RE: [boost] operator++

From: Gary Powell (Gary.Powell_at_[hidden])
Date: 2000-10-11 09:47:57

>  As others have mentioned, it'll break code which
> depends on the implicit conversion of your T; also
> (i++).member will not work 
Now if Bjarne had only listened when I argued for operator.() back in the
late 80's.....

> and & (i++) will not
> produce what you expect -- 
> 
This could be fixed with yet another overload to forward the request. And we
could add operator*() and operator->()....of course this all requires the
usual gamut of type deduction templates.... 

You know this swamp we started to drain could be fixed by a very smart
compiler. One that noticed that either the return value was used or not and
then do the smart thing...

  -gary-

gary.powell_at_[hidden]