FORTRAN floating point accuracy problems

[Werner]

I was waiting till you came up with a problematic case ;-)
Actually, however, depending on how you look at it:

- this is a case of wrong input (99 for seconds)
- the result is correct (try it on Free42 Decimal, for instance)
But I agree that you can't see it as it is rounded in the display..

BTW with x = 1e13/(5e13 + 1), on Free42 Binary, INT(100*x) delivers 19 and not 20 as you stated, and there are no negatives during the computation?

Cheers, Werner