Boost C++ Libraries: Ticket #2228: signed overflow problem in weighted_sum on VC9.0

status changed; resolution set

Eric Niebler — Tue, 19 Aug 2008 16:06:04 GMT

status new → closed
resolution → invalid

This is not a bug. The sample is multiplied by the weight. In C++, and int multiplied by an unsigned int is an unsigned int. (Try it and see.)

If you want the result to be an int, use an int for the weight type.

Pavol Droba — Tue, 19 Aug 2008 18:02:37 GMT

I understand that the current solution is technically correct. However I think that conceptually it is not.

An example: To calculate a weighted mean of signed integers, where weight represents sample cardinality, I would consider it quite natural to expect that the mean will be signed, since I'm accumulating signed integers. Even though the cardinality is natural (i.e. unsigned) number.

Eric Niebler — Tue, 19 Aug 2008 18:27:21 GMT

I understand your point, but I still disagree. I'm not going to get into guessing what a "better" type to hold the result is. It can be argued from the other side ... that using an int to hold the result of an (int*unsigned) can overflow and wrap. There is no good solution. Following the rules of C++ is simply the best (as in least surprising, easiest to explain) thing to do.