On Sunday, 3 July 2016 at 13:41:27 UTC, Ola Fosheim Grøstad wrote:
On Sunday, 3 July 2016 at 11:49:15 UTC, Andrei Alexandrescu
wrote:
Well to be more precise here's what I'm looking for. When you
compare an integral with a floating point number, the integral
is first converted to floating point format. I.e. for long x
and double y, x == y is the same as double(x) == y.
Now, say we want to eliminate the "bad" cases of this
comparison. Those would make it non-transitive. Consider two
distinct longs x1 and x2. If they convert to the same double
y, then x1 == y and x2 == y are true, which is contradictory
with x1 != x2.
If you assume round-to-even rounding mode then you get unique
representations for integers in the ranges as other people have
suggested:
int_to_float : [-((1<<24)-1) , (1<<24)-1]
int_to_double : [-((1<<53)-1) , (1<<53)-1]
Seems to me there are some assumptions being made here that
only the permanent storage bits are in play. You've covered
the implicit leading bit, but what about possible guard digits,
depending on where the values are currently located in the
hardware? Not that I've thought it through, but it's something
to think about.
