Double assignment of the same variable in one expression in C++11

Question

The C++11 standard (5.17, expr.ass) states that

In all cases, the assignment is sequenced after the value computation of the right and left operands, a

Answer 1

Yes, there was a change between C++98 and C++11. I believe your example to be well-defined under C++11 rules, while exhibiting undefined behavior under C++98 rules.

As a simpler example, x = ++x; is undefined in C++98 but is well-defined in C++11. Note that x = x++; is still undefined (side effect of post-increment is unsequenced with the evaluation of the expression, while side effect of pre-increment is sequenced before the same).