Question

§5/4 C++ standard

i = 7, i++, i++;  // i becomes 9
i = ++i + 1;  //the behavior is unspecified

That should be changed to

i = 7, i++, i++;  // the behavior is undefined
i = ++i + 1;  //the behavior is undefined

right?

Answer 1

i = 7, i++, i++; // i becomes 9

is perfectly fine. Operator = has higher precedence than ,so the expression is equivalent to

(i = 7), i++, i++; 

which is perfectly well defined behaviour because , is a sequence point.

As far as

i = ++i + 1; //the behavior is unspecified

is concerned the behaviour is undefined in C++03 but well defined in C++0x. If you have the C++0x draft you can check out in section 1.9/15

i = i++ + 1; // the behavior is undefined