Vineet Soni wrote:
The integral promotion behavior of long bit-fields has changed with g++ 4.4.1 on
x86_64-linux, and appears broken. Consider:
#include <stdio.h>
#define T unsigned long
int main()
{
struct { T f : 33; } s = { 1UL << 32 };
printf("%lx %lx %lx\n", (T)s.f, (T)(s.f << 1), (T)((s.f << 1) >> 1));
struct { T f : 16; } t = { 1UL << 15 };
printf("%lx %lx %lx\n", (T)t.f, (T)(t.f << 17), (T)((t.f << 17) >> 17));
return 0;
}
Based on my interpretation of the C++ standard, the expected output is:
100000000 200000000 100000000
8000 100000000 8000
The actual output is:
$ g++-4.4.1 -Wall -pedantic x.c; ./a.out
100000000 200000000 100000000
8000 0 0
Is this a bug?
Note that 4.0.2 matches the expected output:
$ g++-4.0.2 -Wall -pedantic x.c; ./a.out
100000000 200000000 100000000
8000 100000000 8000
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.
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)
(Note that I am not involved in g++ or gcc development nor
C++ standardization, so all what I say may be totally wrong.)
