++/-- operators

Dan Stone · 02.10.2001, 02:46

I have some problems for you, I'm a Lab Aide 3rd period at Clear Lake High in Hosuton, Texas, and the Comp Sci 1 students are doing the Increment/Decrement operators, and there are some small innacuracys between logic, and what our compilers are doing. On these computers we have Borland C++ 4, and MetroWorks CodeWarrior Pro 5. #include <iostream.h>

int main() {

int x = 10;

++x = ++x + (x++ + ++x); cout << x << endl; return 0; }

do this on paper, and then type it in, totally diffrent answers. any explainations?

--Thx Dan

greka · 02.10.2001, 06:53

how you're performing calculations ? I mean what answer is expected by you :-? <IMG SRC="smile.gif" border="0">

acid · 03.10.2001, 00:23

Well, it was confusing but if you look inside ISO/IEC 9899 you can find chapter like "Expressions - topic 4". Here it is:

 ... 4. Except where noted, the order of evaluation of operands of individual operators and subexpressions of individual expressions, and the order in which side effects take place, is unspecified. 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. [Example: i = v[i++]; // the behavior is undefined i = 7, i++, i++; // ‘i’ becomes 9 i = ++i + 1; // the behavior is undefined i = i + 1; // the value of ’i’ is incremented —end example] ...

 I think it is clear enough that your expression is much more complicated version of i = ++i + 1, so the behavior will be what your compiler can produce <IMG SRC="smile.gif" border="0">

Hope it helps. acid

[ October 02, 2001: Message edited by: acid ]