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