bitfield signedness

[Kalle Olavi Niemitalo]

Wouter Vermaelen <wouter.vermaelen@scarlet.be> writes:

> This is unexpected to me, because the field 'a' is already unsigned, so
> adding an extra cast to unsigned shouldn't change anything.

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:

#include <stdio.h>
#include <typeinfo>

namespace {
	const char *overload(int) { return "int"; }
	const char *overload(unsigned) { return "unsigned"; }
}

int main() {
        struct { unsigned a : 27; } s = {1};
	printf("typeid(unsigned) = %s\n", typeid(unsigned).name());
	printf("typeid(int) = %s\n", typeid(int).name());
	printf("typeid(s.a) = %s\n", typeid(s.a).name());
	printf("typeid(+s.a) = %s\n", typeid(+s.a).name());
	printf("overload(s.a) = %s\n", overload(s.a));
	printf("overload(+s.a) = %s\n", overload(+s.a));
        printf("(s.a < -1) = %d\n", s.a < -1);
        printf("(((unsigned)s.a) < -1) = %d\n", ((unsigned)s.a) < -1);
}

compiled with GCC 4.4.5 on amd64 outputs:

typeid(unsigned) = j
typeid(int) = i
typeid(s.a) = j
typeid(+s.a) = i
overload(s.a) = unsigned
overload(+s.a) = int
(s.a < -1) = 0
(((unsigned)s.a) < -1) = 1

> Is there any way to change the behavior of gcc (so make it output "1 1")?

I don't think there is.  In GCC trunk r108723 (which I happened
to have at hand), it looks like the integer promotions are
implemented in gcc/c-typeck.c (perform_integral_promotions) and
gcc/c-common.c (c_promoting_integer_type_p).  The TYPE_PRECISION
comparisons in these functions do not check any options.