Why must bitfields sum to a multiple of a byte?

[Era Scarecrow]

On Thursday, 2 August 2012 at 12:35:20 UTC, Andrei Alexandrescu 
wrote:

> Please don't. The effort on the programmer side is virtually 
> nil, and keeps things in check. In no case would the use of 
> bitfields() be so intensive that the bloat of one line gets any 
> significance.>

  If you're using a template or something to fill in the sizes, 
then having to calculate the remainder could be an annoyance; but 
those cases would be small in number.

  I'll agree, and it's best leaving it as it is.

  BTW, Wasn't there a new/reserved type of cent/ucent (128bit)?