On 04/29/2011 09:37 PM, Kalle Olavi Niemitalo wrote:
Wouter Vermaelen<wouter.vermaelen@xxxxxxxxxx> writes: In s.a< -1, the< operator performs (C99 6.5.8p3) the usual arithmetic conversions, which include (6.3.1.8p1) the integer promotions (6.3.1.1p2). In unsigned a : 27, the width of the bit-field is part of the type (6.7.2.1p9). If the int type is 32-bit as usual, then all possible values of s.a fit in int, and the result of the integer promotions is an int, even though the bit-field was declared as unsigned. I don't know if C89 had different rules for this. An extended C++ variant of your test program:
> > .... Thanks for your detailed answer. Also thanks for your test program, that typeid() trick is very useful. Wouter
