New Sequence Points in C++11

Question

With the new college year upon us.
We have started to receive the standard why does ++ i ++ not work as expected questions.

After just answering one

Answer 1

I don't think sequencing is relevant to your situation. The expression ++i++ is grouped as ++(i++), so:

• If i is a built-in type, then this is invalid, since i++ is an rvalue.

• If i is of user-defined type and the operators are overloaded, this is a nested function call, such as T::operator++(T::operator++(i), 0), and function arguments are evaluated before the function call is evaluated.