba

1 Answers1