Is cout << (i++) << (i++) Undefined behavior

Asked Jun 05 '20 at 05:15

Active Jun 05 '20 at 05:25

Viewed 58 times

I'm thinking what the output of int i = 0; cout << (i++) << (i++); is in c++11.

As my understanding, this code can be transformed into operator<<(operator<<(cout, i++), i++). Am I right?

I thought it would print 01, however, in my Ubuntu system, it printed 10.

Why? Is it an UB or not? If it's an UB, does it mean that f(g(i++), i++) is UB too? If it's not an UB, why was the output 10, instead of 01?

edited Jun 05 '20 at 05:25

hanie

asked Jun 05 '20 at 05:15

Yves

3

Please don't att irrelevant tags, like the C tag for what clearly can only be a C++ question. – Some programmer dude Jun 05 '20 at 05:17
As for your problem, it's definitely UB. Arguments to functions can be evaluated in any order, and all overloaded operators are implemented as functions. – Some programmer dude Jun 05 '20 at 05:19
Mind you, the previous comment is only correct because you tagged it with the obsolete C++11 tag. – MSalters Jun 05 '20 at 10:01

0 Answers0