undefined behaviour

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

So, things such as:

i=++i; // illegal : i is modified two times
i+(i++); // illegal value of i read not only to determine the value to be stored.
i=i+i; // legal : value of i only read to determine the value to be stored.

So, things such as:

i=++i; // illegal : i is modified two times
i+(i++); // illegal value of i read not only to determine the value to be stored.
i=i+i; // legal : value of i only read to determine the value to be stored.

Where is quote from?
I just tried all these constructs on my compiler (Sun WorkShop 6 C++ 5.3), with full warnings turned on and no warnings or errors appeared whatsoever. The results make sense (to me anyway) as well.

i=++i; // i is pre-incremented and then assigned back to i (albeit unnecessary) but nonetheless, I see no reason why this would be illegal.

i+(i++); // i is post-incremented. The result of this would simply be a 1-up from what it originally was. Of course, if there were an assignment here you could get different results depending on the compiler (but I see nothing illegal about this either). Let's say i started out as 2 and you had the following statement... x=i+(i++); If the 1st i were evaluated 1st then you would have x=2+(2++). The result would be x equal to 4 and i equal to 3. If the 2nd i were evaluated 1st then the result would be x=3+(2). The result, of course, being x equal to 5 and i again being equal to 3. In fact, my compiler has the later as the result.

Also, I may be wrong, but I would be very surprised if operator = would not be a sequence point.

operator = is not a sequence point.

i+(i++);

It just looks like a code which increments i, and does nothing else (an addition whose value is not used, which produces a warning).
The compiler may produces invalid code, which may crash (of course, i don't think any compiler on the earth do), for undefined behavior.
i+(i++) is undefined, because it modifies a variable and reads its value (not only to determine the value to store) in the same expression between the same sequence points.
Even if most compilers will not have any problem with this expression, some compilers may produces a compilation error "undefined behavior".

i=++i;

i is modified two times (incrementation operator + assignment operator) with no sequence point.

Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored. The requirements of this paragraph shall be met for each allowable ordering of the subexpressions of a full expression; otherwise the behavior is undefined.

It is a more strict condition than the simple condition "side effects may occur in any order".
But it probably allow compiler to perform some complex side effects at the same time.

Good point, I hadn't thought about the evaluation of arguments. Also you are right SuperKoko, the assignment operator is not a sequence point.

I understand that :
cout << a << a++;
is undefined behaviour because there are 2 sequence points(2 function calls)
so 'a' will be evaluated seperately and 'a++' will be evaluated sepeartely(and would have incremented the a's value by 1)

but the order in which they are evaluated makes it undefined.

But in the case :
x= a + a++ ;
There is only one sequence point ';' Right?
So the value of 'a' will not be incremented only in the next sequence point

so i don't understand WHY this is undefined behaviour?