sequence-points

I used to think that in C99, even if the side-effects of functions f and g interfered, and although the expression f() + g() does not contain a sequence point, f and g would contain some, so the behavior would be unspecified: either f() would be called before g(), or g() before f(). I am no longer so sure. What if the compiler inlines the functions (which the compiler may decide to do even if the functions are not declared inline ) and then reorders instructions? May one get a result different of the above two? In other words, is this undefined behavior? This is not because I intend to write

Following is the test code: int main() { int a = 3; int b = 4; a = a + b - (b = a); cout << "a :" << a << " " << "b :" << b << "\n"; return 0; } Compiling this gives the following warning: > $ g++ -Wall -o test test.cpp test.cpp: In function ‘int main()’: > test.cpp:11:21: warning: operation on ‘b’ may be undefined > [-Wsequence-point] Why can the operation be undefined? According to my understanding, first the subexpression (b = a) should be evaluated because of higher precedence of (), thus setting b = a. Then, since '+' and '-' have same precedence, the expression would be evaluated left

Can someone tell me how prefix / postfix operators really work? I've been looking online a lot but haven't found anything. From what I can tell prefex first increments, then does the operation and then assigns. Postfix would do the operation first, then assign and then increment. But I'm having a bit of trouble with my code: int x, y; x = 1; y = x + x++; // (After operation y = 2)(x=2) However when I do: y = x++ + x; // (After operation y = 3)(x=2) I'm not sure why these operations would be any different. I have two questions: Could you explain the difference? How does this apply to the other

Sorry for opening this topic again, but thinking about this topic itself has started giving me an Undefined Behavior. Want to move into the zone of well-defined behavior. Given int i = 0; int v[10]; i = ++i; //Expr1 i = i++; //Expr2 ++ ++i; //Expr3 i = v[i++]; //Expr4 I think of the above expressions (in that order) as operator=(i, operator++(i)) ; //Expr1 equivalent operator=(i, operator++(i, 0)) ; //Expr2 equivalent operator++(operator++(i)) ; //Expr3 equivalent operator=(i, operator[](operator++(i, 0)); //Expr4 equivalent Now coming to behaviors here are the important quotes from C++ 0x .