Cedric Roux wrote:
>
> If that is what you refer to:
> 3 An rvalue for an integral bit-field (_class.bit_) can be converted to
> an rvalue of type int if int can represent all the values of the bit-
> field; otherwise, it can be converted to unsigned int if unsigned int
> can represent all the values of the bit-field. If the bit-field is
> larger yet, no integral promotion applies to it. If the bit-field has
> an enumerated type, it is treated as any other value of that type for
> promotion purposes.
> (found on the internet, I don't have the standard)
>
> we can understand that "t" may be promoted to "int" (which is 32b) because
> it occupies 16 bits. And the << arithmetic operator asks for the
> promotion.
>
> Maybe that's why you have 0.
> Maybe that's the way it has to be and gcc 4.0.2 was wrong.
Your citation applies, but my interpretation differs:
In C++ a bit-field has the declared type; the number of bits is not part of its
type [1][2]. The C++ integer promotion rules apply [your citation above]; in
particular the clause: "If the bit-field is larger yet, no integral promotion
applies to it". A bit-field expression of declared type long is unchanged by
integer promotions, since (the number of bits in the bit-field being irrelevant)
long is larger than int and unsigned int. So the expected output is:
100000000 200000000 100000000
8000 100000000 8000
[1] ISO/IEC 14882-1998 [class.bit] The bit-field attribute is not part of the
type of the class member.
[2] C++ Standard Core Language Issue #303
(http://www.open-std.org/JTC1/SC22/WG21/docs/cwg_closed.html#303)
>
> Note that doing:
> printf("%lx %lx %lx\n", (T)t.f, (T)((unsigned long)t.f << 17),
> (T)(((unsigned long)t.f << 17) >> 17));
> prints what you wait (not 0)
> (with a g++ 4.3.2, not even 4.4.1)
Yes, we can cast, but the question is what is the correct behavior in the absence of
a cast. I don't want to find all the places in existing code where a cast might
be necessary, if 4.4.1 behavior is a bug.
-- Vineet
