Support for MIPS r5900

["Jürgen Urban"]

Hello Maciej,

> > I tested the calculation with the type "float".
> > ABI o32 with -mhard-float and -msingle-float produces the following
> results:
> > 1.000000 (0x3f800000) / 0.000000 (0x00000000) = nan (0x7fffffff)
> > 0.000000 (0x00000000) / 0.000000 (0x00000000) = nan (0x7fffffff)
> > 0.000000 (0x00000000) / nan (0x7fc00000) = 0.000000 (0x00000000)
> > 1.000000 (0x3f800000) + 1.000000 (0x3f800000) = 2.000000 (0x40000000)
> > 1.000000 (0x3f800000) + inf (0x7f800000) = inf (0x7f800000)
> > inf (0x7f800000) + inf (0x7f800000) = nan (0x7fffffff)
> > inf (0x7f800000) + -inf (0xff800000) = 0.000000 (0x00000000)
> > nan (0x7fc00000) + nan (0x7fc00000) = nan (0x7fffffff)
> > nan (0x7fc00000) + nan (0xffc00000) = 0.000000 (0x00000000)
> > 
> > The r5900 manual calls the result of 0/0 Fmax. So 0x7fffffff seems to be
> Fmax.
> 
>  So presumably you can get 0x7fffffff as an arithmetic result of a 
> calculation involving regular numbers as well, right?  Say 0x7f7ffffe + 
> 0x74000000 (using the binary-encoded notation)?  That would be beyond the
> IEEE-754 single range.

The FPU of the r5900 calculates the following:
340282306073709652508363335590014353408.000000 (0x7f7ffffd) + 40564819207303340847894502572032.000000 (0x74000000) = 340282346638528859811704183484516925440.000000 (0x7f7fffff)
340282326356119256160033759537265639424.000000 (0x7f7ffffe) + 40564819207303340847894502572032.000000 (0x74000000) = inf (0x7f800000)
340282346638528859811704183484516925440.000000 (0x7f7fffff) + 40564819207303340847894502572032.000000 (0x74000000) = inf (0x7f800000)
inf (0x7f800000) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7f800001)
nan (0x7f800001) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7f800002)
nan (0x7f900000) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7f900001)
nan (0x7f900001) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7f900002)
nan (0x7ffffff1) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7ffffff2)
nan (0x7ffffffc) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7ffffffd)
nan (0x7ffffffd) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7ffffffe)
nan (0x7ffffffe) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7fffffff)
nan (0x7fffffff) + 40564819207303340847894502572032.000000 (0x74000000) = nan (0x7fffffff)

So it seems that it interprets nan and inf as normal numbers, but it stops at 0x7fffffff. So 0x7fffffff should be interpreted as overflow.

Best regards
Jürgen