[Bug c/11892] New: double to int different if intermediate double used and no optimisation

["simon dot marshall at misys dot com" <gcc-bugzilla at gcc dot gnu dot org>]

PLEASE REPLY TO gcc-bugzilla@gcc.gnu.org ONLY, *NOT* gcc-bugs@gcc.gnu.org.

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=11892

           Summary: double to int different if intermediate double used and
                    no optimisation
           Product: gcc
           Version: 3.3.1
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: c
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: simon dot marshall at misys dot com
                CC: gcc-bugs at gcc dot gnu dot org

I hesitate to submit this, but I don't think this merits the usual "what do you 
expect from doubles?" response.  This is with:

Reading specs from /usr/local/lib/gcc-lib/i686-pc-linux-gnu/3.3.1/specs
Configured with: ../configure --prefix=/usr/local --enable-
languages=c,c++,f77,objc,java,ada --enable-threads=posix --enable-shared --
enable-__cxa_atexit --with-system-zlib
Thread model: posix
gcc version 3.3.1

and the program:

#include <stdio.h>

int main ()
{
    double d = 0.1;
    double e = 5;
    double f = e / d;
    int i = (int) f;
    int j = (int) (e / d);

    printf ("f = %lf\n", f);
    printf ("i = %d\n", i);
    printf ("j = %d\n", j);

    return 0;
}

Notice that i and j are the integer result of 5 / 0.1 but that the intermediate 
double f is used to initialise i.  I can just about except that 0.1 is 
represented by 0.10000000000000001 according to gdb.  What does surprise me is 
that i and j can have different values:

$ gcc double.cpp && ./a.out
f = 50.000000
i = 50
j = 49
$ gcc -O double.cpp && ./a.out
f = 50.000000
i = 50
j = 50
$ gcc -O -ffloat-store double.cpp && ./a.out
f = 50.000000
i = 50
j = 49

As you can see, i != j unless I use raw optimisation.  I can just about except 
that the result of the arithmetic is not exactly 50, but surely i should always 
be equal to j?  In other words, I think i==j==50 or i==j==49 are correct 
outcomes but i!=j is not.