[Bug c/14642] New: y<<n where n==32 does not yield zero on x86

[nasix at raytheon dot com]

One would expect y << n, where y is any int and n==sizeof(int)*8, would yield 
zero, but for the x86 target, the result is y. This is due to the x86 shl 
opcode taking the count modulo 32 rather than modulo 64 as with PPC or 68K. The 
code generator ought to limit the shift count generally, and for the case of 
n==sizeof(int)*8, simply return zero. This illustrates the problem:

#include <assert.h>

/* Demonstrates whether C compiler properly shifts
   the operand by the # of bits an integer occupies.
   x86 and MIPS shift count are modulo-32, so they will fail this test.
*/
typedef long INT_T;

/* use function to keep optimizer out of it */
static INT_T shift(const INT_T x,const unsigned count)
{
 return x << count;
}

int main()
{
 assert(shift(1,sizeof(INT_T)*8) == (INT_T)0);
 return 0;
}

-- 
           Summary: y<<n where n==32 does not yield zero on x86
           Product: gcc
           Version: 3.3.1
            Status: UNCONFIRMED
          Severity: normal
          Priority: P1
         Component: c
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: nasix at raytheon dot com
                CC: gcc-bugs at gcc dot gnu dot org
 GCC build triplet: i686-pc-cygwin
  GCC host triplet: i686-pc-cygwin
GCC target triplet: i686-pc-cygwin


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=14642