Marc, are you sure ? It means that the assignment gets evaluated as a boolean expression?? Yes... you've got "port == DPORT" with two equal signs. That's not an assignment. That is going to evaluate to see whether it's true or false. Since DPORT = 0 by default, then it's evaluating "port == 0" which is true, thus its value is 1. So, you've got a statement which is: 1; Essentially, it's a no-op. With DPORT = 0, it doesn't make any difference, save for a compiler warning. If you were to set DPORT to some other number, though, the assignment wouldn't take place as would otherwise be expected. --Marc