Comparison of a bitfield vs a (negative) integer, undefined behavior or compiler bugs?

Question

Here is a small program. Should this print 0 or 1, or does it have undefined behavior ?

#include <stdio.h>
struct S0 {
  unsigned f1 : 1;
};

struct S0 s;

int main (void) {
  int x = -3;
  int y = x >= (0, s.f1);
  printf ("%d\n", y);
  return 0;
}

This is taken from a resent CSmith test, and this case is discussed more here .

In particular, GCC, KCC, and CompCert outputs 0, while MSVC 2010, ICC 12.0.2 and recent Clang outputs 1.

Answer 1

Interesting question.

According to the C99 draft standard 6.5.17.1, the type of (0, s.f1) is the same as that of s.f1, which (per 6.7.2.1.9) is "an unsigned integer type consisting of 1 bit". This is an arithmetic type by virtue of being an integer type, its precision is 1 (per 6.2.6.2.6, and 6.2.6.1.3 implies no padding bits), and therefore its rank is less than that of int (per the second item under 6.3.1.1.1; int has a precision of at least 15, since it must be able to represent values in the range -32767 to 32767 (see 5.2.4.2.1)).

Since both x and the expression (0, s.f1) have arithmetic type, the usual arithmetic conversions are performed (per 6.5.8.3). Since an int can represent the full range of values of s.f1, it is promoted to a (signed) int (per 6.3.1.1.2). Then since both operands are (signed) ints, the common real type is signed int (per 6.3.1.8) and therefore the result of the comparison should be 0.