Question

Possible Duplicate:
Is it allowed to name the parameter in postfix operator ++?

I created an object to hold a list of objects that maintains the current position internally, so I thought this was a great place to overload the pre and post increment operators, which effectively increment this internal position with bounds checking.

What I noticed is, when you call ++ on the object, the argument is 0.

Test code:

#include <stdio.h>
class A {
public:
   A& operator++(int n) { printf("%d  ", n); return *this; }
};
int main() {
   A a;
   a++;
   a.operator++(0);
   a.operator++(1);
   a.operator++(10);
   return 0;
}

This returns 0 0 1 10. From what I understand, this is normal behavior. So, it has made me rethink how operator++ should work. Previously, I was simply calling ++ on my internal position variable if bounds checking passed. But this has the affect of incrementing by 1 no matter what the input argument is. Next, I though of using the += using the argument n as the right hand side, but as you’ll notice, simply calling ++ with no operators (as is customary), gives a zero and the position is not incremented.

Basically, is this something I should even worry about? If so, how do I detect if the user really wanted 0, or if the default behavior (a++) was intended and I should increment by 1?

Answer 1

You should just ignore the parameter.